## Section5.7Curve Sketching

In this section, we discuss how we can tell what the graph of a function looks like by performing simple tests on its derivatives.

### Subsection5.7.1The First Derivative Test and Intervals of Increase/Decrease

The method of Section 5.5.1 for deciding whether there is a relative maximum or minimum at a critical value is not always convenient. We can instead use information about the derivative $f'(x)$ to decide; since we have already had to compute the derivative to find the critical values, there is often relatively little extra work involved in this method.

How can the derivative tell us whether there is a maximum, minimum, or neither at a point? The following so-called First Derivative Test is a procedure for finding relative extrema of a continuous function based on critical points and analyzing behaviour around the critical points:

###### Example5.68. Relative Extrema.

Find all relative maximum and minimum points for $f(x)=\sin x+\cos x$ using the First Derivative Test.

Solution

The domain of $f$ is $D_f = \mathbb{R}\text{.}$ The derivative is $f'(x)=\cos x-\sin x$ and from Example 5.50 the critical values we need to consider are $\pi/4$ and $5\pi/4\text{.}$

To classify these critical points using the First Derivative Test, we need to determine the sign of $f'(x)$ as $x$ approaches each point from the left and from the right.

First pick a test point in the interval $(-\infty,\pi/4)\text{,}$ say $x=0\text{.}$ Evaluate

\begin{equation*} f'(0) = \cos(0)-\sin(0) = 1\text{.} \end{equation*}

The next test point should be in the interval $(\pi/4,5\pi/4)\text{,}$ for example $x=\pi\text{.}$ Here,

\begin{equation*} f'(\pi) = \cos(\pi)-\sin(\pi) = -1\text{.} \end{equation*}

Lastly, we need a test point in the interval $(5\pi/4,\infty)\text{,}$ such as $x=2\pi\text{.}$ Then

\begin{equation*} f'(2\pi) = \cos(2\pi)-\sin(2\pi) = 1\text{.} \end{equation*}

From this information, we can construct the following sign diagram for $f'(x)\text{:}$

We conclude that $f(x)$ must be increasing on the entire interval $(-\infty,\pi/4)\text{,}$ decreasing on $(\pi/4,5\pi/4)\text{,}$ and increasing on $(5\pi/4,\infty)\text{.}$ Therefore, by the First Derivative Test, $f$ has a relative minimum at $5\pi/4$ and a relative maximum at $\pi/4\text{,}$ as shown in the graph below.

###### Example5.69. Relative Extrema.

Find the relative maxima and minima of the function $f(x)=x^{2/3}\text{.}$

Solution

The derivative of $f$ is

\begin{equation*} f'(x)=\frac{2}{3}x^{-1/3}=\frac{2}{3x^{1/3}} \cdot \end{equation*}

The function $f'$ is not defined at $x=0\text{,}$ so $f'$ is discontinuous there. It is continuous everywhere else. Furthermore, $f'$ is not equal to zero anywhere. Thus, $x=0$ is the only critical point of the function $f\text{.}$

Pick a test point (say, $x=-1$) in the interval $\left(-\infty,0\right)$ and compute

\begin{equation*} f'(-1)=-\frac{2}{3} \cdot \end{equation*}

Next, we pick a test point (say $x=1$) in the interval $\left(0,\infty\right)$ and compute

\begin{equation*} f'(1) = \frac{2}{3} \cdot \end{equation*}

We now construct the sign diagram for $f'(x)\text{:}$

We see that the sign of $f'(x)$ changes from negative to positive as we move across $x=0$ from left to right. Thus, an application of the First Derivative Test tells us that $f(0)=0$ is a relative minimum of $f\text{.}$ We confirm these results with the graph of $f\text{,}$ shown in Figure 5.18.

Perhaps you noticed in the graphs from this section that the critical points seem to divide the domain of a function into intervals, where the function either increases or decreases on this interval. We will now formally introduce this concept of increase and decrease.

###### Increasing and Decreasing Functions.

A function $f$ is increasing on an interval $(a,b)$ if for any two numbers $x_1$ and $x_2$ in $(a,b)\text{,}$ $f(x_1) \lt f(x_2)$ whenever $x_1 \lt x_2\text{.}$

A function $f$ is decreasing on an interval $(a,b)$ if for any two numbers $x_1$ and $x_2$ in $(a,b)\text{,}$ $f(x_1) > f(x_2)$ whenever $x_1 \lt x_2\text{.}$

We say that $f$ is increasing at $x=c$ if $c \in (a,b)$ such that $f$ is increasing on the interval $(a,b)\text{,}$ see Figure 5.19. Similarly, we say that $f$ is decreasing at $x=c$ if $c \in (a,b)$ such that $f$ is decreasing on the interval $(a,b)\text{,}$ see Figure 5.20. Increasing functions slope upward from left to right (Figure 5.19) and decreasing functions slope downward from left to right (Figure 5.20). But the slope of a function at a point $x=c$ is the rate of change of the function at $x=c\text{,}$ which is given by the derivative of the function at that point. Therefore, the derivative lends itself naturally as the tool for determining intervals of increase and decrease, provided the function is differentiable on the intervals, see Figures 5.19 and 5.20. This leads us to the following theorem, which we can prove using the Mean Value Theorem from Section 5.6.

We will prove the increasing case. The proof of the decreasing case is similar. Suppose that $f^{\prime }\left( x\right) >0$ on an interval $I\text{.}$ Then $f$ is differentiable, and hence also, continuous on $I\text{.}$ If $x_{1}$ and $x_{2}$ are any two numbers in $I$ and $x_{1}\lt x_{2}\text{,}$ then $f$ is continuous on $\left[ x_{1},x_{2}\right]$ and differentiable on $\left( x_{1},x_{2}\right) \text{.}$ By the Mean Value Theorem, there is some $c$ in $\left( x_{1},x_{2}\right)$ such that

\begin{equation*} f^{\prime }\left( c\right) =\frac{f\left( x_{2}\right) -f\left( x_{1}\right) }{x_{2}-x_{1}}\text{.} \end{equation*}

But $c$ must be in $I\text{,}$ and thus, since $f^{\prime }\left( x\right) >0$ for every $x$ in $I\text{,}$ $f^{\prime }\left( c\right) >0\text{.}$ Also, since $x_{1}\lt x_{2}\text{,}$ we have $x_{2}-x_{1}>0\text{.}$ Therefore, both the left hand side and the denominator of the right hand side are positive. It follows that the numerator of the right hand must be positive. That is, $f\left( x_{2}\right) -f\left( x_{1}\right) >0\text{,}$ or in other words, $f\left( x_{1}\right) \lt f\left( x_{2}\right) \text{.}$ This shows that between $x_{1}$ and $x_{2}$ in $I\text{,}$ the larger one, $x_{2}\text{,}$ necessarily has the larger function value, $f\left( x_{2}\right) \text{,}$ and the smaller one, $x_{1}\text{,}$ necessarily have the smaller function value, $f\left( x_{1}\right) \text{.}$ This means that $f$ is increasing on $I\text{.}$

###### Example5.71. Intervals of Increase and Decrease for $f'(x)=0$.

Determine the intervals where the function $f(x)=x^3-3x^2-24x+32$ is increasing and where it is decreasing.

Solution

The derivative of $f$ is

\begin{equation*} f'(x)=3x^2-6x-24=3(x+2)(x-4)\text{,} \end{equation*}

and it is continuous everywhere. The zeros of $f'(x)$ are $x=-2$ and $x=4\text{,}$ and these points divide the real line into the intervals $(-\infty,-2)\text{,}$$(-2,4)\text{,}$ and $(4,\infty)\text{.}$

To determine the sign of $f'(x)$ in the above intervals, compute $f'(x)$ at a convenient test point within each interval. We find,

\begin{equation*} \begin{array}{lccc} \text{ Interval } \amp \text{ Test Point } \amp f'(c) \amp \text{ Sign of } f'(x) \\ (-\infty,-2) \amp -3 \amp 21 \amp + \\ (-2,4) \amp 0 \amp -24 \amp - \\ (4,\infty) \amp 5 \amp 21 \amp + \end{array} \end{equation*}

Using these results, we obtain the following sign diagram:

We conclude that $f$ is increasing on the intervals $(-\infty,-2)$ and $(4,\infty)$ and is decreasing on the interval $(-2,4)\text{.}$ The graph of $f$ is shown below.

###### Example5.72. Intervals of Increase and Decrease for $f'(x)$ DNE.

Find the interval where the function $f(x)=x^{2/3}$ is increasing and the interval where it is decreasing.

Solution

The derivative of $f$ is

\begin{equation*} f'(x)=\frac{2}{3}x^{-1/3}=\frac{2}{3x^{1/3}} \cdot \end{equation*}

As noted in Example 5.69, $f'$ is not defined at $x=0\text{,}$ is continuous everywhere else, and is not equal to zero in its domain.

Since $f'(-1) \lt 0\text{,}$ we see that $f'(x) \lt 0$ on $\left(-\infty.0\right)\text{.}$ Next, we pick a test point (say $x=1$) in the interval $\left(0,\infty\right)$ and compute

\begin{equation*} f'(1) = \frac{2}{3} \cdot \end{equation*}

Since $f'(1) > 0\text{,}$ we see that $f'(x) > 0$ on $\left(0,\infty\right)\text{.}$ From the consequent sign diagram,

we conclude that $f$ is decreasing on the interval $(-\infty,0)$ and increasing on the interval $(0,\infty)\text{.}$ The graph of $f\text{,}$ shown in Figure 5.18 confirms these results.

###### Example5.73. Intervals of Increase and Decrease.

Find the intervals where the function $f(x)=x+\dfrac{1}{x}$ is increasing and where it is decreasing.

Solution

The derivative of $f$ is

\begin{equation*} f'(x)=1-\frac{1}{x^2}=\frac{x^2-1}{x^2} \cdot \end{equation*}

Since $f'$ is not defined at $x=0\text{,}$ it is discontinuous there. Furthermore, $f'(x)$ is equal to zero when $x^2-1=0\text{,}$ or $x=\pm 1\text{.}$ These values of $x$ partition the domain of $f'$ into the open intervals $(-\infty,-1)\text{,}$$(-1,0)\text{,}$$(0,1)$ and $(1,\infty)\text{,}$ where the sign of $f'$ is different from zero.

To determine the sign of $f'$ in each of these intervals, we compute $f'(x)$ at the test points $x=-2,-\frac{1}{2},\frac{1}{2}$ and $2\text{,}$respectively, obtaining $f'(-2)=\frac{3}{4}\text{,}$$f'(-\frac{1}{2})=-3\text{,}$ $f'(\frac{1}{2}=-3\text{,}$ and $f'(2)=\frac{3}{4}\text{.}$

From the above sign diagram for $f'\text{,}$ we conclude that $f$ is increasing on $(-\infty,-1)$ and $(1,\infty)$ and decreasing on $(-1,0)$ and $(0,1)\text{.}$ The graph of $f$ is shown below. Note that $f'$ does not change sign as we move accros the point of discontinuity, $x=0\text{.}$

Note:Example 5.73 reminds us that we must not automatically conclude that the derivative $f'$ must change sign when we move across a discontinuity or zero of $f'\text{.}$
###### Example5.74. Intervals of Increase and Decrease and Relative Extrema.

Consider the function $f\left( x\right) =x^{4}-2x^{2}\text{.}$ Find where $f$ is increasing and where $f$ is decreasing. Use this information to find the relative maximum and minimum points of $f\text{.}$

Solution

We first compute $f^{\prime }\left( x\right)$ and analyze its sign.

\begin{equation*} f^{\prime }\left( x\right) =4x^{3}-4x=4x\left( x^{2}-1\right) =4x\left( x-1\right) \left( x+1\right)\text{.} \end{equation*}

Thus, $f'(x)=0$ when $x=0,1\text{,}$ and $-1\text{.}$ $f'(x)$ is a continuous function, and so these are the only critical points. This splits the domain into the open intervals $\left(-\infty,-1\right)\text{,}$ $\left(-1,0\right)\text{,}$ $\left(0,1\right)$ and $\left(1,\infty\right)\text{,}$ where $f$ is either increasing or decreasing. Picking appropriate test points, we find

\begin{equation*} f'(-2)=-24 ,\ \ f'(-0.5)=1.5, \ \ f'(0.5)=-1.5, \ \text{ and } \ f'(2)=24\text{.} \end{equation*}

This leads us to the sign diagram,

from which we see that $f$ is increasing on the interval $\left( -1,0\right)$ and on the interval $\left( 1,\infty \right)\text{.}$ Similarly, $f$ is decreasing on the interval $\left( -\infty ,-1\right)$ and on the interval $\left( 0,1\right)\text{.}$

Therefore, at the critical points $-1\text{,}$ $0$ and $1\text{,}$ respectively, $f$ has a relative minimum, a relative maximum and a relative minimum, as shown below.

###### Example5.75. Intervals of Increase and Decrease in Profit Function.

The profit function of Acrosonic is given by

\begin{equation*} P(x)=-0.02x^2+300x-200,000 \end{equation*}

dollars, where $x$ is the number of Acrosonic model F loudspeakers systems produced. Find where the function $P$ is increasing and where it is decreasing.

Solution

The derivative $P'$ is

\begin{equation*} P'(x)=-0.04x+300=-0.04(x-7500)\text{.} \end{equation*}

Thus, $P'(x)=0$ when $x=7500\text{.}$ Furthermore, $P'(x) > 0$ for $x$ in the interval $(0,7500)\text{,}$ and $P'(x) \lt 0$ for $x$ in the interval $(7500,\infty)\text{.}$ This means that the profit function $P$ is increasing on $(0,7500)\text{,}$ and decreasing on $(7500,\infty)$ (see graph below, where both $P$ and $x$ are in units of a thousand).

##### Exercises for Section 5.7.1.

Find all critical points and identify them as relative maximum points, relative minimum points, or neither.

1. $\ds y=x^2-x$

rel min at $x=1/2$
Solution

We have $y' = 2x-1\text{,}$ and so $y' = 0$ precisely when $x=1/2\text{.}$ This is the only critical point of $y\text{.}$ To classify this critical point, we use the First Derivative Test. Since

\begin{equation*} \begin{split} y'(0) \amp = -1 \lt 0\\ y'(1) \amp = 2-1 = 1 > 0, \end{split} \end{equation*}

we can construct the following sign diagram:

Therefore, $f$ must have a relative minimum at $x=1/2\text{.}$

2. $\ds y=2+3x-x^3$

rel min at $x=-1\text{,}$ rel max at $x=1$
Solution

We compute $y'= 3-3x^2\text{.}$ Hence, $y'=0$ when $x=\pm 1\text{.}$ This gives two critical points. Now pick the following test points:

\begin{equation*} \begin{split} y'(-2) \amp \lt 0 \\ y'(0) \amp > 0\\ y'(2) \amp \lt 0 \end{split} \end{equation*}

And so we construct the sign diagram:

Therefore, $y$ has a relative minimum at $x=-1$ and a relative maximum at $x=1\text{.}$

3. $\ds y=x^3-9x^2+24x$

rel max at $x=2\text{,}$ rel min at $x=4$
Solution

Differentiating, we find $y'=3x^2-18x+24\text{.}$ Therefore, $y'=0$ at $x=2$ and $x=4\text{.}$ To classify these two critical points, we pick the following test points:

\begin{equation*} \begin{split} y'(1) \amp > 0\\ y'(3) \amp \lt 0\\ y'(5) \amp >0 \end{split} \end{equation*}

Now construct the following sign diagram:

Thus, by the First Derivative Test, $y$ has a relative minimum at $x=4$ and a relative maximum at $x=2\text{.}$

4. $\ds y=x^4-2x^2+3$

rel min at $x=\pm 1\text{,}$ rel max at $x=0$
Solution

We have $y'(x)=4x^3-4x = 4x(x^2-1)\text{,}$ and so $y'=0$ when $x=-1,0,1\text{.}$ To classify these critical points, we use the first derivative test. We pick the following test points:

\begin{equation*} \begin{split} y'(-2) \amp = -8(4-1) \lt 0 \\ y'(-0.5) \amp = -2(0.25-1) > 0 \\ y'(0.5) \amp = 2(0.25-1) \lt 0 \\ y'(2) \amp = 8(4-1) > 0 \end{split} \end{equation*}

And so we can construct the following sign diagram for $y\text{:}$

Hence, $y$ has relative minima at $x=\pm 1$ and a relative maximum at $x=0\text{.}$

5. $\ds y=3x^4-4x^3$

rel min at $x=1$
Solution

Differentiating, we find $y' = 12x^3-12x^2\text{.}$ Therefore, $y'=0$ when $x=0,1\text{.}$ Since

\begin{equation*} \begin{split} y'(-1) \amp \lt 0 \\ y'(0.5) \amp \lt 0\\ y'(2) \amp >0 \end{split} \end{equation*}

we can construct the following sign diagram:

Hence, $y$ has a relative minimum at $x=1\text{,}$ and neither a relative maximum nor minimum at $x=0\text{.}$

6. $\ds y=(x^2-1)/x$

none
Solution

First notice that the domain of $y$ is $\mathbb{R} \setminus \{0\}\text{.}$ Next, we differentiate:

\begin{equation*} \diff{y}{x} = \diff{}{x} \frac{x^2-1}{x} = \frac{1}{x^2}+1\text{.} \end{equation*}

Since

\begin{equation*} y' = \frac{1}{x^2}+1= 0 \end{equation*}

has no real solutions, and $y'$ is defined everywhere in the domain of $y\text{,}$ $y$ has no critical points.

