## Section5.5Extrema of a Function

In calculus, there is much emphasis placed on analyzing the behaviour of a function $f$ on an interval $I\text{.}$ Does $f$ have a maximum value on $I\text{?}$ Does it have a minimum value? How does the interval $I$ impact our discussion of extrema?

### Subsection5.5.1Relative Extrema

A relative maximum point on a function is a point $(x,y)$ on the graph of the function whose $y$-coordinate is larger than all other $y$-coordinates on the graph at points “close to” $(x,y)\text{.}$ More precisely, $(x,f(x))$ is a relative maximum if there is an interval $(a,b)$ with $a\lt x\lt b$ and $f(x)\ge f(z)$ for every $z$ in $(a,b)\text{.}$ Similarly, $(x,y)$ is a relative minimum point if it has locally the smallest $y$-coordinate. Again being more precise: $(x,f(x))$ is a relative minimum if there is an interval $(a,b)$ with $a\lt x\lt b$ and $f(x)\le f(z)$ for every $z$ in $(a,b)\text{.}$ A relative extremum is either a relative minimum or a relative maximum.

Note:
1. The plural of extremum is extrema and similarly for maximum and minimum.

2. Because a relative extremum is “extreme” locally by looking at points “close to” it, it is also referred to as a local extremum.

###### Definition5.46. Relative Maxima and Minima.

A real-valued function $f$ has a relative maximum at $x_0$ if $f(x_0)\geq f(x)$ for all $x$ in some open interval containing $x_0\text{.}$

A real-valued function $f$ has a relative minimum at $x_0$ if $f(x_0)\leq f(x)$ for all $x$ in some open interval containing $x_0\text{.}$

Relative maximum and minimum points are quite distinctive on the graph of a function, and are therefore useful in understanding the shape of the graph. In many applied problems we want to find the largest or smallest value that a function achieves (for example, we might want to find the minimum cost at which some task can be performed) and so identifying maximum and minimum points will be useful for applied problems as well. Some examples of relative maximum and minimum points are shown in Figure 5.14.

If $(x,f(x))$ is a point where $f(x)$ reaches a relative maximum or minimum, and if the derivative of $f$ exists at $x\text{,}$ then the graph has a tangent line and the tangent line must be horizontal. This is important enough to state as a theorem.

The proof is simple enough and we include it here, but you may accept Fermat's Theorem based on its strong intuitive appeal and come back to its proof at a later time.

We shall give the proof for the case where $f\left( x\right)$ has a relative maximum at $x=a\text{.}$ The proof for the relative minimum case is similar.

Since $f\left( x\right)$ has a relative maximum at $x=a\text{,}$ there is an open interval $\left( c,d\right)$ with $c\lt a\lt d$ and $f\left( x\right) \leq f\left( a\right)$ for every $x$ in $\left( c,d\right) \text{.}$ So, $f\left( x\right) -f\left( a\right) \leq 0$ for all such $x\text{.}$ Let us now look at the sign of the difference quotient $\dfrac{f\left( x\right) -f\left( a\right) }{x-a}\text{.}$ We consider two cases according as $x>a$ or $x\lt a\text{.}$

If $x>a\text{,}$ then $x-a>0$ and so, $\dfrac{f\left( x\right) -f\left( a\right) }{x-a}\leq 0\text{.}$ Taking limit as $x$ approach $a$ from the right, we get

\begin{equation*} \lim_{x\rightarrow a^{+}}\frac{f\left( x\right) -f\left( a\right) }{x-a}\leq 0\text{.} \end{equation*}

On the other hand, if $x\lt a\text{,}$ then $x-a\lt 0$ and so,$\dfrac{f\left( x\right) -f\left( a\right) }{x-a}\geq 0\text{.}$ Taking limit as $x$ approach $a$ from the left, we get

\begin{equation*} \lim_{x\rightarrow a^{-}}\frac{f\left( x\right) -f\left( a\right) }{x-a}\geq 0\text{.} \end{equation*}

Since $f$ is differentiable at $a\text{,}$

\begin{equation*} f^{\prime }\left( a\right) =\lim\limits_{x\rightarrow a^{+}}\dfrac{f\left( x\right) -f\left( a\right) }{x-a}=\lim\limits_{x\rightarrow a^{-}}\dfrac{f\left( x\right) -f\left( a\right) }{x-a}\text{.} \end{equation*}

Therefore, we have both $f^{\prime }\left( a\right) \leq 0$ and $f^{\prime }\left( a\right) \geq 0\text{.}$ So, $f^{\prime }\left( a\right) =0\text{.}$

Thus, the only points at which a function can have a relative maximum or minimum are points at which the derivative is zero, as in the left-hand graph in Figure 5.14, or the derivative is undefined, as in the right-hand graph. This leads us to define these special points.