7. $\ds y=3x^2-(1/x^2)$

none
Solution

Notice that the domain of $y$ is $\mathbb{R} \setminus \{0\}\text{.}$ Next, we differentiate:

\begin{equation*} y' = 6x + \frac{2}{x^3}\text{.} \end{equation*}

Then

\begin{equation*} 6x + \frac{2}{x^3} = 0 \implies x^4 = -\frac{3}{2}\text{,} \end{equation*}

and so this gives no critical points. Although $y'$ is undefined at $x=0\text{,}$ this is not a point in the domain and so $y$ has no critical points, and hence no relative extrema.

8. $\ds y=\cos(2x)-x$

rel min at $x=7\pi/12+k\pi\text{,}$ rel max at $x=-\pi/12+k\pi\text{,}$ for integer $k$
Solution

The domain of $y$ is all real numbers. Now differentiate:

\begin{equation*} y' = -2\sin(2x)-1\text{.} \end{equation*}

Hence, the critical points of $y$ are given by the solutions to

\begin{equation*} y'=0 \implies \sin(2x) = \frac{1}{2}\text{.} \end{equation*}

The solution to

\begin{equation*} \sin^{-1} \frac{1}{2} = 2x \end{equation*}

is $x = \frac{\pi}{12}\text{.}$ This gives the solution in the interval $[0, \pi/2]\text{.}$ Thus, $x = \frac{\pi}{2} - \frac{\pi}{12} = \frac{5\pi}{12}$ is the second solution over the entire interval $[0,\pi]\text{.}$ Therefore, there are an infinite number of critical points:

\begin{equation*} x = \frac{\pi}{12} + k\pi \text{ and } x = \frac{5\pi}{12} + k\pi\text{,} \end{equation*}

for any integer $k\text{.}$ We classify the critical points over one period of $\cos 2x\text{,}$ $[0,\pi]\text{.}$ Since

\begin{equation*} \begin{split} f'(0) \amp \lt 0 \\ f'(3\pi/12) \amp > 0 \\ f'(\pi) \amp \lt 0 \end{split} \end{equation*}

we conclude that $y$ has relative minima at $x=\frac{\pi}{12}+k\pi$ and relative maxima at $x=\frac{5\pi}{12}+k\pi\text{.}$

9. $\ds f(x) = (5-x)/(x+2)$

none
Solution

The domain of $y$ is all real numbers such that $x \neq -2\text{.}$ Now differentiate:

\begin{equation*} y' = \frac{-7}{(x+2)^2}\text{.} \end{equation*}

The derivative $y'$ is defined over the entire domain of $y\text{,}$ so the critical points of $y$ are given by

\begin{equation*} y'=0 \implies -7 = 0\text{,} \end{equation*}

which has no solutions. Therefore, $y$ has no relative maxima or minima.

10. $\ds f(x) = |x^2 - 121|$

rel max at $x=0\text{,}$ rel min at $x=\pm 11$
Solution

We first write $f$ using the piecewise definition of the absolute value function.

\begin{equation*} f(x) = \begin{cases}121 - x^2 \amp \text{ if } -11 \leq x \leq 11 \\ x^2 - 121 \amp \text{ otherwise } \end{cases} \end{equation*}

The graph of $f$ will have two corners: one at $x=11\text{,}$ and one at $x=-11\text{.}$ We find

\begin{equation*} f'(x) = \begin{cases}-2x \amp \text{ if } -11 \lt x \lt 11 \\ 2x \amp \text{ otherwise } \end{cases} \end{equation*}

where $f'$ is undefined at $x= \pm 11\text{.}$ Since $f'(x) = 0$ at $x=0\text{,}$ we therefore have 3 critical points, namely $x=-11, 0$ and $11\text{.}$ We will classify these critical points using the first derivative test. We pick the following test points and construct a sign diagram for $f'\text{:}$

\begin{equation*} \begin{split} f'(-12) \amp = 2(-12) \lt 0 \\ f'(-10) \amp = -2(-10) > 0 \\ f'(10) \amp = -2(10) \lt 0 \\ f'(12) \amp = 2(12) > 0 \end{split} \end{equation*}

Therefore, $x=0$ corresponds to a relative maximum, and $x=\pm 11$ correspond to relative minima of $f\text{.}$

11. $\ds f(x) = x^3/(x+1)$

rel min at $x=-3/2\text{,}$ neither at $x=0$
Solution
12. $\ds f(x) = \sin ^2 x$

rel min at $n\pi\text{,}$ rel max at $\pi/2+n\pi$
Solution

First, we note that the domain of $f(x)=\dfrac{x^3}{(x+1)}$ is $(-\infty,-1) \cup (-1,\infty)\text{.}$ Now,

\begin{equation*} f'(x)=\frac{x^2(2x+3)}{(x+1)^2}\text{,} \end{equation*}

so $f'$ is also undefined at $x=-1\text{,}$ but this is not a critical point as it is not in the domain of $f\text{.}$ To find the critical points of $f\text{,}$ we solve

\begin{equation*} \begin{split} f'(x) \amp = 0 \\ x^2(2x+3) \amp = 0 \\ x \amp = 0,-3/2. \end{split} \end{equation*}

We pick appropriate test points at which to evaluate $f'\text{,}$ for example

\begin{equation*} f'(-2)= -4, \ \ f'(-0.5)= 2, \ \ \text{ and } \ \ f'(1)=5/4\text{.} \end{equation*}

Thus, we find the following sign diagram for $f'\text{:}$

Thus, $f$ has a relative minimum at $x=-3/2\text{,}$ and the critical point $x=0$ is neither a relative maximum nor a relative minimum.

13. $\ds f(x)=\sec x$

rel min at $2n\pi\text{,}$ rel max at $(2n+1)\pi\text{,}$ for integer $n$
Solution

The function $f(x)=\sin^2(x)$ is defined for all real numbers and has a period of $\pi\text{.}$ Its derivative,

\begin{equation*} f'(x)=2\sin x \cos x \end{equation*}

is also defined for all real numbers and has infinitely many zeroes as follows.

\begin{equation*} f'(x) = 0 \implies \sin x = 0 \ \ \text{ or } \ \ \cos x = 0 \implies x = \frac{n\pi}{2}\text{,} \end{equation*}

for all integers $n\text{.}$ It suffices to classify the critical points of $f$ over one period, for example $[0,\pi]\text{.}$ We find the sign diagram of $f'$ to be

We see that $x=0$ and $x=\pi$ are relative minima, and $x=\pi/2$ is a relative maximum. To generalize these results, we split the critical points $x=\frac{n\pi}{2}$ into two cases: $n$ odd, and $n$ even. We can represent $n$ even as $n=2m$ for all integers $m\text{,}$ and $n$ odd as $n=2m+1$ for all integers $m\text{.}$ Thus, $x=m\pi$ are relative minima and $x = \pi/2+m\pi$ are relative maxima.

Let $\ds f(\theta) = \cos^2(\theta) - 2\sin(\theta)\text{.}$ Find the intervals where $f$ is increasing and the intervals where $f$ is decreasing in $[0,2\pi]\text{.}$ Use this information to classify the critical points of $f$ as either relative maximums, relative minimums, or neither.

min at $\pi/2+2n\pi\text{,}$ max at $3\pi/2+2n\pi$

Solution

We differentiate:

\begin{equation*} f'(\theta) = \diff{}{\theta} \left(\cos^2\theta - 2\sin\theta\right) = -2 \cos\theta (\sin \theta + 1) \end{equation*}

So for $\theta \in [0,2\pi]\text{,}$ $f'(\theta) = 0$ when

\begin{equation*} \cos \theta = 0 \implies \theta = \frac{\pi}{2}, \frac{3\pi}{2} \end{equation*}

or when

\begin{equation*} \sin\theta = -1 \implies \theta = \frac{3\pi}{2}\text{.} \end{equation*}

Therefore, there are only two critical points, $\frac{\pi}{2}$ and $\frac{3\pi}{2}\text{.}$ We use the following test points:

\begin{equation*} f'(0) = -2, f'(\pi) = 2, f'(2\pi) = -2\text{.} \end{equation*}

Hence, $f$ is decreasing on $[0, \pi/2) \cup (3\pi/2, 2\pi]$ and increasing on $(\pi/2,3\pi/2)\text{.}$ The function therefore has a relative minimum at $\theta = \frac{\pi}{2}$ and a relative maximum at $\theta = \frac{3\pi}{2}\text{.}$

Let $r>0\text{.}$ Find the relative maxima and minima of the function $\ds f(x) =\sqrt{r^2 -x^2 }$ on its domain $[-r,r]\text{.}$

Relative max at $x = 0\text{.}$
Solution

We first differentiate:

\begin{equation*} f'(x) = -\frac{2x}{\sqrt{r^2-x^2}}\text{,} \end{equation*}

which is undefined at $x=\pm r\text{.}$ Since these are endpoints of the domain of $f\text{,}$ we do not consider them to be critical points. Now solve:

\begin{equation*} f'(x) = -\frac{2x}{\sqrt{r^2-x^2}} = 0 \implies x = 0\text{.} \end{equation*}

Hence, $f$ has one critical points. Since

\begin{equation*} \begin{split} f(r) \amp = 0 \\ f(0) \amp = r \\ f(-r) \amp = 0, \end{split} \end{equation*}

we see that $f$ has a relative maximum at $x=0\text{.}$

Given the graph of a function $f\text{,}$ determine the intervals where $f$ is increasing, constant, and decreasing.

decreasing on $(-\infty,0)\text{,}$ increasing on $(0,\infty)$
increasing on $(-1,1)\cup(3,\infty)$ and decreasing on $(-\infty,-1)\cup(1,3)\text{.}$
increasing on $(-2,-1)\text{,}$ decreasing on $(1,2)\text{,}$ constant on $(-1,1)$
decreasing on $(-\infty,1)\cup(1,2)$ and increasing on $(2,\infty)\text{.}$

Given the graph of a function $f\text{,}$ determine the relative maxima and relative minima, if any.

rel min at -1,1 and rel max at 0
rel min at -1,2 and rel max at 0
rel min at -1,1 and rel max at 0
rel min at 4 and rel max at -2

A subsidiary of ThermoMaster manufactures an indoor-outdoor thermometer. Management estimates that the profit (in dollars) realizable by the company for the manufacture and sale of $x$ units of thermometers each week is

\begin{equation*} P(x)=-0.001x^2+8x-5000\text{.} \end{equation*}

Find the intervals where the profit function $P$ is increasing and the intervals where $P$ is decreasing.

Increasing on $(0,4000)$ and decreasing on $(4000,\infty)\text{.}$

Solution

The domain of $P$ is $[0,\inf]\text{.}$ We differentiate the profit function:

\begin{equation*} P'(x) = -0.002x + 8\text{.} \end{equation*}

Therefore,

\begin{equation*} P'(x) = 0 \implies x = 4000\text{.} \end{equation*}

There are no points in the domain of $P$ for which $P'$ is undefined, and so we conclude that $x=4000$ is the only critical point. Since

\begin{equation*} P'(0) = 8 > 0, \text{ and } P'(10,000) = -12 \lt 0\text{,} \end{equation*}

we conclude that the company's profit is increasing on the interval $[0,4000)$ and decreasing on the interval $(4000,\infty)\text{.}$

Based on data from the Central Provident Fund of a certain country (a government agency similar to the Canada Pension Plan), the estimated cash in the fund in 1995 is given by
\begin{equation*} A(t)=-96.6t^4+403.6t^3+660.9t^2+250 \ \ \ 0 \leq t \leq 5 \end{equation*}
where $A(t)$ is measure in billions of dollars and $t$ is measured in decades, with $t=0$ corresponding to 1995. Find the interval where $A$ is increasing and the interval where $A$ is decreasing and interpret your results. Hint

Increasing from 1995 to 2035 and decreasing from 2035 to 2045.

Solution

The domain of $A(t)$ is $[0,5]\text{.}$ Now differentiate:

\begin{equation*} A'(t) = -386.4t^3+1,210.8t^2+1,321.8t = t(-386.4t^2+1,210.8t+1,321.8)\text{.} \end{equation*}

There are no points in the domain of $A$ for which $A'$ is undefined, and so to find the critical points, we set $A'=0\text{:}$

\begin{equation*} 0=t(-386.4t^2+1,210.8t+1,321.8) \implies t = 0, -0.86, 3.99\text{.} \end{equation*}

We reject the negative solution, and so the critical points of $A$ are $t=0$ and $t \approx 4\text{.}$ Since

\begin{equation*} A'(1) >0 \text{ and } A'(5) \lt 0\text{,} \end{equation*}

we conclude that fund is increasing from 1995 to 2035, and decreasing from 2035 to 2045.

Sales in the Web-hosting industry are projected to grow in accordance with the function

\begin{equation*} f(t)=-0.05t^3+0.56t^2+5.47t+7.5 \ \ \ 0 \leq t \leq 6 \end{equation*}

where $f(t)$ is measured in billions of dollars and $t$ is measured in years, with $t=0$ corresponding to 1999.

1. Find the interval where $f$ is increasing and the interval where $f$ is decreasing. Answer
Increasing on $(0,6)$
Solution

To find the intervals of increase and decrease, we need to find the zeros of $f'\text{.}$

\begin{equation*} f'(x) = -0.15t^2+1.12t+5.47 = 0\text{,} \end{equation*}

which gives $t=-3.36628$ and $t=10.8329\text{.}$ $f$ therefore has no critical points since neither of these two solutions are in its domain. We check the sign of $f'$ by picking any point in $[0,6]\text{,}$ e.g $f(0) = 5.47 > 0\text{,}$ and so $f$ is increasing on the entire interval $[0,6]\text{.}$

2. What does your result tell you about the sales in the Web-hosting industry from 1999 through 2005? Answer

We conclude that sales were increasing from 1999 to 2005.

According to a study conducted in 1997, the number of subscribers (in thousands) to the Canadian cellular market in the next 6 years is approximated by the function
\begin{equation*} N(t)=0.09444t^3-1.44167t^2+10.65695t+52 \ \ \ 0 \leq t \leq 6 \end{equation*}
where $t$ is measured in years, with $t=0$ corresponding to 1997.
1. Find the interval where $N$ is increasing and the interval where $N$ is decreasing. Answer
Increasing on $(0,6)\text{.}$
Solution

We first compute $f'(t)\text{:}$

\begin{equation*} N'(t) = 3(0.09444)t^2-2(1.44167)t+10.65695 = 0.28332t^2-2.8833tt+10.65695 \end{equation*}

Using the quadratic formula, we find that $N'(t) = 0$ has no real solutions. Since $N'(0) = 10.65695 > 0\text{,}$ we see that $N(t)$ is increasing on the entire interval $[0,6]\text{.}$

2. What does your result tell you about the number of subscribers in the Canadian cellular market in the years under consideration? Answer

This means that the number of subscribers is increasing from 1997 to 2003.

Let $\ds f(x) =a x^2 + bx + c$ with $a\neq 0\text{.}$ Show that $f$ has exactly one critical point using the First Derivative Test. Give conditions on $a$ and $b$ which guarantee that the critical point will be a maximum.

Solution

We first differentiate:

\begin{equation*} f'(x) = \diff{}{x} \left(ax^2+bx+c\right) = 2ax + b, a \neq 0\text{.} \end{equation*}

First, let $b = 0\text{.}$ Then

\begin{equation*} f'(x) = 0 \implies x = 0\text{.} \end{equation*}

Since there are no points such that $f'(x)$ is undefined, this is the only critical point of $f\text{.}$

Next, assume that $b \neq 0\text{.}$ Then

\begin{equation*} f'(x) = 0 \implies x = -\frac{2a}{b}\text{.} \end{equation*}

Since there are no points such that $f'(x)$ is undefined, this is the only critical point of $f\text{.}$

Hence, $f$ has exactly one critical point.

### Subsection5.7.2Concavity and Inflection Points

We know that the sign of the derivative tells us whether a function is increasing or decreasing; for example, when $f'(x)>0\text{,}$ $f(x)$ is increasing. However, a function can increase like this

or like this
And similarly, a function can decrease like this
or like this

How can you determine which way it is? For example, as you can see in Figure 5.21, the function $y=x^3$ increases as we move from left to right, but the sections of the curve defined on the intervals $\left(-\infty,0\right)$ and $\left(0,\infty\right)$ increase in different ways. As we approach the origin from the left along the curve, the curve turns clockwise and stays below its tangents as shown in Figure 5.21. In other words, the slopes of the tangents are decreasing on the interval $\left(-\infty,0\right)\text{.}$ As we move away to the right of the origin along the curve, the curve turns counterclockwise and stays above its tangents as shown in Figure 5.21. In other words, the slopes of the tangents are increasing on the interval $\left(0,\infty\right)\text{.}$ This manner of curving defines the concavity of the curve.

###### Definition5.76. Concavity.

Suppose that $y=f(x)$ is a differentiable function on the open interval $(a,b)\text{,}$ then $f$ is

1. concave up on $(a,b)$ if $f'$ is increasing on $(a,b)\text{;}$ and

2. concave down on $(a,b)$ if $f'$ is decreasing on $(a,b)\text{.}$

Suppose that a function is twice differentiable. This means that we can get information from the sign of $f''$ even when $f'$ is not zero. Suppose that $f''(a)>0\text{.}$ This means that near $x=a\text{,}$ $f'$ is increasing. If $f'(a)>0\text{,}$ this means that $f$ slopes up and is getting steeper; if $f'(a)\lt 0\text{,}$ this means that $f$ slopes down and is getting less steep. The two situations of concave up are shown in Figure 5.22.

Now suppose that $f''(a)\lt 0\text{.}$ This means that near $x=a\text{,}$ $f'$ is decreasing. If $f'(a)>0\text{,}$ this means that $f$ slopes up and is getting less steep; if $f'(a)\lt 0\text{,}$ this means that $f$ slopes down and is getting steeper. The two situations of concave down are shown in Figure 5.23.

These observations lead us to the following theorem on concavity.

We now have all the tools to determine the intervals, where the curve of a function is either concave up or concave down.

###### Guideline for Determining Intervals of Concavity.

Suppose that $y=f(x)$ is a twice differentiable function on its domain.

1. Find all critical points of $y=f'(x)$ (note that these are not necessarily the same as for the function $f$) and all $x$-values where $f'$ is undefined.

2. These $x$-values section the domain into open intervals.

3. Choose a number $n$ in each interval. Then apply the Second Derivative Test for Concavity to these numbers $n$ to decide if $f$ is concave up or concave down on these intervals.

If we are trying to understand the shape of the graph of a function, knowing where it is concave up and concave down helps us to get a more accurate picture. Of particular interest are points at which the concavity changes from up to down or down to up; such points are called inflection points .

###### Definition5.78. Inflection Point.

Any value of $x$ in the domain of $f$ where the tangent line exists or is vertical and where the concavity changes is called an inflection point.

If the concavity changes from up to down at $x=a\text{,}$ $f''$ changes from positive to the left of $a$ to negative to the right of $a\text{,}$ and usually $f''(a)=0\text{.}$ We can identify such points by first finding where $f''(x)$ is zero and then checking to see whether $f''(x)$ does in fact go from positive to negative or negative to positive at these points. Note that it is possible that $f''(a)=0$ but the concavity is the same on both sides; $\ds f(x)=x^4$ at $x=0$ is an example. These observations lead us to the following guideline for determining inflection points.

###### Guideline for Determining Inflection Points.

Suppose that $y=f(x)$ is a twice differentiable function on its domain.

1. Compute $f'(x)$ and $f''(x)\text{.}$

2. Find all critical points $c$ of the function $y=f'(x)$ (note that these are not necessarily the same as for the function $f$).

3. Determine the sign of $f''$ to the left and to the right of $c\text{.}$

4. If there is a change in signs, then $c$ is an inflection point.

###### Example5.79. Concavity and Inflection Point.

Describe the concavity of $\ds f(x)=x^3-x$ using intervals, and determine if there are any inflection points.

Solution

We find the first two derivatives of $f$ to be

\begin{equation*} \ds f'(x)=3x^2-1 \ \text{ and } \ f''(x)=6x\text{.} \end{equation*}

Next, we create the sign diagram for $f''\text{:}$

From this sign diagram, we see that $f$ is concave up for $x \in (-\infty,0)$ and concave down for $x \in (0,\infty)\text{.}$ The concavity changes as we move through $x=0$ and so the point $(0,f(0)) = (0,0)$ is an inflection point of $f\text{.}$ The graph of $f$ is shown below.

###### Example5.80. Concavity and Inflection Point.

Describe the concavity of

\begin{equation*} f(x)=\frac{1}{x^2+1} \end{equation*}

using intervals, and determine if there are any inflection points.

Solution

We find the first and second derivatives of $f$ to be

\begin{equation*} f'(x) = \frac{d}{dx}(x^2+1)^{-1} = -2x(x^2+1)^{-2} = -\frac{2x}{(x^2+1)^2}\text{,} \end{equation*}

and

\begin{equation*} \begin{split} f''(x) \amp = \frac{(x^2+1)^2(-2)+2(2x)(x^2+1)(2x)}{(x^2+1)^4} \\ \amp = \frac{(x^2+1)(-2(x^2+1)+8x^2)}{(x^2+1)^4} \\ \amp = \frac{(x^2+1)(6x^2-2)}{(x^2+1)^4}\\\amp = \frac{2(3x^2-1)}{(x^2+1)^3} \cdot \end{split} \end{equation*}

The critical points of $f''$ will occur when $f''= 0$ since $f''$ is continuous everywhere. We then solve

\begin{equation*} \begin{split} 3x^2-1 \amp =0 \\ x^2 \amp = \frac{1}{3} \\ x \amp = \pm \sqrt{3}. \end{split} \end{equation*}

The sign diagram for $f''$ is shown below:

Therefore, $f$ is concave upward on $(-\infty, \sqrt{3}/3) \cup (\sqrt{3}/3,\infty)$ and concave downward on $(-\sqrt{3}/3,\sqrt{3}/3)\text{.}$ Furthermore, $f''(x)$ changes sign as we move across the points $x= \pm \sqrt{3}/3\text{.}$ Since

\begin{equation*} f\left(\pm \frac{\sqrt{3}}{3} \right) = 3/4\text{,} \end{equation*}

the points $(\pm \sqrt{3}/3, 3/4)$ are inflection points of $f\text{.}$ The graph of $f$ is shown below.

###### Example5.81. Concavity.

Using intervals, describe the concavity of $\ds{f(x)=x+\frac{1}{x}}\text{.}$

Solution

We compute $\ds{f'(x) = 1-\frac{1}{x^2}}$ and $\ds{f''(x) = \frac{2}{x^3}}\text{.}$ From the sign diagram for $f''\text{,}$

we see that $f$ is concave downward for $x\in (-\infty,0)$ and concave upward for $x\in(0,\infty)\text{.}$ The graph of $f$ is shown below.

#### Subsubsection5.7.2.1Law of Diminishing Returns

The law of diminishing returns in economics is related to concavity. The graph of the function shown in Figure 5.24 gives the output $y$ from a given input $x\text{.}$ For example, if the input were advertising costs for some product, then the output might be the corresponding revenue from sales. The graph shows an inflection point at $(c,f(c))\text{.}$ For $x\lt c\text{,}$ the graph is concave up, which means that the rate of change of the slope is increasing. Therefore, the output $y$ is increasing at a faster rate with each additional dollar spent. For $x>c\text{,}$ the graph is concave down, which means that the rate of change of the slope is decreasing. Therefore, the output $y$ is decreasing at a faster rate with each additional dollar spent, i.e. producing a diminishing return. Therefore, the point $(c,f(c))$ is referred to as the point of diminishing return. Another example for diminishing returns comes from agriculture, where there is a fixed amount of land, machinery, etc.. Here, adding workers increases production a lot at first, then less and less with each additional worker.

###### Example5.82. Consumer Price Index.

Suppose a certain country's consumer price index (CPI) between the years 1990 and 1999 is approximated by

\begin{equation*} I(t)=-0.2t^3+3t^2+50 \ \ \ \ 0 \leq t \leq 9\text{.} \end{equation*}

Determine the inflection point of $I\text{.}$

Solution

We calculate the first and second derivatives of $I\text{:}$

\begin{equation*} I'(t)=-0.6t^2+6 \text{ , and } \end{equation*}
\begin{equation*} I''(t)=-1.2t+6=-1.2(t-5)\text{.} \end{equation*}

So $I''$ is continuous everywhere and $I''(t)=0 \implies t=5\text{.}$ This gives the only critical point of $I''\text{.}$ Since

\begin{equation*} I''(t) > 0 \text{ for } t \lt 5 \text{ , and } I''(t) \lt 0 \text{ for } t > 5\text{,} \end{equation*}

the point $(5, I(5)) = (5,100)$ is an inflection point of $I\text{.}$ The graph of $I$ is sketched below:

This result reveals that the inflation rate only started decelerating when $t=5\text{,}$ that is, in the year 1995.

##### Exercises for Section 5.7.2.

Describe the concavity of the functions below.

1. $\ds y=x^2-x$

CCU everywhere
Solution

Differentiating twice, we find

\begin{equation*} \begin{split} y''(x) \amp = \diff[2]{}{{x^2}} \left(x^2-x\right)\\ \amp = \diff{}{x} \left(2x-1\right)\\ \amp = 2 \end{split} \end{equation*}

Therefore, $y''> 0$ for all real numbers. That is, $y$ is concave up on $(-\infty,\infty)\text{.}$

2. $\ds y=2+3x-x^3$

CCU when $x\lt 0\text{,}$ CCD when $x>0$
Solution

We compute:

\begin{equation*} \begin{split} y''(x) \amp = \diff[2]{}{{x^2}} \left(2+3x-x^3\right)\\ \amp = \diff{}{x} \left(3-3x^2\right)\\ \amp = -6x \end{split} \end{equation*}

Hence, $y''=0$ at $x=0$ and there are no points where $y''$ is undefined. Using the following test points,

\begin{equation*} y''(-1) = 6 > 0, \ \ y''(1) = -6 \lt 0\text{,} \end{equation*}

we arrive at the sign diagram for $y''\text{:}$

We conclude that $y$ is concave up from $(-\infty, 0)$ and concave down from $(0, \infty)\text{.}$

3. $\ds y=x^3-9x^2+24x$

CCD when $x\lt 3\text{,}$ CCU when $x>3$
Solution

We first find the second derivative of $y\text{:}$

\begin{equation*} \begin{split} y''(x) \amp = \diff[2]{}{{x^2}} \left(x^3-9x^2+24x\right)\\ \amp = \diff{}{x} \left(3x^2-18x+24\right)\\ \amp = 6x-18 \end{split} \end{equation*}

Therefore, $y''$ is defined for all $x$ and $y''=0$ at $x=3\text{.}$ Since

\begin{equation*} y''(0) = -18 \lt 0, \ \ y''(4) = 24-18 > 0\text{,} \end{equation*}

we arrive at the following sign diagram for $y''\text{:}$

Hence, $y$ is concave down on the interval $(-\infty,3)$ and concave up on the interval $(3,\infty)\text{.}$

4. $\ds y=x^4-2x^2+3$

CCU when $\ds x\lt -1/\sqrt3$ or $\ds x>1/\sqrt3\text{,}$ CCD when $\ds -1/\sqrt3\lt x\lt 1/\sqrt3$
Solution

We first compute the second derivative:

\begin{equation*} \begin{split} y''(x) \amp = \diff[2]{}{{x^2}} \left(x^4-2x^2+3\right)\\ \amp = \diff{}{x} \left(4x^3-4x\right)\\ \amp = 12x^2-4 \end{split} \end{equation*}

Then

\begin{equation*} y''=0 \implies x = \pm \frac{1}{\sqrt{3}}\text{.} \end{equation*}

We then use the following test points:

\begin{equation*} y''(-1) = 8 > 0, \ \ y''(0) = -4 \lt 0, \ \ y''(1) = 8 > 0 \end{equation*}

And so our sign diagram for the second derivative is the following:

Hence, $y$ is concave up on the interval $\left(-\infty, - \frac{1}{\sqrt{3}}\right)\cup\left(\frac{1}{\sqrt{3}},\infty\right)$ and concave down on the interval $\left(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)\text{.}$

5. $\ds y=3x^4-4x^3$

CCU when $x\lt 0$ or $x>2/3\text{,}$ CCD when $0\lt x\lt 2/3$
Solution

We first compute $y''\text{:}$

\begin{equation*} \begin{split} y''(x) \amp = \diff[2]{}{x} \left(3x^4-4x^3\right)\\ \amp = \diff{}{x} \left(12x^3-12x^2\right)\\ \amp = 36x^2-24x \end{split} \end{equation*}

There are no points at which $y''$ is undefined, so it remains to determine the points where $y''=0\text{:}$

\begin{equation*} \begin{split} y''(x) \amp = 0 \\ 36x^2-24x \amp = 0 \\ 12x (3x-2) \amp = 0 \\ x \amp = 0, \frac{2}{3} \end{split} \end{equation*}

Now, since

\begin{equation*} y''(-1) = 36 + 24 > 0, \ \ y''(1/2) = -3 \lt 0, \ \ y''(1) = 12 > 0\text{,} \end{equation*}

we create the sign diagram for $y''\text{:}$

We conclude that $y$ is concave down on $(0, 2/3)$ and concave up on $(-\infty, 0) \cup (2/3,\infty)\text{.}$

6. $\ds y=(x^2-1)/x$

CCU when $x\lt 0\text{,}$ CCD when $x>0$
Solution

First notice that $y$ has a discontinuity at $x=0\text{.}$ We calculate $y''\text{:}$

\begin{equation*} \begin{split} y''(x) \amp = \diff[2]{}{x} \left(\frac{x^2-1}{x}\right)\\ \amp = \diff{}{x} \left(\frac{1}{x^2}+1\right)\\ \amp = -\frac{2}{x^3} \end{split} \end{equation*}

And so $y'' = 0$ has no solutions, and $y''$ has a discontinuity at $x=0\text{.}$ Since

\begin{equation*} y''(-1) = -\frac{2}{-1} = 2, \text{ and } y''(1)=-\frac{2}{1} = -2\text{,} \end{equation*}

we can make the following sign diagram for $y''\text{:}$

We conclude that $y(x)$ is concave up on the interval $(-\infty, 0)$ and concave down on the interval $(0,\infty)\text{,}$ and has no inflection point.