###### Definition5.48. Critical Point.

Any value of $x$ in the domain of $f$ for which $f'(x)$ is zero or undefined is called a critical point of $f\text{.}$

The $x$-values $a\text{,}$ $b$ and $c$ above are places for which $f'(x)$ is zero, and the $x$-values $d$ and $e$ above are places for which $f'(x)$ is undefined.

Note: When looking for relative maximum and minimum points, you are likely to make two sorts of mistakes.

1. You may forget that a maximum or minimum can occur where the derivative does not exist. You should therefore check whether the derivative exists everywhere.

2. You might also assume that any place that the derivative is zero is a relative maximum or minimum point, but this is not true. A portion of the graph of $\ds f(x)=x^3$ is shown in Figure 5.15. The derivative of $f$ is $f'(x)=3x^2\text{,}$ and $f'(0)=0\text{,}$ but there is neither a maximum nor minimum at $(0,0)\text{.}$ In other words, the converse of Fermat's Theorem — if $f'(a) = 0$ at some point $x=a\text{,}$ then $f$ must have a relative extremum at that point — is not true.

Since the derivative is zero or undefined at both relative maximum and relative minimum points, we need a way to determine which, if either, actually occurs. The most elementary approach, but one that is often tedious or difficult, is to test directly whether the $y$-coordinates “near” the potential maximum or minimum are above or below the $y$-coordinate at the point of interest. Of course, there are too many points “near” the point to test, but a little thought shows we need only test two provided we know that $f$ is continuous (recall that this means that the graph of $f$ has no jumps or gaps).

Suppose, for example, that we have identified three points at which $f'$ is zero or nonexistent: $\ds (x_1,y_1)\text{,}$ $\ds (x_2,y_2)\text{,}$ $\ds (x_3,y_3)\text{,}$ and $\ds x_1\lt x_2\lt x_3$ (see Figure 5.16). Suppose that we compute the value of $f(a)$ for $\ds x_1\lt a\lt x_2\text{,}$ and that $\ds f(a)\lt f(x_2)\text{.}$ What can we say about the graph between $a$ and $\ds x_2\text{?}$ Could there be a point $\ds (b,f(b))\text{,}$ $\ds a\lt b\lt x_2$ with $\ds f(b)>f(x_2)\text{?}$ No: if there were, the graph would go up from $(a,f(a))$ to $(b,f(b))$ then down to $\ds (x_2,f(x_2))$ and somewhere in between would have a relative maximum point. (This is not obvious; it is a result of the Extreme Value Theorem stated in the next section.) But at that relative maximum point the derivative of $f$ would be zero or nonexistent, yet we already know that the derivative is zero or nonexistent only at $\ds x_1\text{,}$ $\ds x_2\text{,}$ and $\ds x_3\text{.}$ The upshot is that one computation tells us that $\ds (x_2,f(x_2))$ has the largest $y$-coordinate of any point on the graph near $\ds x_2$ and to the left of $\ds x_2\text{.}$ We can perform the same test on the right. If we find that on both sides of $\ds x_2$ the values are smaller, then there must be a relative maximum at $\ds (x_2,f(x_2))\text{;}$ if we find that on both sides of $\ds x_2$ the values are larger, then there must be a relative minimum at $\ds (x_2,f(x_2))\text{;}$ if we find one of each, then there is neither a relative maximum or minimum at $\ds x_2\text{.}$

It is not always easy to compute the value of a function at a particular point. The task is made easier by the availability of calculators and computers, but they have their own drawbacks—they do not always allow us to distinguish between values that are very close together. Nevertheless, because this method is conceptually simple and sometimes easy to perform, you should always consider it.