7. $\ds y=3x^2-(1/x^2)$

CCU when $x\lt -1$ or $x>1\text{,}$ CCD when $-1\lt x\lt 0$ or $0\lt x\lt 1$
Solution

Notice that the domain of $y$ is $x \in \mathbb{R}-\{0\}\text{.}$ We now compute the second derivative of $y\text{:}$

\begin{equation*} \begin{split} y''(x) \amp = \diff[2]{}{x} \left(3x^2-\frac{1}{x^2}\right) \\ \amp = \diff{}{x} \left(6x+\frac{2}{x^3}\right)\\ \amp = 6 - \frac{6}{x^4} \end{split} \end{equation*}

So we have that $y'' = 0$ at $x=\pm 1\text{.}$ We now evaluate $y''$ at the following test points:

\begin{equation*} y''(-2) = \frac{45}{8} > 0, \ \ y''(-1/2) = -\frac{1}{2}\lt 0, \ \ y''(1/2) = -\frac{1}{2} \lt 0, \ \ y''(2) = \frac{45}{8} > 0\text{.} \end{equation*}

And so the sign diagram for $y''$ is

We conclude that $y$ is concave down on $(-1,0)\cup(1,0)$ and concave up on $(-\infty,-1)\cup(1,\infty)\text{,}$ with inflection points at $x = \pm 1\text{.}$

8. $y=\sin x + \cos x$

CCD on $((8n-1)\pi/4,(8n+3)\pi/4)\text{,}$ CCU on $((8n+3)\pi/4,(8n+7)\pi/4)\text{,}$ for integer $n$
Solution

We compute the second derivative of $y\text{:}$

\begin{equation*} \begin{split} y''(x) \amp = \diff[2]{}{x} \left(\sin x+ \cos x\right) \\ \amp = \diff{}{x} \left(\cos x - \sin x\right)\\ \amp = -\sin x - \cos x \end{split} \end{equation*}

And so $y''(x) = 0$ when

\begin{equation*} \sin x = - \cos x \implies x = n\pi - \frac{\pi}{4}, n \in \mathbb{Z} \end{equation*}

We know that $y(x)$ is periodic on $[0,2\pi]\text{,}$ and so we only need to consider the critical points at $\frac{3\pi}{4}$ and $\frac{7\pi}{4}\text{.}$ Hence, we take the following test points:

\begin{equation*} y''(0) = -1, \ \ y''(3\pi/2) = 1, \ \ y''(2\pi) = - 1 \end{equation*}

This gives us the following sign diagram for $y''\text{:}$

Therefore, on $[0,2\pi]\text{,}$ $y$ is concave up on the interval $\left(-\frac{3\pi}{4},\frac{7\pi}{4}\right)\text{.}$ We conclude that $y$ is concave up on the intervals $\left(\frac{3\pi}{4} + 2\pi n, \frac{7\pi}{4} + 2\pi n\right)$ and concave down on the intervals $\left(-\frac{\pi}{4} + 2\pi n, \frac{3\pi}{4}+ 2\pi n \right)$ for any integer $n\text{.}$

9. $\ds y = 4x+\sqrt{1-x}$

CCD everywhere
10. $\ds y = (x+1)/\sqrt{5x^2 + 35}$

CCU on $\ds (-\infty,(21-\sqrt{497})/4)$ and $\ds (21+\sqrt{497})/4,\infty)$
11. $\ds y= x^5 - x$

CCU on $(0,\infty)$
12. $\ds y = 6x + \sin 3x$

CCD on $(2n\pi/3,(2n+1)\pi/3)$
13. $\ds y = x+ 1/x$

CCU on $(0,\infty)$
14. $\ds y = x^2+ 1/x$

CCU on $(-\infty,-1)$ and $(0,\infty)$
15. $\ds y = (x+5)^{1/4}$

CCD everywhere
16. $\ds y = \tan^2 x$

CCU everywhere
17. $\ds y =\cos^2 x - \sin^2 x$

CCU on $(\pi/4+n\pi,3\pi/4+n\pi)$ for $n$ integer
18. $\ds y = \sin^3 x$

inflection points at $n\pi\text{,}$ $\ds \pm\arcsin(\sqrt{2/3})+n\pi$ for $n$ integer

Describe the concavity of the graphs shown below using intervals, and determine if there are any inflection points.

From the graph above, we see that the graph is CCU on $(-\infty,0)$ and CCD on $(0,\infty)\text{.}$ There is an inflection point at $(0,0)\text{,}$ since the concavity changes.
From the graph above, we see that the graph is CCU on $(-\infty,1) \cup (5,\infty)\text{,}$ CCD on $(1,5)\text{.}$ There are inflection points at $(1,0)$ and $(5,0)$ where the concavity changes.
From the graph above, we see that the graph is CCU on $(-\infty,-1) \cup (1,\infty)$ and CCD on $(-1,1)\text{.}$ There are no inflection points: although the concavity changes around $\pm 1\text{,}$ these points are not in the domain.
From the graph above, we see that the graph is CCU on $(-\infty,0)\text{,}$ CCD on $(0,\infty)\text{.}$ There is an inflection point at $(0,0)$ where the conacvity changes.

Determine if there are any inflection points.

1. $f(x)=x^3+4$

$(0,4)$
Solution

Notice that the domain of $f$ is $(-\infty,\infty)\text{.}$ We now compute the first and second derivatives of $f\text{:}$

\begin{equation*} f'(x) = 3x^2, \ \ f''(x) = 6x\text{.} \end{equation*}

So $f''(x) = 0$ at $x=0$ only, and there are no points at which $f''(x)$ is undefined. Therefore, $x=0$ is the only critical point of $f'(x)\text{.}$ Since

\begin{equation*} f''(-1) = -6 \lt 0, \ \ \text{ and } \ \ f''(1) = 6 > 0\text{,} \end{equation*}

we conclude that $(0, f(0))= (0,4)$ must be an inflection point of $f\text{.}$

2. $f(t)=6t^3-18t^2+12t-15$

$(1,-15)$
Solution

Since $f(t)$ is a polynomial, the domain of $f(t)$ is all real numbers. Now compute:

\begin{equation*} f'(t) = 18t^2-36t+12, \ \ \text{ and } \ \ f''(t) = 36t - 36\text{.} \end{equation*}

Hence, $t=1$ is the only critical point of $f'(t)\text{.}$ We now evaluate:

\begin{equation*} f''(0) = - 36 \lt 0, \ \ \text{ and } \ \ f''(2) = 36 > 0\text{,} \end{equation*}

so the point $(1,f(1)) = (1,-15)$ is an inflection point of $f\text{.}$

3. $g(x)=3x^4-4x^3+1$

$(0,1)$ and $(2/3,11/27)$
Solution

First compute:

\begin{equation*} g'(x) = 12x^3 - 12x^2, \ \ \text{ and } \ \ g''(x) = 12(3x^2-2x)\text{.} \end{equation*}

There are no points for which $g''$ is undefined, and so

\begin{equation*} g''(x) = 0 \implies x = 0, 2/3 \end{equation*}

are the only critical points of $g'(x)\text{.}$ To determine if these are inflection points of $g\text{,}$ we use the following test points:

\begin{equation*} g''(-1) > 0, \ \ g''(1/2) \lt 0 , \ \ g''(1) > 0\text{.} \end{equation*}

We conclude that the points $(0,1)$ and $(2/3, 11/27)$ are both inflection points of $g\text{.}$

4. $h(t)=\sqrt[3]{t}+12$

$(0,12)$
Solution

The domain of $h(t)$ is all real numbers. We first differentiate:

\begin{equation*} h'(t) = \frac{1}{3t^{2/3}}, \ \ \text{ and } \ \ h''(t) = \frac{-2}{9t^{5/3}}\text{.} \end{equation*}

Therefore, there are no critical points of $h'(t)$ but $h'(t)$ is undefined at $t=0\text{.}$ So we check:

\begin{equation*} h''(-1) > 0, \ \ \text{ and } h''(1) \lt 0\text{.} \end{equation*}

Therefore, $h(t)$ has an inflection point at $(0, 12)\text{.}$

5. $p(q)=(q-1)^3-3$

$(1,-1)$
Solution

We calculate

\begin{equation*} p'(q)=3(q-1)^2 \ \ \text{ and } \ \ \ p''(q)=6(q-1)\text{.} \end{equation*}

So $p''(q)=0$ at $q=1\text{.}$ To determine if this is an inflection point, we must check that the sign of the second derivative changes as we move across this point. Since

\begin{equation*} p''(0) = -6 \lt 0 \ \ \text{ and } \ \ p''(2)=6(1) > 0\text{,} \end{equation*}

we conclude that $(1,p(1)) = (1,2)$ is an inflection point of $p\text{.}$

6. $h(s)=\frac{2}{1+s^2}$

$(-\sqrt{3}/3,3/2)$ and $(\sqrt{3}/3,3/2)$
Solution

We have

\begin{equation*} h'(s)=\frac{-4s}{(s^2+1)^2} \ \ \text{ and } \ \ h''(s)=\frac{4(3s^2-1)}{(s^2+1)^3}\text{,} \end{equation*}

therefore $h''(s) = 0$ has solutions $s = \pm \frac{1}{\sqrt{3}}\text{.}$ Again, we need to check the sign of the second derivative around these points. We pick appropriate test points and calculate,

\begin{equation*} h''(-1) = 1, \ \ \text{ and } \ \ h''(0) = -4, \ \ \text{ and } \ \ h''(1) = 1\text{.} \end{equation*}

Thus, we determine that $h$ has two inflection points, $(\pm 1/\sqrt{3}, 3/2)\text{.}$

7. $p(q)=qe^{-4q}$

$(1/2,1/2e^{-2})$
Solution

First calculate

\begin{equation*} p'(q)=e^{-4q}(1-4q) \ \ \text{ and } \ \ p''(q)=8e^{-4q}(2q-1) \end{equation*}

and find that $p''(q) = 0$ at $q=1/2\text{.}$ Since

\begin{equation*} p''(0) = 8 > 0 \ \ \text{ and } \ \ p''(1)=-8/e^4 \lt 0\text{,} \end{equation*}

we see that $f''$ changes from negative to positive as we move through $q=1/2\text{.}$ Thus, $(1/2,1/2e^{2})$ is an inflection point of $p\text{.}$

8. $f(a)=\cos(3a)-\frac{1}{2}$

$(\pi/4+n\pi,-1/2)$ for n integer

Identify the intervals on which the graph of the function $\ds f(x) = x^4-4x^3 +10$ is of one of these four shapes: concave up and increasing; concave up and decreasing; concave down and increasing; concave down and decreasing.

up/incr: $(3,\infty)\text{,}$ up/decr: $(-\infty,0)\text{,}$ $(2,3)\text{,}$ down/decr: $(0,2)$

Solution

The function $f(x)$ is a polynomial, and so is defined for all real numbers. Next, we differentiate:

\begin{equation*} f'(x) = 4x^3 - 12 x^2, \ \ \text{ and } \ \ f''(x)= 12x^2-24x\text{.} \end{equation*}

We first determine the intervals of increase and decrease. We find that

\begin{equation*} f'(x) = 0 \implies x = 0,3\text{.} \end{equation*}

This gives us the critical points of $f(x)\text{.}$ We now use the test points:

\begin{equation*} f'(-1) \lt 0, \ \ f'(1) > 0, \ \ f'(4) >0\text{.} \end{equation*}

We now create the sign diagram for $f'\text{:}$

Hence, $f(x)$ is increasing on the interval $(0,3)$ and decreasing on $(-\infty,0)\cup(3,\infty)\text{.}$

Next, we look at the concavity of the function. The critical points of $f'$ are given by

\begin{equation*} f''(x) = 0 \implies x = 0, 2\text{.} \end{equation*}

Since

\begin{equation*} f''(-1) > 0, \ \ f''(1) \lt 0, \ \ f''(3) > 0\text{,} \end{equation*}

we arrive at the following sign diagram for $f''\text{:}$

We conclude that

• $f(x)$ is concave up and increasing on $(3,\infty)\text{;}$

• $f(x)$ is concave up and decreasing on $(-\infty,0)\cup(2,3)\text{;}$

• $f(x)$ is concave down and increasing on $(0,2)\text{;}$

• and that $f(x)$ is concave down and decreasing nowhere.

Describe the concavity of $\ds y = x^3 + bx^2 + cx + d\text{.}$ You will need to consider different cases, depending on the values of the coefficients.

Solution

Notice that, for any choice of coefficients, the domain of $f(x)$ will be all real numbers. Now differentiate:

\begin{equation*} f'(x) = 3x^2+2bx+c, \ \ \text{ and } \ \ f''(x) = 6x+2b\text{.} \end{equation*}

We need to determine the point(s) where $f''(x) = 0\text{.}$ To do so, we consider three cases.

CASE 1: $b = 0$

Then $f''(x) = 6x$ and therefore $f'(x)$ has one critical point at $x=0\text{.}$ Since

\begin{equation*} f''(-1) = -6 \lt 0, \ \ \text{ and } \ \ f''(1) = 6 > 0\text{,} \end{equation*}

the point $(0,d)$ is an inflection point of $f(x)\text{.}$ We conclude that $f(x)$ is concave up on $(0,\infty)$ and concave down on $(-\infty,0)\text{.}$

CASE 2: $b>0$

Then $f''(x)=6x+2b = 0$ at $x=-\frac{b}{3}\text{.}$ Since $b > 0\text{,}$ we have that

\begin{equation*} -b \lt -\frac{b}{3} \lt 0\text{.} \end{equation*}

So we evaluate:

\begin{equation*} f''(-b) = -4b \lt 0, \ \ f''(0) = 2b > 0\text{.} \end{equation*}

Hence, $f''(x)$ is concave down from $(-\infty, -b/3)$ and concave up on $(-b/3,\infty)\text{.}$

CASE 3: $b \lt 0$

Again, we have that $x=-\frac{b}{3}$ is a critical point of $f'(x)\text{,}$ but now since $b \lt 0$

\begin{equation*} 0 \lt -\frac{b}{3} \lt -b\text{.} \end{equation*}

Again, we evalaute:

\begin{equation*} f''(0) = 2b \lt 0, \ \ f''(-b) = -4b > 0\text{.} \end{equation*}

Hence, $f''(x)$ is concave up from $(-\infty, -b/3)$ and concave down on $(-b/3,\infty)\text{.}$

Let $n$ be an integer greater than or equal to two, and suppose $f$ is a polynomial of degree $n\text{.}$ How many inflection points can $f$ have? Hint
Use the Second Derivative Test and the Fundamental Theorem of Algebra.
Solution

Let

\begin{equation*} f(x) = P_n(x) = a_nx^n + a_{n-1}x^{n-1} + \dots a_0, n \geq 2\text{.} \end{equation*}

Therefore,

\begin{equation*} f''(x) = n(n-1)a_nx^{n-2} + (n-1)(n-2)a_{n-1}x^{n-3}+ \dots + 2a_2\text{.} \end{equation*}

All inflection points of $f$ must satisfy $f''(x) = 0\text{.}$ Since $f''(x)$ is a polynomial of degree $n-2$ (where $n-2 \geq 0$), by the Fundamental Theorem of Algebra, $f''$ has at most $n-2$ roots. These roots are the only potential critical points of $f'(x)\text{,}$ and so $f(x)$ has at most $n-2$ inflection points.

A retailer determines that their revenue $R$ as a function of the amount $q$ (in thousands of dollars) spent on advertising can be approximated by

\begin{equation*} R(q)=-0.003q^3+1.35q^2+2q+8000 \ \ \ 0\leq q \leq 400 \end{equation*}

thousands of dollars.

1. Determine the concavity of $R$ using intervals and determine if there are any inflection points.

$(150,28850)$
Solution

The domain of $R$ is $[0,400]\text{.}$ We first differentiate:

\begin{equation*} R'(q) = -0.009q^2+2.7q+2 \text{ and } R''(q) = -0.018q+2.7\text{.} \end{equation*}

We now solve for the critical points of $R'(q)\text{.}$ We set $R''(q) = 0\text{:}$

\begin{equation*} -0.018q+2.7 = 0 \implies q = 150\text{.} \end{equation*}

Since $R'(q)$ is a polynomial (and so is defined over all real numbers), and $q=150$ is in the domain of $R\text{,}$ we conclude that $q=150$ is the only critical point of $R'(q)\text{.}$ We arrive at the following sign diagram for $R''(q)\text{:}$

Hence, the graph of the revenue function $R(q)$ is concave up on $[0,150)$ and concave down on $(150,400]\text{.}$ The point $(150, 28850)$ is an inflection point of $R\text{.}$

2. If the retailer currently spends $140,000 on advertising, should they consider increasing this amount? Answer The revenue function $R(q)$ is increasing on the interval $[0,300]\text{.}$ Therefore, the point $q=150$ indicated the point of diminishing returns. Hence, one would not recommend increasing the amount spent from$140,000.

A grocery store determines that their sales $S$ as a function of the amount $q$ (in thousands of dollars) spent on advertising can be approximated by

\begin{equation*} S(q) = -0.002q^3+0.6q^2+q+500 \ \ \ \ 0\leq q \leq 200 \end{equation*}

thousands of dollars. Determine if there are any inflection points and, if there is, discuss its significance.