###### Example5.49. Testing for Relative Extrema in Cubic Function.

Find all relative maximum and minimum points for the function $\ds f(x)=x^3-x\text{.}$

Solution

The derivative is $\ds f'(x)=3x^2-1\text{.}$ This is defined everywhere and is zero at $\ds x=\pm \sqrt{3}/3\text{.}$ Looking first at $\ds x=\sqrt{3}/3\text{,}$ we see that $\ds f(\sqrt{3}/3)=-2\sqrt{3}/9\text{.}$ Now we test two points on either side of $\ds x=\sqrt{3}/3\text{,}$ choosing one point in the interval $(-\sqrt{3}/3,\sqrt{3}/3)$ and one point in the interval $(\sqrt{3}/3,\infty)\text{.}$ Since $\ds f(0)=0>-2\sqrt{3}/9$ and $\ds f(1)=0>-2\sqrt{3}/9\text{,}$ there must be a relative minimum at $\ds x=\sqrt{3}/3\text{.}$ For $\ds x=-\sqrt{3}/3\text{,}$ we see that $\ds f(-\sqrt{3}/3)=2\sqrt{3}/9\text{.}$ This time we can use $x=0$ and $x=-1\text{,}$ and we find that $\ds f(-1)=f(0)=0\lt 2\sqrt{3}/9\text{,}$ so there must be a relative maximum at $\ds x=-\sqrt{3}/3\text{.}$

Of course this example is made very simple by our choice of points to test, namely $x=-1\text{,}$ $0\text{,}$ $1\text{.}$ We could have used other values, say $-5/4\text{,}$ $1/3\text{,}$ and $3/4\text{,}$ but this would have made the calculations considerably more tedious, and we should always choose very simple points to test if we can.

###### Example5.50. Testing for Relative Extrema in Trigonometric Function.

Find all relative maximum and minimum points for $f(x)=\sin x+\cos x\text{.}$

Solution

The derivative is $f'(x)=\cos x-\sin x\text{.}$ This is always defined and is zero whenever $\cos x=\sin x\text{.}$ Recalling that the $\cos x$ and $\sin x$ are the $x$- and $y$-coordinates of points on a unit circle, we see that $\cos x=\sin x$ when $x$ is $\pi/4\text{,}$ $\pi/4\pm\pi\text{,}$ $\pi/4\pm2\pi\text{,}$ $\pi/4\pm3\pi\text{,}$ etc. Since both sine and cosine have a period of $2\pi\text{,}$ we need only determine the status of $x=\pi/4$ and $x=5\pi/4\text{.}$ We can use $0$ and $\pi/2$ to test the critical value $x= \pi/4\text{.}$ We find that $\ds f(\pi/4)=\sqrt{2}\text{,}$ $\ds f(0)=1\lt \sqrt{2}$ and $\ds f(\pi/2)=1\text{,}$ so there is a relative maximum when $x=\pi/4$ and also when $x=\pi/4\pm2\pi\text{,}$ $\pi/4\pm4\pi\text{,}$ etc. We can summarize this more neatly by saying that there are relative maxima at $\pi/4\pm 2k\pi$ for every integer $k\text{.}$

We use $\pi$ and $\frac{3\pi}{2}$ to test the critical value $x=5\pi/4\text{.}$ The relevant values are $\ds f(5\pi/4)=-\sqrt2\text{,}$ $\ds f(\pi)=-1>-\sqrt2\text{,}$ $\ds f(3\pi/2)=1>-\sqrt2\text{,}$ so there is a relative minimum at $x=5\pi/4\text{,}$ $5\pi/4\pm2\pi\text{,}$ $5\pi/4\pm4\pi\text{,}$ etc. More succinctly, there are relative minima at $5\pi/4\pm 2k\pi$ for every integer $k\text{.}$

###### Example5.51. Testing for Relative Extrema in Power Function.

Find all relative maximum and minimum points for $g\left( x\right) =x^{2/3}\text{.}$

Solution

The derivative is $g^{\prime }\left( x\right) =\frac{2}{3}x^{-1/3}\text{.}$ This is undefined when $x=0$ and is not equal to zero for any $x$ in the domain of $g^{\prime }\text{.}$ Now we test two points on either side of $x=0\text{.}$ We use $x=-1$ and $x=1\text{.}$ Since $g\left( 0\right) =0\text{,}$ $g\left( -1\right) =1>0$ and $g\left( 1\right) =1>0\text{,}$ there must be a relative minimum at $x=0\text{.}$

##### Exercises for Section 5.5.1.

Find all relative maximum and minimum points $(x,y)$ by the method of this section.

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

min at $x=1/2$
Solution

Let $y = f(x)\text{.}$ Then $f(x) = x^2-x$ is a polynomial, and so its domain is all real numbers. Differentiating, we find:

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

Hence, $f'(x) = 0$ when $x=0.5\text{,}$ and there are no points in the domain of $f$ for which $f'(x)$ is undefined. Therefore, the only critical point is at $x=0.5\text{.}$ We use the test points $x=0$ and $x=1\text{:}$

\begin{equation*} \begin{array}{c|c} x \amp f(x) \\ \hline 0 \amp 0\\ 0.5 \amp -0.75 \\ 1 \amp 0 \end{array} \end{equation*}

And so $f$ has a relative minimum at the point $(0.5,-0.75)\text{.}$

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

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

The function $2+3x-x^3$ is a polynomial, and so has domain $(-\infty, \infty)\text{.}$ Now differentiate:

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

Then $y'=0$ has two solutions: $x=\pm 1\text{.}$ There are no points for which $y'$ is undefined, so $y$ has two critical points at $x=1$ and $x=-1\text{.}$ Since

\begin{equation*} \begin{array}{c|c} x \amp y \\ \hline -2 \amp 4\\ -1 \amp 0 \\ 0 \amp 2\\ 1 \amp 4\\ 2 \amp 0 \end{array} \end{equation*}

we must have that $x=-1$ is a relative minimum, and $x=1$ is a relative maximum.

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

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

The function $x^3-9x^2+24x$ is a polynomial, and so its domain is $(-\infty,\infty)\text{.}$ We now calculate $y'\text{:}$

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

Therefore,

\begin{equation*} \begin{split} y' \amp = 0 \\ 3x^2-18x+24 \amp = 0 \\ 3 (x^2-6x+8) \amp = 0 \\ 3(x-2)(x-4) \amp = 0 \\ x \amp = 2, 4. \end{split} \end{equation*}

There are no points in the domain of $y$ such that $y'$ is undefined, and so $x=2$ and $x=4$ are the only critical points of $y\text{.}$ Since

\begin{equation*} \begin{array}{c|c} x \amp y \\ \hline 1 \amp 16\\ 2 \amp 20 \\ 3 \amp 18\\ 4 \amp 16\\ 5 \amp 20 \end{array} \end{equation*}

we see that $y$ has a relative maximum at $x=2$ and a relative minimum at $x=4\text{.}$

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

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

Let $y=f(x)\text{.}$ The domain of $f$ is all real numbers. Calculate $f'(x)\text{:}$

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

Therefore, $f'(x)=0$ when $x=-1,0,1$ and is defined over the entire domain of $f\text{.}$ Notice that $f(x)$ is an even function (that is, $f(x)=f(-x)$). Therefore, we need only classify the relative maxima and minima for $x \leq 0\text{.}$ Now select appropriate test points and evaluate $f\text{:}$

\begin{equation*} \begin{array}{c|c} x \amp f(x) \\ \hline -2 \amp 11\\ -1 \amp 2 \\ -0.5 \amp 2.5625\\ 0 \amp 3 \end{array} \end{equation*}

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

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

min at $x=1$
Solution

Let $y=f(x)\text{.}$ The domain of $f$ is $(-\infty,\infty)\text{.}$ Now differentiate:

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

Hence, $f'(x) = 0$ at $x=0$ and $x=1\text{.}$ Since $f'(x)$ is defined for all real numbers, the only critical points of $f$ are at $x=0$ and $x=1\text{.}$ We now evaluate $f(x)=3x^4-4x^3$ at the following test points:

\begin{equation*} \begin{array}{c|c} x \amp f(x) \\ \hline -1 \amp 7 \\ 0 \amp 0 \\ 1 \amp -1 \\ 2 \amp 16 \end{array} \end{equation*}

Therefore, $f(x)$ has a relative minimum at $x=1\text{,}$ and neither a relative maximum nor a relative minimum at $x=0\text{.}$

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

none
Solution

Let $y=f(x)\text{.}$ The domain of $f$ is $\{x \in \mathbb{R} | x \neq 0\}\text{.}$ Calculate $f'(x)\text{:}$

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

We see that $f'(x)$ is undefined at $x=0\text{,}$ which is not in the domain of $f\text{.}$ So $x=0$ cannot be a critical point of $f\text{.}$ Further,

\begin{equation*} f'(x) = 1+x^2= 0 \end{equation*}

has no real solutions. We conclude that $f$ has no relative maxima or minima.