$(100,4600)$

Solution

We first differentiate:

\begin{equation*} S'(q) = -0.006q^2+1.2q+1 \text{ and } S''(q) = -0.012q+1.2\text{.} \end{equation*}

We now solve for the critical points of $S'(q)\text{:}$

\begin{equation*} S''(q) = 0 \implies q = 100\text{.} \end{equation*}

Since there are no points at which $S'$ is undefined and $q=100$ is in the domain of $S\text{,}$ we conclude that this is the only critical point of $S'\text{.}$ We now draw the sign diagram for $S''\text{:}$

Therefore, the point $(100, 4600)$ is an inflection point of $S(q)\text{.}$ Since $S(q)$ is increasing on its entire domain, and is concave up from $[0,100)$ and concave down from $(100,200]\text{,}$ we conclude that the point $(100,4600)$ is the point of diminishing returns. Therefore, it is not recommended to increase the quantity spent on advertising beyond \$100,000.

A company wants to determine whether or not their cost-cutting measures (implemented at time $t \equiv 0$) will be effective. Suppose the profit $P$ (in hundreds of dollars) of the company over the next 8 years can be approximated by

\begin{equation*} P(t)=t^3-9t^2+40t+50 \ \ \ 0\leq t \leq 8\text{.} \end{equation*}

By studying the concavity of $P\text{,}$ deduce whether or not these measures will be effective.

After declining the first 3 years, the growth rate of the company's profit will once again rise.

Solution

Differentiating twice, we find

\begin{equation*} P''(t) = 6(t-3)\text{.} \end{equation*}

Hence, $P''(t) = 0 \implies t = 3\text{.}$ This gives the only critical point of $P'(t)\text{.}$ We now construct the sign diagram for the second derivative:

Therefore, the graph of the profit function is concave down on $[0,3)$ and concave up on $(3,8]\text{.}$ Therefore, after declining during the first 3 years, the growth rate of the company's profit starts to rise after the third year of implementing the cost-cutting measures.

### Subsection5.7.3The Second Derivative Test for Relative Extrema

The basis of the First Derivative Test is that if the derivative changes from positive to negative at a critical point then there is a relative maximum at the point, and similarly for a relative minimum. However, we can also use our knowledge from concavity to test for relative extrema at a critical point. In Figure 5.(a), we observe that the graph of the function $f$ shown here has a relative minimum at the critical point $x=c$ and that the graph is concave upward at that point. Similarly, in Figure 5.(b), we observe that the graph of the function $f$ shown here has a relative maximum at the critical point $x=c$ and that the graph is concave downward at that point. Furthermore, in Section 5.7.2 we just learned that given a twice differentiable function $f$ and a critical point $x=c$ of $f$ the function $f$ is concave upward at $x=c$ if $f''(c) > 0$ and concave downward at $x=c$ if $f''(c) \lt 0\text{.}$ We therefore introduce the so-called Second Derivative Test for relative extrema as an alternative technique to the First Derivative Test for checking critical points for relative extrema, as long as the second derivative exists.

Note: The above theorem indicates that the Second Derivative Test for relative extrema is inconclusive if $f''(c)=0$ for a critical point $c$ as required, but furthermore, this test also tells us nothing when $f''(c)$ does not exist and also when there are critical points such that $f'(c)$ does not exist. You may wonder why we should bother with the Second Derivative Test. Well, if the second derivative is easy to compute, then this test is an efficient method for finding relative extrema.
###### Example5.84. Second Derivative Test for Relative Extrema.

Consider again $f(x)=\sin x + \cos x\text{,}$ with $f'(x)=\cos x-\sin x$ and $f''(x)=-\sin x -\cos x\text{.}$ Use the Second Derivative Test to determine which critical points are relative maxima or minima.

Solution

Since $\ds f''(\pi/4)=-\sqrt{2}/2-\sqrt2/2=-\sqrt2\lt 0\text{,}$ we know there is a relative maximum at $\pi/4\text{.}$ Since $\ds f''(5\pi/4)=-(-\sqrt{2}/2)-(-\sqrt2/2)=\sqrt2>0\text{,}$ there is a relative minimum at $5\pi/4\text{.}$

###### Example5.85. Second Derivative Test for Relative Extrema.

Let $\ds f(x)=x^4$ and $\ds g(x)=-x^4\text{.}$ Classify the critical points of $f(x)$ and $g(x)$ as either maximum or minimum.

Solution

The derivatives for $f(x)$ are $f'(x)=4x^3$ and $\ds f''(x)=12x^2\text{.}$ Zero is the only critical value, but $f''(0)=0\text{,}$ so the Second Derivative Test tells us nothing. However, $f(x)$ is positive everywhere except at zero, so clearly $f(x)$ has a relative minimum at zero.

On the other hand, for $g(x)=-x^4\text{,}$ $g'(x)=-4x^3$ and $g''(x)=-12x^2\text{.}$ So $g(x)$ also has zero as its only only critical value, and the second derivative is again zero, but $g(x)$ has a relative maximum at zero.

We conclude this section with a summary of the graphical information gained from both the first and second derivatives of a function $f\text{.}$ The first derivative tells us about the monotonicity of the graph of $f\text{,}$ i.e. where the graph goes up ($f$ is increasing) or down ($f$ is decreasing). The second derivative tells us about the concavity of the graph of $f\text{,}$ i.e. where the graph curves up ($f$ is concave up) or down ($f$ is concave down). Together, these properties tell us how the function $f$ is increasing or decreasing as is shown in Table 5.26.

##### Exercises for Section 5.7.3.

Find all relative maximum and minimum points by the Second Derivative Test.

1. $\ds y=x^2-x$

rel min at $x=1/2$
Solution

We first differentiate:

\begin{equation*} y' = 2x -1 \\ \text{ and } \\ y'' = 2\text{.} \end{equation*}

Therefore, $y' = 0 \implies x = 1/2$ is the only critical point of $y\text{.}$ Since $y''> 0$ for all $x\text{,}$ by the Second Derivative Test, $x=1/2$ must be a relative minimum.

2. $\ds y=2+3x-x^3$

rel min at $x=-1\text{,}$ rel max at $x=1$
Solution

We will use to second derivative test to determine the relative extrema of $y=2+3x-x^3\text{.}$ First, we calculate

\begin{equation*} y' = 3-3x^2 \ \ \text{ and } \ \ y''=-6x\text{.} \end{equation*}

As before, we find the critical points of $y$ by setting the first derivative to zero.

\begin{equation*} 3-3x^2 = 0 \implies x = \pm 1\text{.} \end{equation*}

We now evaluate the second derivative at each critical point to determine the concavity of $y$ at these points:

\begin{equation*} y''(-1) = -6 \lt 0 \ \ \text{ and } \ \ y''(1) = 6 > 0\text{.} \end{equation*}

Therefore, $y$ is concave downward at $x=-1\text{,}$ and so this corresponds to a relative maximum. Since $y$ is concave upward at $x=1\text{,}$ this corresponds to a relative minimum.

3. $\ds y=x^3-9x^2+24x$

rel max at $x=2\text{,}$ rel min at $x=4$
Solution

Differentiating twice, we find

\begin{equation*} y' = 3x^2-18x + 24 \ \ \text{ and } \ \ y'' = 6x-18\text{.} \end{equation*}

To find the critical points of $y\text{,}$ we set $y' = 0\text{:}$

\begin{equation*} 3x^2-18x+24 = 0 \implies x=2,4\text{.} \end{equation*}

Now since

\begin{equation*} y''(2) = 12-18 \lt 0, \ \ \text{ and } \ \ y''(4) = 24-18 > 0\text{,} \end{equation*}

by the Second Derivative Test, we conclude that $x=2$ is a relative maximum of $y$ and $x=4$ is a relative minimum.

4. $\ds y=x^4-2x^2+3$

rel min at $x=\pm 1\text{,}$ rel max at $x=0$
Solution

We compute $y'(x)$ and $y''(x)\text{:}$

\begin{equation*} \begin{split} y'(x) \amp = \diff{}{x} \left(x^4-2x^2+3\right)\\ \amp = 4x^3-4x\\ y''(x) \amp = 12x^2 - 4 \end{split} \end{equation*}

Hence, the critical points of $y$ occur when $y'(x) = 0\text{:}$

\begin{equation*} 4x(x^2-1) = 0 \implies x = 0, \pm 1 \end{equation*}

We now determine the concavity of $y$ at each of the three critical points found above:

\begin{equation*} y''(-1) = y''(1) = 8 > 0, \text{ and } y''(0) = -4 \lt 0 \end{equation*}

Therefore, $y$ has a relative maximum at $x=0$ and relative minima at $x=\pm 1\text{.}$

5. $\ds y=3x^4-4x^3$

rel min at $x=1$
Solution

We differentiate twice:

\begin{equation*} y' = 12x^3 - 12x^2, \ \ \text{ and } \ \ y'' = 36x^2-24x\text{.} \end{equation*}

There are no points at which $y'$ is undefined, and so to find all critical points of $y\text{,}$ we set $y'=0\text{:}$

\begin{equation*} 12x^3 - 12x^2 = 0 \implies x = 0, 1\text{.} \end{equation*}

We now evaluate the second derivative at each critical point:

\begin{equation*} y''(0) = 0, \ \ \text{ and } \ \ y''(1) = 36-24 > 0\text{.} \end{equation*}

Therefore, by the Second Derivative Test, $y$ has a relative minimum at $x=1\text{,}$ and is inconclusive at $x=0\text{.}$ (The First Derivative Test tells us that this is neither a relative maximum nor a relative minimum).

6. $\ds y=(x^2-1)/x$

none
Solution

First, notice that $y$ is undefined when $x = 0\text{.}$ Next, we calculate

\begin{equation*} y'(x) = \frac{1}{x^2} +1, \text{ and } y''(x) = -\frac{2}{x^3}\text{.} \end{equation*}

Therefore, $y$ has no critical points since $y'(x) = 0$ has no real solutions, and $y'(x)$ is undefined only at $x=0\text{,}$ which is not in the domain of $y\text{.}$ Hence, $y$ has no relative maxima or minima.

7. $\ds y=3x^2-(1/x^2)$

none
Solution

The domain of $y$ is $(-\infty,0)\cup(0,\infty)\text{.}$ Now differentiate:

\begin{equation*} y' = 6x+\frac{2}{x^3}, \ \ \text{ and } \ \ y'' = 6 - \frac{6}{x^4}\text{.} \end{equation*}

Hence, $y'$ is undefined at $x=0\text{,}$ but this is not in the domain of $y$ and so is not a critical point. We solve:

\begin{equation*} y' = 0 \implies x^4 = -\frac{1}{3}\text{,} \end{equation*}

which has no real solutions. Therefore, $y$ has no relative extrema.

8. $y=\cos(2x)-x$

rel min at $x=7\pi/12+n\pi\text{,}$ rel max at $x=-\pi/12+n\pi\text{,}$ for integer $n$
9. $\ds y = 4x+\sqrt{1-x}$

rel max at $x=63/64$
10. $\ds y = \frac{x+1}{\sqrt{5x^2 + 35}}$

rel max at $x=7$
11. $\ds y= x^5 - x$

rel max at $\ds -5^{-1/4}\text{,}$ rel min at $\ds 5^{-1/4}$
12. $\ds y = 6x +\sin 3x$

none
13. $\ds y = x+ 1/x$

rel max at $-1\text{,}$ rel min at $1$
14. $\ds y = x^2+ 1/x$

rel min at $\ds 2^{-1/3}$
15. $\ds y = (x+5)^{1/4}$

none
16. $\ds y = \tan^2 x$

rel min at $n\pi\text{,}$ for integer $n$
17. $\ds y =\cos^2 x - \sin^2 x$

rel max at $n\pi\text{,}$ rel min at $\pi/2+n\pi$ for integer $n$
18. $\ds y = \sin^3 x$

rel max at $\pi/2+2n\pi\text{,}$ rel min at $3\pi/2+2n\pi$ for integer $n$

Suppose a function $y=f(x)$ has the following properties:

\begin{equation*} f(2)=1, f'(2)>0, \text{ and } f''(2) \lt 0\text{.} \end{equation*}

Decide which graph below corresponds to the function $f\text{.}$

Solution

All three graphs satisfy $f(2)=1\text{.}$ However, at $x=2\text{,}$ the first derivative of graph (c) is undefined, and so we can eliminate this option. Further, graph (b) has an inflection point at $x=2\text{.}$ We are left with graph (a), which is increasing and concave down at $x=2\text{.}$

Suppose a function $y=f(x)$ has the following properties:

\begin{equation*} f(1)=2, f'(x)> 0 \text{ on } (-\infty,1) \cup (1,\infty), \text{ and } f''(1) = 0\text{.} \end{equation*}

Decide which graph below corresponds to the function $f\text{.}$

Solution

All three graphs satisfy the condition $f(1)=2$ and are increasing on $(-\infty,1)\text{.}$ However, only graphs (b) an (c) are increasing on $(1,\infty)\text{.}$ Both the graphs (b) and (c) have an inflection point at $x=1\text{,}$ however, we see that $f''(1)$ is undefined in (b). Therefore, the function must correspond to graph (c).

Suppose a function $y=f(x)$ has the following properties:

\begin{equation*} f'(0) \text{ undefined } , f'(x) \lt 0 \text{ on } (-\infty,0), f''(x) \lt 0 \text{ on } (0,3)\text{,} \end{equation*}

and $f$ has an inflection point at $x=3\text{.}$ Decide which graph below corresponds to the function $f\text{.}$

Solution

First look at $x=0\text{.}$ In (a), the graph has a cusp at $x=0\text{,}$ and so $f'(0)$ is undefined. In (b) and (c), $f$ is undefined at $x=0\text{,}$ and so $f'(0)$ is also undefined. We see also that all three graphs are decreasing on the interval $(-\infty, 0)$ and concave downward on $(0,3)\text{.}$ The remaining condition is that there is an inflection point at $x=3\text{.}$

The graph (a) is concave downward on $(0,\infty)$ and so cannot have an inflection point at $x=3\text{.}$ The graph (c) changes concavity around $x=3\text{,}$ but the point itself is not defined, and so cannot be an inflection point. We deduce that the desired graph must be (b), and indeed see that $x=3$ is an inflection point.

A busy coffee shop determines that the number $N$ of transactions processed $t$ hours after opening at $6$ am can be described by

\begin{equation*} N(t) = -t^3+5t^2+25t \qquad 0 \leq t \leq 8\text{.} \end{equation*}

What is the shop's busiest hour?

11 am
Solution

Differentiating twice, we find

\begin{equation*} N'(t) = -3t^2+10t+25, \ \ \text{ and } \ \ N''(t) = -6t+10\text{.} \end{equation*}

To find the critical points of $N\text{,}$ we set $N'(t) = 0\text{:}$

\begin{equation*} -3t^2+10t+25 = 0 \implies t = -5/3, 5\text{.} \end{equation*}

Since the domain of $N$ is $[0,8]\text{,}$ we conclude that $t=5$ is the only critical point of $N\text{.}$ We now compute

\begin{equation*} N''(5) = -30+10 \lt 0\text{,} \end{equation*}

and so by the Second Derivative Test, the point $t=5$ corresponds to a relative maximum. Since this is the only relative extremum in the domain of $N\text{,}$ this must be the absolute maximum of $N\text{.}$

Therefore, the shop is busiest 5 hours after opening, at 11 am.

A manufacturer determines that the daily cost $C$ (in dollars) of producing $q$ units is given by

\begin{equation*} C(q)=q^3-30q^2+300q+50\text{.} \end{equation*}

Determine if there are any inflection points and interpret your result.

$(10,1050)$

Solution

We find the first and second derivatives of $C\text{:}$

\begin{equation*} C'(q) = 3q^2-60q+300, \ \ C''(q) = 6q-60\text{.} \end{equation*}

We now find the critical points of both $C$ and $C'\text{:}$

\begin{equation*} C'(q) = 0 \implies 3q^2-60q+300 = 0 \implies q = 10\text{,} \end{equation*}

and

\begin{equation*} C''(q) = 0 \implies 6q-6=0 \implies q =10\text{.} \end{equation*}

On the interval $[0,10)\text{,}$ we see that $C'>0$ and $C''\lt 0\text{.}$ On the interval $(10,\infty)\text{,}$ we find $C'>0$ and $C''\lt 0\text{.}$ Hence, the point $(10,1050)$ is an inflection point of $C\text{,}$ and $C$ is increasing over its entire domain.

Since $C$ moves from increasing and concave up to increasing and concave down, we conclude that $q=10$ is the point of diminishing returns.