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

none
Solution

Let $y=f(x)\text{.}$ Notice that the domain of $f$ is all real numbers such that $x \neq 0\text{.}$ To find the relative extrema, we first calculate $f'(x)\text{:}$

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

$f'(x)$ is undefined at $x=0\text{,}$ but this cannot be a relative extremum since it is not in the domain of $f\text{.}$ We next try solving $f'(x)=0\text{:}$

\begin{equation*} \begin{split} 0 \amp = f'(x) = 6x + \frac{2}{x^3} \\ 0 \amp = 6x^4 + 2 \\ x^4 \amp = -\frac{1}{3}, \end{split} \end{equation*}

which gives no solutions. We conclude that $f$ has no relative maxima or minima.

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

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

Let $y=f(x)\text{.}$ We first notice that the domain of $f(x)$ is all real numbers. Differentiating, we have

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

Since $f'(x)$ is defined for all real numbers, all critical points of $f$ will be solutions to $f'(x) = 0\text{:}$

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

So in the interval $[-\pi/2,\pi/2]\text{,}$ the solution to this equation is

\begin{equation*} x = \frac{1}{2} \sin^{-1} \left(\frac{1}{2}\right) = \frac{1}{2}\left(\frac{\pi}{6}\right)\text{.} \end{equation*}

Therefore, since $\sin(2x)$ has a period of $\pi\text{,}$ we get

\begin{equation*} f'(x) = 0 \implies x = \frac{\pi}{12} + k\pi, \ k \in \mathbb{Z}\text{.} \end{equation*}

This gives an infinite number of critical points.

9. $\ds f(x) = \begin{cases}x-1 \amp x \lt 2 \\ x^2 \amp x\geq 2 \end{cases}$

none
Solution

The graph of $f$ has a corner at $x=2\text{,}$ and so $f'(2)$ is undefined. For $x \neq 2\text{,}$ we find

\begin{equation*} f'(x) = \begin{cases}1 \amp \text{ for } x \lt 2 \\ 2x \amp \text{ for } x > 2 \end{cases} \end{equation*}

So $f'(x) = 0$ has no solutions: $f'(x)$ is clearly nonzero for all $x \lt 2\text{,}$ and $f'(x) = 2x = 0$ has no solutions for $x > 2\text{.}$ Therefore $x=2$ is our only critical point. We now evaluate $f(x)$ at appropriate test points to determine if $x=2$ corresponds to a relative extremum or not. Since

\begin{equation*} f(1) = (1)-1 = 0, \ \ f(2)=(2)^2 = 4, \ \ \text{ and } \ \ f(3)=(3)^2 = 9\text{,} \end{equation*}

we conclude that $f$ has no relative extrema.

10. $\ds f(x) = \begin{cases}x-3 \amp x \lt 3 \\ x^3 \amp 3\leq x \leq 5\\ 1/x \amp x>5 \end{cases}$

relative max at $x=5$
Solution

The domain of $f(x)$ is all real numbers. We compute:

\begin{equation*} f'(x) = \begin{cases}1 \amp \text{ for } x \lt 3 \\ 3x^2 \amp \text{ for } 3\lt x \lt 5 \\ -\frac{1}{x^2} \amp \text{ for } x > 5 \end{cases} \end{equation*}

with $f'(x)$ undefined at $x=3$ and $x=5\text{,}$ only. Now set $f'(x) = 0\text{:}$ For $x \lt 3\text{,}$ $f(x) = x-3 = 0 \implies x = 3\text{,}$ so no solutions (since $x \neq 3$). For $3\lt x\lt 5\text{,}$ $x^3 = 0 \implies x = 0\text{,}$ and so again no solutions. There are furthermore no solutions for $x > 5\text{,}$ and so the only critical points are at $x=3$ and $x=5\text{.}$ Check:

\begin{equation*} \begin{array}{c|c} x \amp f(x) \\ \hline 2 \amp 2-3 = -1 \\ 3 \amp 3^3 = 27 \\ 4 \amp 4^3 = 64 \\ 5 \amp 5^3 = 125\\ 6 \amp 1/6 \end{array} \end{equation*}

Therefore, $f(x)$ has a relative maximum at $x=5\text{.}$

11. $\ds f(x) = x^2 - 98x + 4$

relative min at $x=49$
Solution

$f(x)$ is a polynomial, and so has domain $(-\infty,\infty)\text{.}$ Differentiating, we get

\begin{equation*} f'(x) = 2x - 98\text{.} \end{equation*}

Since $f'(x) = 0$ at $x=49$ and $f'(x)$ is defined for all real numbers, $x=49$ is the only critical point of $f\text{.}$ Now use the test points:

\begin{equation*} \begin{array}{c|c} x \amp f(x) \\ \hline 0 \amp 4 \\ 49 \amp -2397 \\ 100 \amp 204 \end{array} \end{equation*}

Therefore, $f(x)$ has a relative minimum at $x=49\text{.}$

12. $\ds f(x) = \begin{cases}-2 \amp x = 0 \\ 1/x^2 \amp x \neq 0 \end{cases}$

relative min at $x=0$
Solution

We compute:

\begin{equation*} f'(x) = \begin{cases}\text{ undefined } \amp x = 0 \\ -\frac{2}{x^3} \amp x \neq 0 \end{cases} \end{equation*}

Therefore, $x=0$ is the only critical point of $f\text{.}$ Since

\begin{equation*} \begin{array}{c|c} x \amp f(x) \\ \hline -1 \amp 1/(-1)^2 =1\\ 0 \amp -2 \\ 1 \amp 1/1^2 = 1 \end{array} \end{equation*}

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

For any real number $x$ there is a unique integer $n$ such that $n \leq x \lt n +1\text{,}$ and the greatest integer function is defined as $\ds\lfloor x\rfloor = n\text{.}$ Where are the critical values of the greatest integer function? Which are relative maxima and which are relative minima?

none
Solution

Notice that the domain of the greatest integer (or floor) function, $f(x) = \left \lfloor{x}\right \rfloor$ is $\mathbb{R}\text{,}$ while the range is $\mathbb{Z}\text{.}$

We visualize $\left \lfloor{x}\right \rfloor$ below:

Hence, $f(x) = \left \lfloor{x}\right \rfloor$ has no relative extrema.

Explain why the function $f(x) =1/x$ has no relative maxima or minima.

Solution

The function $f(x) = \dfrac{1}{x}$ has domain $\{x \in \mathbb{R} | x \neq 0\}\text{.}$ We calculate $f'(x)\text{:}$

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

Therefore, $f'(x)$ is undefined only at $x=0\text{,}$ which is not in the domain of $f\text{,}$ and is never equal to zero. Thus, $f(x)$ has no relative maxima or minima.

How many critical points can a quadratic polynomial function have?

one

Solution

Any polymomial $P(x)$ has domain $(-\infty,\infty)\text{.}$ Since the derivative of a polynomial is once again a polynomial, $P'(x)$ is defined over all real numbers. Since the general form of a quadratic polynomial is:

\begin{equation*} P_2(x) = ax^2 + bx + c\text{,} \end{equation*}

for some real constants $a\text{,}$$b\text{,}$$c\text{,}$ such that $a \neq 0\text{.}$ The general form of the derivative is then:

\begin{equation*} P'_2(x) = 2ax + b\text{.} \end{equation*}

This means that

\begin{equation*} P'_2(x) = 2ax + b = 0 \implies x = -\frac{b}{2a}\text{.} \end{equation*}

Hence, any quadratic polynomial can has exactly one critical point.

Show that a cubic polynomial can have at most two critical points. Give examples to show that a cubic polynomial can have zero, one, or two critical points.

Solution

Any polymomial $P(x)$ has domain $(-\infty,\infty)\text{.}$ Since the derivative of a polynomial is once again a polynomial, $P'(x)$ is defined over all real numbers. Since the general form of a cubic polynomial is:

\begin{equation*} P_3(x) = ax^3 + bx^2 + cx + d\text{,} \end{equation*}

with $a\neq 0\text{.}$ Then the general form of the derivative is

\begin{equation*} P'_3(x) = 3ax^2 + 2bx + c\text{.} \end{equation*}

Hence,

\begin{equation*} P'_3(x) = 3ax^2 + 2bx + c = 0 \end{equation*}

has solutions which are given by the quadratic formula:

\begin{equation*} x= \frac{-2b \pm \sqrt{4b^2 - 12ac}}{6a}\text{.} \end{equation*}

Hence, $P_3(x)$ can have zero, one or two critical points. For example:

\begin{equation*} y=x^3-2x+1 \end{equation*}

has two critical points,

\begin{equation*} y= x^3 \end{equation*}

has one critical point and

\begin{equation*} y = x^3+x^2+x+1 \end{equation*}

has no critical points.