A busy coffee shop determines that the number $N$ of transactions processed $t$ hours after opening at $6$ am can be described by

\begin{equation*} N(t) = -t^3+5t^2+25t \qquad 0 \leq t \leq 8\text{.} \end{equation*}
1. Describe the rate of change of the number of transactions between 6 am and 11 am, and between 11 am and 2 pm.

The number of transactions is increasing from 6 to 11 am; the rate of transactions is increasing the fastest at around 8am, and then starts to decrease. From 11 am to 2 pm, the number of transactions is decreasing; the rate of transactions is decreasing until the shop closes.
Solution

Differentiating, we find

\begin{equation*} N'(t) = -3t^2+10t+25\text{.} \end{equation*}

To find the critical points of $N\text{,}$ we set $N'(t) = 0\text{:}$

\begin{equation*} -3t^2+10t+25 = 0 \implies t = -5/3, 5\text{.} \end{equation*}

Since the domain of $N$ is $[0,8]\text{,}$ we conclude that $t=5$ is the only critical point of $N\text{.}$ We now determine the sign daigram for $N'\text{:}$

Therefore, the number of transactions are increasing between 6 am and 11 am, and decreasing between 11 am and 2 pm.

2. When is the rate of change of the number of transactions maximal?

By the First Derivative Test, the rate of change of the number of transactions is maximal at 11 am.

### Subsection5.7.4Asymptotes and Other Things to Look For

#### Subsubsection5.7.4.1Vertical Asymptotes Revisited

Vertical Asymptotes were introduced in Section 3.5.3. Since they play an important role in curve sketching, we provide a short summary as a reminder of the idea. A vertical asymptote is a place where the function approaches infinity, typically because the formula for the function has a denominator that becomes zero. For example, the reciprocal function $f(x)=1/x$ has a vertical asymptote at $x=0\text{,}$ and the function $\tan x$ has a vertical asymptote at $x=\pi/2$ (and also at $x=-\pi/2\text{,}$ $x=3\pi/2\text{,}$ etc.). Whenever the formula for a function contains a denominator it is worth looking for a vertical asymptote by checking to see if the denominator can ever be zero, and then checking the limit at such points. Note that there is not always a vertical asymptote where the derivative is zero: $f(x)=(\sin x)/x$ has a zero denominator at $x=0\text{,}$ but since $\ds \lim_{x\to 0}(\sin x)/x=1$ there is no asymptote there. Note also that the graph of a function can have any number of vertical asymptotes from none (e.g., any polynomial function, Figure 5.(a)) to infinitely many (e.g., tangent function) Figure 5.(c).

#### Subsubsection5.7.4.2Horizontal Asymptotes Revisited

In Section 3.5.4 we discussed horizontal asymptotes. These too are a fundamental feature when sketching the graph of a function, and so we offer a short summary about them. A horizontal asymptote is a horizontal line to which $f(x)$ gets closer and closer as $x$ approaches $\infty$ (or as $x$ approaches $-\infty$). Hence, the graph of a function can have at most two horizontal asymptotes. Horizontal asymptotes can be identified by computing the limits $\ds \lim_{x \to \infty}f(x)$ and $\ds \lim_{x \to -\infty}f(x)\text{.}$ For example, a polynomial function has no horizontal asymptotes as shown in Figure 5.(a). Since $\ds \lim_{x \to \infty}1/x=\lim_{x \to -\infty}1/x=0\text{,}$ the line $y=0$ (that is, the $x$-axis) is a horizontal asymptote in both directions; for example, see the reciprocal function in Figure 5.(b). The arctangent is an example of a function with two horizontal asymptotes (Figure 5.(c)).

#### Subsubsection5.7.4.3Slant Asymptotes and other Asymptotic Behaviour

Some functions have straight asymptotes that are neither horizontal nor vertical, but slanted.These too are a fundamental feature when sketching the graph of a function, and so we offer a short summary about slant asymptotes from our introduction in Section 3.5.5. A slant asymptote is again a line that $f(x)$ gets closer and closer to as $x$ approaches $\infty$ (or as $x$ approaches -$\infty$). If $y=mx+b$ describes a slant asymptote, then $\lim\limits_{x \to \infty}\left(f(x)-(mx+b)\right)=0$ or $\lim\limits_{x\to -\infty}\left(f(x)-(mx+b)\right)=0\text{.}$ In the case of rational functions, slant asymptotes occur when the degree of the polynomial in the numerator is one more than the degree of the polynomial in the denominator, see Figure 5.29.

If the degree between the polynomials of the numerator and denominator are higher than one, say $n\text{,}$ then the rational function exhibits asymptotic behaviour towards a polynomial with degree $n\text{.}$ This polynomial can be found by long division and taking a similar limit approach as for slant asymptotes, but this is beyond the scope of this material. Other functions could also exhibit different asymptotic behaviour, but such asymptotes are somewhat more difficult to identify and we will ignore them.

#### Subsubsection5.7.4.4Endpoints or Other Special Points of Domain

If the domain of the function does not extend out to infinity, we should also ask what happens as $x$ approaches the boundary of the domain. For example, the function $\ds y=f(x)=1/\sqrt{r^2-x^2}$ has domain $-r\lt x\lt r\text{,}$ and $y$ becomes infinite as $x$ approaches either $r$ or $-r$ (see Figure 5.30). In this case we might also identify this behaviour because when $x=\pm r$ the denominator of the function is zero.

If there are any points where the derivative fails to exist (a cusp or corner), then we should take special note of what the function does at such a point.

#### Subsubsection5.7.4.5Function Symmetry

Finally, it is worthwhile to notice any symmetry. A function $f(x)$ that has the same value for $-x$ as for $x\text{,}$ i.e., $f(-x)=f(x)\text{,}$ is called an even function. Its graph is symmetric with respect to the $y$-axis. Some examples of even functions are: $\ds x^n$ when $n$ is an even number, $\cos x\text{,}$ and $\ds \sin^2x\text{,}$ see Figure 5.(a). On the other hand, a function that satisfies the property $f(-x)=-f(x)$ is called an odd function. Its graph is symmetric with respect to the origin. Some examples of odd functions are: $x^n$ when $n$ is an odd number, $\sin x\text{,}$ and $\tan x\text{,}$ see Figure 5.(b). Of course, most functions are neither even nor odd, and do not have any particular symmetry.

##### Exercises for Section 5.7.4.

State the domain and determine the vertical and horizontal asymptotes, if any, of the following functions.

1. $f(x)=\dfrac{9x-1}{x^2-9}$

$D=(-\infty,-3)\cup(-3,3)\cup(3,\infty)\text{;}$$x=\pm 3\text{;}$$y=0\text{.}$
Solution

The domain of $f$ is all real numbers such that

\begin{equation*} x^2-9 \neq 0 \implies x \neq \pm 3\text{.} \end{equation*}

Therefore, the domain of $f$ is $(-\infty,-3)\cup(-3,3)\cup(3,\infty)\text{.}$ We now look for vertical asymptotes:

\begin{equation*} \lim_{x\to-3^-} \frac{9x-1}{x^2-9} = \lim_{x\to-3^-} \frac{9x-1}{(x-3)(x+3)} = -\infty\text{,} \end{equation*}

and similarly

\begin{equation*} \lim_{x\to-3^+} \frac{9x-1}{x^2-9} = \lim_{x\to-3^+} \frac{9x-1}{(x-3)(x+3)} = \infty\text{.} \end{equation*}

Therefore, $f(x)$ has a vertical asymptote at $x=-3\text{.}$ We now check the behaviour of the function around $x=3\text{:}$

\begin{equation*} \lim_{x\to 3^-} \frac{9x-1}{x^2-9} = \lim_{x\to 3^-} \frac{9x-1}{(x-3)(x+3)} = -\infty\text{,} \end{equation*}

and similarly

\begin{equation*} \lim_{x\to 3^+} \frac{9x-1}{x^2-9} = \lim_{x\to 3^+} \frac{9x-1}{(x-3)(x+3)} = \infty\text{.} \end{equation*}

So $f(x)$ has a second vertical asymptote at $x=3\text{.}$ We now look for horizontal asymptotes. First, check

\begin{equation*} \lim_{x\to\infty} \frac{9x-1}{x^2-9} = \lim_{x\to\infty} \frac{9x}{x^2} = \lim_{x\to\infty} \frac{9}{x} = 0\text{.} \end{equation*}

Similarly, we find

\begin{equation*} \lim_{x\to-\infty} \frac{9x-1}{x^2-9} = \lim_{x\to-\infty} \frac{9x}{x^2} = \lim_{x\to-\infty} \frac{9}{x} = 0\text{.} \end{equation*}

Therefore, the graph of the function $f(x)$ has one horizontal asymptote at $y=0\text{.}$

2. $f(x)=\dfrac{8x^2+9}{x^2-16}$

$D=(-\infty,-4)\cup(-4,4)\cup(4,\infty)\text{;}$$x=\pm 3\text{;}$$y=8\text{.}$
Solution

The domain of $f(x)$ is $(-\infty,-4)\cup(-4,4)\cup(4,\infty)\text{.}$ We first check for vertical asymptotes. Since

\begin{equation*} \lim_{x\to-4^-} \frac{8x^2+9}{(x-4)(x+4)} = -\infty\text{,} \end{equation*}

and

\begin{equation*} \lim_{x\to-4^+} \frac{8x^2+9}{(x-4)(x+4)} = \infty\text{,} \end{equation*}

we conclude that $f(x)$ has a vertical asymptote at $x=-4\text{.}$ Similarly,

\begin{equation*} \lim_{x\to 4^-} \frac{8x^2+9}{(x-4)(x+4)} = -\infty\text{,} \end{equation*}

and

\begin{equation*} \lim_{x\to 4^+} \frac{8x^2+9}{(x-4)(x+4)} = \infty\text{,} \end{equation*}

and so $f(x)$ has a second vertical asymptote at $x=4\text{.}$ To look for any horizontal asymptotes, we compute:

\begin{equation*} \lim_{x\to \infty} \frac{8x^2+9}{x^2-16} = \lim_{x\to \infty} \frac{8x^2}{x^2} = 8\text{,} \end{equation*}

and

\begin{equation*} \lim_{x\to -\infty} \frac{8x^2+9}{x^2-16} = \lim_{x\to -\infty} \frac{8x^2}{x^2} = 8\text{.} \end{equation*}

Therefore, the graph of $f(x)$ has one horizontal asymptote at $y=8\text{.}$

3. $f(x)=\dfrac{6(x^2+10)}{x^2+2x-15}$

$D=(-\infty,-5)\cup(-5,3)\cup(3,\infty)\text{;}$$x=-5,3\text{;}$$y=6\text{.}$
Solution

The domain of $f(x)$ is $(-\infty,-5)\cup(-5,3)\cup(3,\infty)\text{.}$ First, we will check for any vertical asymptotes:

\begin{equation*} \lim_{x\to -5^-} \frac{6(x^2+10)}{(x-3)(x+5)} = -\infty \end{equation*}

and

\begin{equation*} \lim_{x\to -5^+} \frac{6(x^2+10)}{(x-3)(x+5)} = +\infty\text{.} \end{equation*}

Similarly, we find that

\begin{equation*} \lim_{x\to 3^-} \frac{6(x^2+10)}{(x-3)(x+5)} = -\infty \end{equation*}

and

\begin{equation*} \lim_{x\to 3^+} \frac{6(x^2+10)}{(x-3)(x+5)} = +\infty\text{.} \end{equation*}

Therefore, the graph of $f(x)$ has two vertical asymptotes, one at $x=3$ and another at $x=-5\text{.}$ Next, we compute the following limits:

\begin{equation*} \lim_{x\to \infty} \frac{6(x^2+10)}{x^2+2x-15} = \lim_{x\to\infty} \frac{6x^2}{x^2} = 6\text{,} \end{equation*}

and

\begin{equation*} \lim_{x\to -\infty} \frac{6(x^2+10)}{x^2+2x-15} = \lim_{x\to-\infty} \frac{6x^2}{x^2} = 6\text{.} \end{equation*}

Thus, the graph of $f(x)$ has one horizontal asymptote at $y=6\text{.}$

Match the graphs labelled A through F with one of the rational functions labelled $a$ through $f\text{.}$

\begin{equation*} \begin{array}{lll} a(x)=\dfrac{x^2-25}{x^2+3x-10} \amp b(x)=\dfrac{5}{x^2-2x-8} \amp c(x)=\dfrac{3x^2-5}{x^2-2x+1} \\[1ex] d(x)=\dfrac{x^2}{x^2+x-12}\amp e(x)= \dfrac{7}{x^2+x-12} \amp f(x)=\dfrac{x^2-2x}{x^2-7x+10} \end{array} \end{equation*}

$e(x)\text{,}$ $a(x)\text{,}$ $f(x)$

$c(x)\text{,}$ $d(x)\text{,}$ $b(x)$

Solution

We first write each of the functions in their factorized form. This gives,

\begin{equation*} \begin{array}{ll} a(x)=\dfrac{(x-5)(x+5)}{(x-5)(x-2)} \amp b(x)=\dfrac{5}{(x+2)(x-4)} \\ c(x)=\dfrac{3x^2-5}{(x-1)^2} \amp d(x)=\dfrac{x^2}{(x+4)(x-3)}\\ e(x)= \dfrac{7}{(x+4)(x-3)} \amp f(x)=\dfrac{x(x-2)}{(x-2)(x-5)} \end{array} \end{equation*}

From this, we see that $a(x)$ must correspond to the graph $B\text{,}$ since it is the only function with a vertical asymptote at $x=2\text{.}$ Next, we see that $b(x)$ has vertical asymptotes at $x=-2$ and $x=4$ and so must correspond to the graph $F\text{.}$ Similarly, $c(x)$ has a vertical asymptote at $x=1$ and must correspond to the graph $D\text{.}$

Both $e(x)$ and $d(x)$ have vertical asymptotes at $x=-4$ and $x=3\text{,}$ and so we need to check the horizontal asymptote(s) of each. Since

\begin{equation*} \lim_{x\to \infty} d(x) = \lim_{x\to \infty} \dfrac{x^2}{(x+4)(x-3)} = 1, \text{ and } \end{equation*}
\begin{equation*} \lim_{x\to \infty} e(x) = \lim_{x\to \infty} \dfrac{7}{(x+4)(x-3)} = 0\text{,} \end{equation*}

we conclude that $d(x)$ corresponds to the graph $E$ and $e(x)$ corresponds to the graph $A\text{.}$

Finally, we see that $f(x)$ must correspond to the graph $C$ since it has a vertical asymptote at $x=5\text{.}$

Draw and label the vertical and horizontal asymptotes, if any, directly on the graphs.

HA at $y=0$
Solution

Since

\begin{equation*} \lim_{x\to\infty} \left(2-\frac{1}{x^2}\right) = 2\text{,} \end{equation*}

and

\begin{equation*} \lim_{x\to 0 } \left(2-\frac{1}{x^2}\right) = - \infty\text{,} \end{equation*}

$y$ has a horizontal asymptote at $y=2$ and a vertical asymptote at $x=0\text{.}$

HA at $y=0$
Solution

Since $\sqrt{x^2+1} = 0$ has no solutions, $y$ has no vertical asymptotes. Now, we see that

\begin{equation*} \lim_{x\to -\infty} \frac{3x}{\sqrt{x^2+1}} = -3, \text{ and }\text{,} \end{equation*}
\begin{equation*} \lim_{x\to \infty} \frac{3x}{\sqrt{x^2+1}} = 3\text{.} \end{equation*}

Hence, $y$ has two horizontal asymptoptes: at $y=\pm 3\text{.}$

VA at $x=\pm 1\text{,}$ HA at $y=0$
VA at $x=0\text{,}$ HA at $y=2$
HA at $y=\pm 3$

Determine the horizontal and vertical asymptotes of the following functions.