Explore the family of functions $\ds f(x) = x^3 + cx +1$ where $c$ is a constant. How many and what types of relative extremes are there? Your answer should depend on the value of $c\text{,}$ that is, different values of $c$ will give different answers.

Solution

First, note that the domain of $f(x)=x^3+cx+1$ is $(-\infty,\infty)\text{,}$ and that $f'(x)=3x^2 + c$ is defined everywhere ($c$ is some constant). We now try to find solutions to $f'(x) = 0\text{.}$

\begin{equation*} \begin{split} 3x^2 + c \amp = 0\\ 3x^2 \amp = -c \\ x^2 \amp = -\frac{c}{3} \end{split} \end{equation*}

We need to consider three cases for the value of the constant $c\text{:}$

CASE 1: $c > 0$

Then $x^2 = -\frac{c}{3} \lt 0$ has no solutions, and $f$ has no relative extrema.

CASE 2: $c = 0$

Then our only critical point is $x = 0\text{.}$ To classify this point, we consider the following values of $f(x)= x^3+1\text{,}$

\begin{equation*} f(-1)=0, \ \ f(0) = 1, \ \ f(1) = 2\text{,} \end{equation*}

and conclude that this is neither a relative maximum nor minimum.

CASE 3: $c \lt 0$

Then $x^2 - \frac{c}{3} > 0$ has two unique solutions,

\begin{equation*} x = \pm \sqrt{\frac{-c}{3}} = \pm \sqrt{\frac{|c|}{3}} \cdot \end{equation*}

By once again comparing points on either side of each critical point, we find that $- \sqrt{\frac{-c}{3}}$ is a relative maximum, and $\sqrt{\frac{-c}{3}}$ is a relative minimum (see graph below for examples).

We generalize the preceding two questions. Let $n$ be a positive integer and let $f$ be a polynomial of degree $n\text{.}$ How many critical points can $f$ have? Hint

Recall the Fundamental Theorem of Algebra, which says that a polynomial of degree $n$ has at most $n$ roots.

Solution

Any polymomial $P(x)$ has domain $(-\infty,\infty)\text{.}$ Since the derivative f a polynomial is once again a polynomial, $P'(x)$ is defined over all real numbers. Now, if $f(x)$ is a polynomial of degree $n\text{,}$ it has the form

\begin{equation*} f(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1 x + a_0\text{.} \end{equation*}

Hence,

\begin{equation*} f'(x) = na_nx^{n-1} + (n-1)a_{n-1}x^{n-2} + ... + 2 a_2 x + a_1\text{.} \end{equation*}

This is a polynomial of degree $(n-1)\text{.}$ So by the fundamental theorem of algebra, $f'(x)$ has at most $n-1$ real roots. Hence, $f(x)$ has at most $n-1$ critical points.

### Subsection5.5.2Absolute Extrema

Unlike a relative extremum, which is only “extreme” relative to points “close to” it, an absolute extremum is “extreme” compared to all other points in the interval under consideration. Some examples of absolute maximum and minimum points are shown in Figure 5.17. This leads us to the following definitions.

###### Definition5.52. Absolute Maxima and Minima.

A real-valued function $f$ has an absolute maximum at $x_0$ if $f(x_0)\geq f(x)$ for all $x$ in the domain of $f\text{.}$

A real-valued function $f$ has an absolute minimum at $x_0$ if $f(x_0)\leq f(x)$ for all $x$ in the domain of $f\text{.}$

Note:
1. Notice that the definition of absolute extrema entails that an absolute extremum, unlike a relative extremum, can fall on an endpoint as shown in Figure 5.17.

2. Because of the “global” nature of an absolute extremum it is also often referred to as a global extremum.

###### Example5.53. Absolute Extrema Using a Graph.

Find the absolute extrema of the following functions using their graphs.