1. $f(x)=\dfrac{1}{x}$

HA at $y=0\text{;}$ VA at $x=0$
Solution

Since

\begin{equation*} \lim_{x\to 0^-} \frac{1}{x} = -\infty\text{,} \end{equation*}

and

\begin{equation*} \lim_{x\to 0^+} \frac{1}{x} = \infty\text{,} \end{equation*}

we see that $f(x)$ has a vertical asymptote at $x=0\text{.}$ Furthermore,

\begin{equation*} \lim_{x\to\infty} \frac{1}{x} = \lim_{x\to-\infty}\frac{1}{x} = 0\text{,} \end{equation*}

and so $f(x)$ has a horizontal asymptote at $y=0\text{.}$

2. $f(q) = -\dfrac{2}{q^2}$

HA at $y=0\text{;}$ VA at $x=0$
Solution

Since

\begin{equation*} \lim_{q\to 0^-} \frac{-2}{q^2} =\lim_{q\to 0^+} \frac{-2}{q^2} = -\infty\text{,} \end{equation*}

$f(q)$ has a vertical asymptote at $q=0\text{.}$ Next, since

\begin{equation*} \lim_{q\to\infty} \frac{-2}{q^2} = \lim_{q\to-\infty}\frac{-2}{q^2} = 0\text{,} \end{equation*}

and so $f(q)$ has a horizontal asymptote at $y=0\text{.}$

3. $g(t) = \dfrac{t-1}{t+1}$

HA at $y=1\text{;}$ VA at $x=-1$
Solution

We investigate the behaviour of $g$ as $t$ appraoched $-1\text{:}$

\begin{equation*} \lim_{t\to-1^+} \frac{t-1}{t+1} = -\infty\text{,} \end{equation*}

and

\begin{equation*} \lim_{t\to-1^-} \frac{t-1}{t+1} = \infty\text{.} \end{equation*}

Hence, $g(t)$ has a vertical asymptote at $t=-1\text{.}$ We now compute

\begin{equation*} \lim_{t\to\infty} \frac{t-1}{t+1} = \lim_{t\to\infty} \frac{t}{t} = 1\text{.} \end{equation*}

Similarly, we find

\begin{equation*} \lim_{t\to -\infty} \frac{t-1}{t+1} = \lim_{t\to-\infty} \frac{t}{t} = 1\text{.} \end{equation*}

Therefore, $g(t)$ has a horizontal asymptote at $t=1\text{.}$

4. $h(s) = s^3-3s^2+s+1$

none
Solution

Since the function $h(s)$ is a polynomial, it has no horizontal nor vertical asymptotes.

5. $f(t) = \dfrac{t^2}{t^2-9}$

HA at $y=1\text{;}$ VA at $t= \pm 3$
Solution

First, we notice that we can rewrite

\begin{equation*} f(t) = \frac{t^2}{(t-3)(t+3)}\text{.} \end{equation*}

So we investigate:

\begin{equation*} \lim_{t\to 3^-} \frac{t^2}{(t-3)(t+3)} = -\infty\text{,} \end{equation*}

and

\begin{equation*} \lim_{t\to 3^+} \frac{t^2}{(t-3)(t+3)} = \infty\text{.} \end{equation*}

Similarly,

\begin{equation*} \lim_{t\to -3^-} \frac{t^2}{(t-3)(t+3)} = \infty\text{,} \end{equation*}

and

\begin{equation*} \lim_{t\to -3^+} \frac{t^2}{(t-3)(t+3)} = -\infty\text{.} \end{equation*}

We conclude that $f(t)$ has two vertical asymptotes at $t=\pm 3\text{.}$ We now look for any horizontal asymptotes:

\begin{equation*} \lim_{t\to\infty} \frac{t^2}{t^2-9} = \lim_{t\to\infty} \frac{t^2}{t^2} = 1\text{,} \end{equation*}

and

\begin{equation*} \lim_{t\to-\infty} \frac{t^2}{t^2-9} = \lim_{t\to-\infty} \frac{t^2}{t^2} = 1\text{,} \end{equation*}

and so the graph of $f$ as one horiztonal asymptote at $y=1\text{.}$

6. $f(x) = \dfrac{3x}{x^2-x-6}$

HA at $y=0\text{;}$ VA at $x=-2,3$
Solution

We first factor the denominator:

\begin{equation*} f(x) = \frac{3x}{(x+2)(x-3)}\text{.} \end{equation*}

Now investigate the behaviour as $x$ approches $-2\text{:}$

\begin{equation*} \lim_{x\to-2^-} \frac{3x}{(x+2)(x-3)} = -\infty\text{,} \end{equation*}

and

\begin{equation*} \lim_{x\to-2^+} \frac{3x}{(x+2)(x-3)} = \infty\text{.} \end{equation*}

Hence, the graph of $f(x)$ has a vertical asymptote at $x=-2\text{.}$ Next, since

\begin{equation*} \lim_{x\to 3^-} \frac{3x}{(x+2)(x-3)} = -\infty\text{,} \end{equation*}

and

\begin{equation*} \lim_{x\to 3^+} \frac{3x}{(x+2)(x-3)} = \infty\text{,} \end{equation*}

we conclude that $f(x)$ has a second vertical asymptote at $x=3\text{.}$ We now look for any horizontal asymptotes:

\begin{equation*} \lim_{x\to\infty} \frac{3x}{x^2-x-6} = \lim_{x\to\infty} \frac{3x}{x^2} = 0\text{,} \end{equation*}

and

\begin{equation*} \lim_{x\to-\infty} \frac{3x}{x^2-x-6} = \lim_{x\to-\infty} \frac{3x}{x^2} = 0\text{.} \end{equation*}

Therefore, the graph of $f(x)$ has one horizontal asymptote at $y=0\text{.}$

7. $h(t)=2+\dfrac{5}{(t-2)^2}$

HA at $y=2\text{;}$ VA at $t=2$
Solution

We first investigate the behaviour of $h$ as $t$ approached $2\text{.}$

\begin{equation*} \lim_{t\to 2} \left(2+\frac{5}{(t-2)^2}\right) = 2+ 5\lim_{t\to 2} \frac{1}{(t-2)^2} = \infty\text{.} \end{equation*}

Therefore, the graph of $h(t)$ has a vertical asymptote at $t=2\text{.}$ Since

\begin{equation*} \lim_{t\to\infty} \left(2+\frac{5}{(t-2)^2}\right) = 2\text{,} \end{equation*}

and

\begin{equation*} \lim_{t\to-\infty} \left(2+\frac{5}{(t-2)^2}\right) = 2\text{,} \end{equation*}

we see that the graph of $h(t)$ has one horizontal asymptote at $y=2\text{.}$

8. $f(s)=\dfrac{s^2-2}{s^2-4}$

HA at $y=1\text{;}$ VA at $x= \pm 2$
Solution

First, notice that

\begin{equation*} f(s) = \frac{s^2-2}{s^2-4} = \frac{s^2-2}{(s+2)(s-2)}\text{.} \end{equation*}

So as $s$ approaches $\pm 2\text{,}$ we find:

\begin{equation*} \lim_{s\to -2^-} \frac{s^2-2}{(s+2)(s-2)} = \infty\text{,} \end{equation*}

and

\begin{equation*} \lim_{s\to -2^+} \frac{s^2-2}{(s+2)(s-2)} = -\infty\text{,} \end{equation*}

which means that the graph of $f$ has a vertical asymptote at $s=-2\text{.}$ Furthermore, we find that

\begin{equation*} \lim_{s\to 2^-} \frac{s^2-2}{(s+2)(s-2)} = -\infty\text{,} \end{equation*}

and

\begin{equation*} \lim_{s\to 2^+} \frac{s^2-2}{(s+2)(s-2)} = +\infty\text{,} \end{equation*}

and so the graph of $f$ has a second vertical asymptote at $s=2\text{.}$ We now look for any horizontal asymptotes. To do this, we compute

\begin{equation*} \lim_{s\to\infty} \frac{s^2-2}{s^2-4} = \lim_{s\to\infty} \frac{s^2}{s^2} = 1\text{,} \end{equation*}

\begin{equation*} \lim_{s\to-\infty} \frac{s^2-2}{s^2-4} = \lim_{s\to-\infty} \frac{s^2}{s^2} = 1\text{.} \end{equation*}

We conclude that $f$ has one horizontal asymptote at $y=1\text{.}$

9. $g(x) = \dfrac{x^3-x}{x(x+1)}$

none
Solution

We first notice that we can factor $g(x)$ as follows:

\begin{equation*} g(x) = \frac{x^3-x}{x(x+1)} = \frac{x(x-1)(x+1)}{x(x+1)}\text{.} \end{equation*}

Therefore, $g$ has removable (or hole) discontinuities at $x=0$ and $x=-1\text{,}$ and hence no vertical asymptotes. Furthermore,

\begin{equation*} \lim_{x\to \pm \infty} g(x) = \lim_{x\to \pm \infty} (x-1) = \lim_{x \to \pm \infty} x\text{,} \end{equation*}

and so $g(x)$ either grows or decreases without bound as $x \to \pm \infty\text{,}$ respectively. Thus, $g$ has no horizontal asymptotes.

Find the slant asymptote of the following functions:

1. $f(x)=\dfrac{5x^2-3x+1}{x+2}$

$y=5x+13$
Solution

We wish to find a line $y=mx+b$ such that

\begin{equation*} \lim_{x\to \pm \infty} \left[\frac{5x^2-3x+1}{x+2} - (mx+b)\right] = 0\text{.} \end{equation*}

We write:

\begin{equation*} \begin{split} \lim_{x\to \pm \infty} \left[\frac{5x^2-3x+1}{x+2} - (mx+b)\right] \amp = \lim_{x\to \pm \infty} \frac{5x^2-3x+1 - mx^2-(2m+b)x-2b}{x+2}\\ \amp = \lim_{x\to \pm \infty} \frac{5x-3+1/x - mx-(2m+b)-2b/x}{1+2/x}. \end{split} \end{equation*}

Therefore, we need

\begin{equation*} 5-m = 0, \ \ \text{ and } \ \ 3+2m+b = 0\text{.} \end{equation*}

Solving, we find that $m=5$ and $b=13\text{.}$ With this choice, we see that

\begin{equation*} \lim_{x\to \pm \infty} \left[\frac{5x^2-3x+1}{x+2} - (5x+13)\right]] = 0\text{,} \end{equation*}

and so $y=5x+13$ is the slant asymptote of $f(x)\text{.}$

2. $f(x)=\dfrac{x^2}{x-1}$

$y=x+1$
Solution

We wish to find a line $y=mx+b$ such that

\begin{equation*} \lim_{x\to \pm \infty} \left[\frac{x^2}{x-1} - (mx+b)\right] = 0\text{.} \end{equation*}

Therefore, we write

\begin{equation*} \begin{split} \lim_{x\to \pm \infty} \left[\frac{x^2}{x-1} - (mx+b)\right] \amp = \lim_{x\to \pm \infty} \frac{x^2-mx^2+(b-m)x-b}{x-1} \\ \amp = \lim_{x\to \pm \infty} \frac{x-mx+b-m-b/x}{1-1/x}. \end{split} \end{equation*}

In order for this limit to equal zero, we require $m=1$ and $b=1\text{.}$ With this choice, we find

\begin{equation*} \lim_{x\to \pm \infty} \left[\frac{x^2}{x-1} - (x+1)\right] = 0\text{,} \end{equation*}

and so the line $y=x+1$ is the slant asymptote of $f(x)\text{.}$

3. $f(x)=\dfrac{x^2+3x+2}{x-1}$

$y=x+4$
Solution

Consider

\begin{equation*} \begin{split} \lim_{x\to \pm \infty} \left[\frac{(x^2+3x+2)-(x-1)(mx+b)}{x-1}\right] \amp = \lim_{x\to \pm \infty} \frac{(x^2+3x+2)-mx^2-(b-m)x+b}{x-1}\\ \amp = \lim_{x\to \pm \infty} \frac{x+3+2/x-mx-(b-m)+b/x}{1-1/x}. \end{split} \end{equation*}

Therefore, if we take $m=1$ and $b=4\text{,}$ we see that

\begin{equation*} \lim_{x\to \pm \infty} \frac{x+3+2/x-x-3-1/x}{1-1/x} = \lim_{x\to\pm \infty} \frac{2/x-1/x}{1-1/x} = 0\text{.} \end{equation*}

Hence, the line $y=x+4$ is the slant asymptote of $f(x)\text{.}$

4. $f(x)=\dfrac{x^2-5x+4}{x-3}$

$y=x-1$
Solution

We wish to find a line $y=mx+b$ such that

\begin{equation*} \lim_{x\to \pm \infty} \left[\frac{x^2-5x+4}{x-3} - (mx+b)\right] = 0\text{.} \end{equation*}

Therefore, we write

\begin{equation*} \begin{split} \lim_{x\to \pm \infty} \left[\frac{x^2-5x+4}{x-3} - (mx+b)\right] \amp = \lim_{x\to \pm \infty} \frac{x^2-5x+4 - mx^2 + (3m-b)x - 3b}{x-3} \\ \amp = \lim_{x\to \pm \infty} \frac{x-5+4/x - mx +(3m-b) - 3b/x}{1-3/x}. \end{split} \end{equation*}

So if we take $m=1$ and $b=-1\text{,}$ we see that

\begin{equation*} \begin{split} \lim_{x\to \pm \infty} \frac{x-5+4/x - mx +(3m-b) - 3b/x}{1-3/x} \amp = \lim_{x\to \pm \infty} \frac{x-5+4/x-x+5+3/x}{1-3/x}\\ \amp = 0. \end{split} \end{equation*}

Therefore, the line $y=x-1$ is the slant asymptote of $f(x)\text{.}$

### Subsection5.7.5Summary of Curve Sketching

The following is a guideline for sketching a curve $y=f(x)$ by hand. Each item may not be relevant to the function in question, but utilizing this guideline will provide all information needed to make a detailed sketch of the function.

###### Guideline for Curve Sketching.

Given a function $y=f(x)\text{,}$ follow these steps to sketch the graph of $f\text{.}$

1. Determine the domain, and the $x$-values $u_1,u_2,...,u_m$ where the function is undefined. Graph a set of coordinate axes that is suitable.

2. Find all intercepts. Graph them, if any.

3. Determine if $f$ has any symmetry.

4. Determine asymptotes:

1. Horizontal Asymptotes ($x \to \infty$ and $x \to -\infty$). Graph them, if any.

2. Vertical Asymptotes ($x \to u_i$ for $i=1,2,...,m$). Graph them, if any.

5. Calculate $f'\text{:}$

1. Determine the critical points $c_1,c_2,...,c_n$ of $f$ and use these numbers along with $u_1,u_2,...,u_m$ to create a sign chart for $f'\text{.}$

2. Determine the intervals of increase and decrease.

3. Apply the First Derivative Test to determine relative extrema. Graph them, if any.

6. Calculate $f''\text{:}$

1. Determine the critical points $c'_1,c'_2,...,c'_n$ of $f'$ and use these numbers along with $u_1,u_2,...,u_m$ to create a sign chart for $f''\text{.}$

2. Determine the intervals of concave up and concave down.

3. Determine all inflection points. Graph them, if any.

7. Use both sign charts to complete the sketch of the graph and don't forget to label the graph with relevant information.

###### Example5.86. Curve Sketching.

Sketch the graph of the function $\ds{f(x)=\frac{2x^2}{x^2-1}} \cdot$

Solution
1. The domain is $\{x: x^2-1 \neq 0\}= \{x: x \neq \pm 1\}=(- \infty,-1) \cup (-1,1) \cup (1, \infty)$

2. There is an $x$-intercept at $x=0\text{.}$ The $y$ intercept is $y=0\text{.}$

3. $f(-x)=f(x)\text{,}$ so $f$ is an even function (symmetric about $y$-axis)

4. $\ds{ \lim_{x \to \pm \infty} \frac{2x^2}{x^2-1} =\lim_{x \to \pm \infty} \frac{2}{1-1/x^2} = 2 }\text{,}$ so $y=2$ is a horizontal asymptote. Now the denominator is 0 at $x =\pm1\text{,}$ so we compute:

\begin{equation*} \lim_{x \to 1^+} \frac{2x^2}{x^2-1} = + \infty, \; \lim_{x \to 1^-} \frac{2x^2}{x^2-1} = - \infty, \; \lim_{x \to -1^+} \frac{2x^2}{x^2-1} = - \infty, \; \lim_{x \to -1^-} \frac{2x^2}{x^2-1} = + \infty\text{.} \end{equation*}

So the lines $x=1$ and $x=-1$ are vertical asymptotes.

5. For critical values we take the derivative:

\begin{equation*} f'(x) = \frac{4x(x^2-1)-2x^2 \cdot 2x}{(x^2-1)^2} = \frac{-4x}{(x^2-1)^2}\text{.} \end{equation*}

Note that $f'(x)=0$ when $x=0$ (the top is zero). Also, $f'(x)=DNE$ when $x=\pm 1$ (the bottom is zero). As $x=\pm 1$ is not in the domain of $f(x)\text{,}$ the only critical point is $x=0$ (recall that to be a critical point we need it to be in the domain of the original function). Drawing a number line and including all of the split points of $f'(x)$ we have:

Thus $f$ is increasing on $(-\infty, -1)\cup(-1,0)$ and decreasing on $(0,1)\cup(1, \infty)\text{.}$ By the First Derivative Test, $x=0$ is a relative max.

6. For possible inflection points we take the second derivative:

\begin{equation*} f''(x)= \frac{12x^2+4}{(x^2-1)^3} \end{equation*}

The top is never zero. Also, the bottom is only zero when $x=\pm 1$ (neither of which are in the domain of $f(x)$). Thus, there are no possible inflection points to consider. Drawing a number line and including all of the split points of $f''(x)$ we have:

Hence $f$ is concave up on $(- \infty,-1)\cup(1, \infty)\text{,}$ concave down on $(-1,1)\text{.}$

7. We put this information together and sketch the graph. We combine some of this information on a single number line to see what shape the graph has on certain intervals:

Note that there is a horizontal asymptote at $y=2$ and that the curve has $x$-int of $x=0$ and $y$-int of $y=0\text{.}$ Therefore, a sketch of $f(x)$ is as follows:

###### Example5.87. Curve Sketching.

Sketch the graph of the function

\begin{equation*} f(x)=x^3-6x^2+9x+2\text{.} \end{equation*}
Solution

Obtain the following information on the graph of $f\text{.}$

1. The domain of $f$ is $(-\infty,\infty)\text{.}$

2. By setting $x=0\text{,}$ we find that the $y$-intercept is 2. The $x$-intercept is found by setting $y=0\text{,}$ which in this case leads to a cubic equation. Since the solution is not readily found, we will not use this information.

3. Since $f$ is a cubic polynomial, we expect odd symmetry. This will become more obvious once we analyze $f'$ and $f''\text{.}$

4. We now look for any asymptotes of $f\text{:}$

\begin{equation*} \lim_{x\to \infty} f(x) = \lim_{x\to\infty}\left(x^3-6x^2+9x+2\right) = \infty \end{equation*}
\begin{equation*} \lim_{x\to -\infty} f(x) = \lim_{x\to -\infty}\left(x^3-6x^2+9x+2\right) =-\infty \end{equation*}

We see that $f$ decreases without bound as $x$ decreases and increases without bound as $x$ increases. Therefore, $f$ has no horizontal asymptotes. Since $f$ is a polynomial,there are no vertical asymptotes.

5. \begin{equation*} f'(x)=3x^2-12x+9 = 3(x-3)(x-1) \end{equation*}

Setting $f'(x)=0$ gives $x=1$ or $x=3$ as our only critical points. The following sign diagram for $f$ shows that $f$ is increasing on the intervals $(-\infty,1)$ and $(3,\infty)$ and decreasing on the interval $(1,3)\text{.}$

Since the sign of $f'$ changes as we move across the critical point $x=1\text{,}$ a relative maximum occurs at $(1,f(1)) = (1,6)\text{.}$ Similarly, a relative minimum of $f$ occurs at $(3,2)\text{.}$

6. We find

\begin{equation*} f''(x)=6x-12=6(x-2)\text{,} \end{equation*}

which is equal to zero when $x=2\text{.}$ The sign diagram for $f''\text{,}$

shows that $f$ is concave downward on $(-\infty,2)$ and concave upward on $(2,\infty)\text{.}$ Since the sign of $f''$ changes sign as we move across $x=2\text{,}$ $f$ must have an inflection point at $(2,f(2))=(2,4)\text{.}$ In fact, we can show that $f$ exhibits odd symmetry about this point.

7. Putting all of the above information together, we arrive at the following graph of $f(x)\text{.}$

Suppose we are not given the definition of a function as in Examples 5.86 and 5.87, but rather, we are being provided information about its domain and other behaviours. Sometimes, these pieces of information may still allow us to sketch the curve of such a function as shown in the next example.

###### Example5.88. Curve Sketching with Pieces of Information.

Suppose the following properties of a function $f$ are given.

1. Domain: $(-\infty,1)\cup(1,\infty)\text{.}$

2. Intercepts $(0,-1)$ and $(-1,0)\text{.}$

3. $f$ is symmetric about the line $y=-x+2\text{.}$

4. $x=1$ is a vertical asymptote and $y=1$ is a horizontal asymptote.

5. $f$ is decreasing on $(-\infty,1)\cup(1,\infty)\text{.}$

6. $f$ has no relative extrema.

7. $f$ is concave downward on $(-\infty,1)$ and concave upward on $(1,\infty)\text{.}$

8. $f$ has no points of inflection.

Solution

We initialize our graph with the $x$- and $y$- intercepts (item 2), along with the given horizontal and vertical asymptotes (item 4).

Let's first look at the bottom left-hand corner of the graph. We are given that, here, $f$ is decreasing and concave downward (items 5 and 7). We can therefore connect the two intercepts like so:

The line of symmetry $y=-x+2$ (item 3) is drawn in black. We reflect the curve across this line, ensuring that our final curve is concave up and decreasing on the interval $(1,\infty)$ (items 5 and 7). Our final result is below.

##### Exercises for Section 5.7.

Follow the Curve Sketching Guideline provided in this section to sketch the graphs of the following functions.

1. $\ds y=x^5-5x^4+5x^3$

We wish to sketch the graph of $y=f(x)=x^5-5x^4+5x^3\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$
2. $\ds y=x^3-3x^2-9x+5$

We wish to sketch the graph of $y=f(x)=x^3-3x^2-9x+5\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$

3. $\ds y=(x-1)^2(x+3)^{2/3}$

We wish to sketch the graph of $y=f(x)=(x-1)^2(x+3)^{2/3}\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$

4. $\ds x^2+x^2y^2=a^2y^2\text{,}$ $a>0\text{.}$

We wish to sketch the graph of $x^2+x^2y^2=a^2y^2\text{,}$ $a>0\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$

5. $\ds y= x^5 - x$

We wish to sketch the graph of $y=f(x)=x^5-x\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$

6. $\ds y=x(x^2+1)$

We wish to sketch the graph of $y=f(x)=x(x^2+1)\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$

7. $\ds y=x^3+6x^2 + 9x$

We wish to sketch the graph of $y=f(x)=x^3+6x^2+9x\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$

Follow the Curve Sketching Guideline provided in this section to sketch the graphs of the following functions.

1. $\ds y = 4x+\sqrt{1-x}$

Solution

We follow the Curve Sketching Guideline to sketch the graph of $y=f(x)=4x+\sqrt{1-x}\text{:}$

1. The domain of $y$ is all $x$ such that $1-x \geq 0 \implies x \leq 1\text{.}$ We further notice that $f(1) = 4\text{.}$

2. The function has a $y$-intercept at

\begin{equation*} y = f(0) = 1\text{.} \end{equation*}

To find any $x$-intercepts, we set $f(x)=0\text{:}$

\begin{equation*} \begin{split} 0 \amp = 4x + \sqrt{1-x} \\ -4x \amp = \sqrt{1-x} \\ 16x^2 + x -1 \amp = 0 \ \ (x \lt 0)\\ x\amp = -\frac{1}{32}-\frac{\sqrt{65}}{32} \approx -0.28 \end{split} \end{equation*}

And so $f$ has one $x$-intercept. We provide a sketch of our work so far below:

3. There is no apparent symmetry.

4. There are no horizontal or vertical asymptotes.

5. We compute

\begin{equation*} f'(x) = 4 - \frac{1}{2\sqrt{1-x}}\text{.} \end{equation*}

Therefore, $f(x)$ has critical points at $x=1$ (where $f'(x)$ undefined) and where

\begin{equation*} f'(x) = 0 \implies \sqrt{1-x} = \frac{1}{8} \implies x = \frac{63}{64} \approx 0.98\text{.} \end{equation*}

Since

\begin{equation*} f'(0) = 4 - \frac{1}{2} > 0 \ \ \text{ and } \ \ f'(0.99) = -1 \lt 0\text{,} \end{equation*}

we conclude that $f$ is increasing on the interval $\left(-\infty, \frac{63}{64}\right)$ and decreasing on $\left(\frac{63}{64}, 1\right)\text{.}$ By the First Derivative Test, the point $\left(\frac{63}{54}, \frac{65}{16}\right)$ is a local maximum. The point $(1,4)$ is not a local extrema.

6. The second derivative of $f$ is

\begin{equation*} f''(x) = \frac{1}{4(1-x)^{3/2}}\text{.} \end{equation*}

Therefore, $f''(x)$ DNE at $x=1$ and $f''(x)=0$ has no solutions. Since

\begin{equation*} f''(0) = -\frac{1}{2} \lt 0\text{,} \end{equation*}

we conclude that $y$ is concave down on $(-\infty, 1)\text{.}$ A final sketch is provided below.

2. $\ds y = (x+1)/\sqrt{5x^2 + 35}$

We wish to sketch the graph of $y=f(x)=(x+1)/\sqrt{5x^2+35}\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$

3. $\ds y = x+ 1/x$

We wish to sketch the graph of $y=f(x)=x+1/x\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$

4. $\ds y = x^2+ 1/x$

We wish to sketch the graph of $y=f(x)=x^2+1/x\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$

5. $\ds y = (x+5)^{1/4}$

We wish to sketch the graph of $y=f(x)=(x+5)^{1/4}\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$

6. $\ds y=x/(x^2-9)$

Solution

We wish to sketch the graph of $y=f(x)=\dfrac{x}{x^2-9}\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We see that $f(x)$ is undefined at $x = \pm 3$ and that $f(0)= 0\text{.}$ Next, we look for asymptotes of $f\text{.}$ Since

\begin{equation*} \begin{array}{lll} \lim\limits_{x\to\infty} f(x) = 0 \amp \lim\limits_{x\to 3^+} f(x) = +\infty \amp \lim\limits_{x\to -3^+} f(x) = +\infty \\ \lim\limits_{x\to -\infty}f(x) = 0 \amp \lim\limits_{x \to 3^-} f(x) = -\infty \amp \lim\limits_{x\to -3^-} f(x) = -\infty, \end{array} \end{equation*}

we see that $y=0$ is a horizontal asymptote and $x=\pm 3$ are vertical asymptotes:

Next, compute

\begin{equation*} f'(x) = -\frac{x^2+9}{(x^2-9)^2}\text{.} \end{equation*}

Since $f'(x)=0$ has no solutions, and $f'(x)$ is defined for all $x \neq \pm 3\text{,}$ we conclude that $f$ has no relative extrema. Additionally, $f'(0) \lt 0$ and so $f$ is decreasing on its entire domain. We now investigate the concavity of $f\text{.}$

\begin{equation*} f''(x) = \frac{2x(x^2+27)}{(x^2-9)^3}\text{,} \end{equation*}

therefore $f''(x)=0$ at $x=0\text{,}$ and $f''(x)$ is defined for all $x \neq \pm 3\text{.}$ By picking appropriate test points, we find that $f$ is concave down on $(-\infty,-3)\cup(0,3)$ and concave up on $(-3,0)\cup(3,\infty)\text{.}$ Thus, $x=0$ is an inflection point of $f\text{.}$ We summarize these findings in the following diagram.

We arrive at the following graph of $f(x)\text{.}$

7. $\ds y=x^2/(x^2+9)$

We wish to sketch the graph of $y=f(x)=x/(x^2+9)\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$

8. $\ds y=2\sqrt{x} - x$

We wish to sketch the graph of $y=f(x)=2\sqrt{x}-x\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$

9. $\ds y=(x-1)/(x^2)$

We wish to sketch the graph of $y=f(x)=(x-1)/(x^2)\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$

Follow the Curve Sketching Guideline provided in this section to sketch the graphs of the following functions.

1. $\ds y=xe^x$

We wish to sketch the graph of $y=f(x)=xe^x\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$

2. $\ds y=(e^x+e^{-x})/2$

We wish to sketch the graph of $y=f(x)=(e^x+e^{-x})/2\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$

3. $\ds y=e^{-x}\cos x$

We wish to sketch the graph of $y=f(x)=e^{-x}\cos x\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$

4. $\ds y=e^x-\sin x$

We wish to sketch the graph of $y=f(x)=e^{x} - \sin x\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$

5. $\ds y=e^x/x$

We wish to sketch the graph of $y=f(x)=e^x/x\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$

6. $\ds y = \tan^2 x$

We wish to sketch the graph of $y=f(x)=\tan^2(x)\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$

7. $\ds y =\cos^2 x - \sin^2 x$

We wish to sketch the graph of $y=f(x)=\cos^2(x)-\sin^2(x)\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$

8. $\ds y = \sin^3 x$

We wish to sketch the graph of $y=f(x)=\sin^3(x)\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$

9. $\ds y = 6x + \sin 3x$

We wish to sketch the graph of $y=f(x)=6x+\sin(3x)\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$

10. $\ds y=3\sin(x) - \sin^3(x)\text{,}$ for $x\in[0,2\pi]$

We wish to sketch the graph of $y=f(x)=3\sin(x)-\sin^3(x)$ for $x\in [0,2\pi]\text{.}$ Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of $f(x)\text{.}$

Sketch the graph of $y=f(x)$ using the following information:

Domain: $(-\infty,\infty)$

$y$-intercept: 2

Asymptotes: none

Increasing on: $(-\infty,0)\cup(4,\infty)$

Decreasing on: $(0,4)$

Relative Extrema: relative max at $(0,2)\text{,}$ relative min at $(4,-30)$

Concavity: Downward on $(-\infty,2)\text{,}$ upward on $(2,\infty)$

Inflection point: $(2,-14)$

Solution

We are given three points on the graph of $f$ which we can use to initialize our set of axes. Plot,

Additionally, we are told that the point $(0,2)$ is a relative maximum and $(4,-30)$ is a relative minimum. We can add this information to our sketch as follows:

Next, we summarize the information on the shape of $f$ in the following diagram:

Using this diagram, we can piece together the final graph:

Sketch the graph of $y=f(x)$ using the following information:

Domain: $(-\infty,\infty)$

$y$-intercept: 0

$x$-intercepts: 0,2

Asymptotes: none

Increasing on: $(1.5,\infty)$

Decreasing on: $(-\infty,0)\cup(0,3/2)$

Relative Extrema: relative min at $(3/2,-27/80)$

Concavity: Downward on $(0,1)\text{,}$ upward on $(-\infty,0)\cup(1,\infty)$

Inflection points: $(0,0)$ and $(1,-1/5)$

Solution

We are given four points on the graph of $f$ which we can use to initialize our set of axes. Plot,

Additionally, we are told that the point $(3/2,-27/80)$ is a relative minimum. We can add this information to our sketch as follows:

Next, we summarize the information on the shape of $f$ in the following diagrams:

Using this diagram, we can piece together the final graph:

Sketch the graph of $y=f(x)$ using the following information:

Domain: $(-\infty,0)\cup(0,\infty)$

$x$-intercept: 1

Asymptotes: $x$-axis and $y$-axis

Increasing on: $(0,2)$

Decreasing on: $(-\infty,0)\cup(2,\infty)$

Relative Extrema: relative max at $(2,1/4)$

Concavity: Downward on $(-\infty,0)\cup(0,3)\text{,}$ upward on $(3,\infty)$

Inflection point: $(3,2/9)$

Solution

We are given three points on the graph of $f$ which we can use to initialize our set of axes. Plot,

Additionally, we are told that the point $(2,1/4)$ is a relative maximum. We can add this information to our sketch as follows:

Next, we summarize the information on the shape of $f$ in the following diagrams:

Using this diagram, we can piece together the final graph:

Sketch the graph of $y=f(x)$ using the following information:

Domain: $(-\infty,\infty)$

Symmetry: Odd $x$-intercepts: $\pm 5\sqrt{5}\text{,}$0

Asymptotes: none

Increasing on: $(-\infty,-2.15)\cup(2.15,\infty)$

Decreasing on: $(-2.15,2.15)$

Relative Extrema: relative max at $(-2.15,4.3)\text{,}$ relative min at $(2.15,-4.3)$

Concavity: Upward on $(0,\infty)\text{,}$ Downward on $(-\infty, 0)$

We wish to plot the graph of $y=f(x) = x-5x^{1/3}\text{.}$ We first plot the given points:
Next, we use the fact that $y$ has relative extrema at $x = \pm 2.15\text{:}$