1. $f(x)=x^2$ on the interval $(-\infty,\infty)\text{.}$

2. $f(x)=|x|$ on the interval $[-1,2]\text{.}$

3. $f(x)=\cos x$ on the interval $[0,\pi]\text{.}$

Solution
1. This parabola has an absolute minimum at $x=0\text{.}$ However, it does not have an absolute maximum.

2. This graph looks like a check mark. It has an absolute minimum at $x=0$ and an absolute maximum at $x=2\text{.}$

3. $f(x)=\cos(x)$ has an absolute minimum at $x=\pi$ and an absolute maximum at $x=0$ on the interval $[0,\pi]\text{.}$

Like Fermat's Theorem, the following theorem has an intuitive appeal. However, unlike Fermat's Theorem, the proof relies on a more advanced concept called compactness, which will only be covered in a course typically entitled Analysis. So, we will be content with understanding the statement of the theorem.

Although this theorem tells us that an absolute extremum exists, it does not tell us what it is or how to find it.

Note that if an absolute extremum is inside the interval (i.e. not an endpoint), then it must also be a relative extremum. This immediately tells us that to find the absolute extrema of a function on an interval, we need only examine the relative extrema inside the interval, and the endpoints of the interval. We can devise a method for finding absolute extrema for a function $f$ on a closed interval $[a,b]\text{.}$

###### Guideline for Finding Absolute Extrema Given Continuity of $f$ and Closed Interval.
1. Verify the function is continuous on $[a,b]\text{.}$

2. Find the derivative and determine all critical values of $f$ that are in $(a,b)\text{.}$

3. Evaluate the function at the critical values found in Step 2 and the endpoints $x=a$ and $x=b$ of the interval.

4. The absolute maximum value and absolute minimum value of $f$ correspond to the largest and smallest $y$-values respectively found in Step 3.

Why must a function be continuous on a closed interval in order to use this theorem? Consider the following example.

###### Example5.55. Absolute Extrema and Continuity.

Find any absolute extrema for $f(x)=1/x$ on the interval $[-1,1]\text{.}$

Solution

The function $f$ is not continuous at $x=0\text{.}$ Since $0\in [-1,1]\text{,}$ $f$ is not continuous on the closed interval:

\begin{align*} \lim_{x\to 0^+}f(x)\amp =+\infty\\ \lim_{x\to 0^-}f(x)\amp =-\infty\, \text{,} \end{align*}

so we are unable to apply the Extreme Value Theorem. Therefore, $f(x)=1/x$ does not have an absolute maximum or an absolute minimum on $[-1,1]\text{.}$

However, if we consider the same function on an interval where it is continuous, the theorem will apply. This is illustrated in the following example.

###### Example5.56. Absolute Extrema and Continuity.

Find any absolute extrema for $f(x)=1/x$ on the interval $[1,2]\text{.}$

Solution

The function $f$ is continous on the interval, so we can apply the Extreme Value Theorem. We begin with taking the derivative to be $f'(x)=-1/x^2$ which has a critical value at $x=0\text{,}$ but since this critical value is not in $[1,2]$ we ignore it. The only points where an extrema can occur are the endpoints of the interval. To find the maximum or minimum we can simply evaluate the function: $f(1)=1$ and $f(2)=1/2\text{,}$ so the absolute maximum is at $x=1$ and the absolute minimum is at $x=2\text{.}$

Why must an interval be closed in order to use the above theorem? Recall the difference between open and closed intervals. Consider a function $f$ on the open interval $(0,1)\text{.}$ If we choose successive values of $x$ moving closer and closer to $1\text{,}$ what happens? Since 1 is not included in the interval we will not attain exactly the value of 1. Suppose we reach a value of 0.9999 — is it possible to get closer to 1? Yes: There are infinitely many real numbers between 0.9999 and 1. In fact, any conceivable real number close to 1 will have infinitely many real numbers between itself and 1. Now, suppose $f$ is decreasing on $(0,1)\text{:}$ As we approach 1, $f$ will continue to decrease, even if the difference between successive values of $f$ is slight. Similarly if $f$ is increasing on $(0,1)\text{.}$

Consider a few more examples:

###### Example5.57. Determining Absolute Extrema.

Determine the absolute extrema of $f(x)=x^3-x^2+1$ on the interval $[-1,2]\text{.}$

Solution

First, notice $f$ is continuous on the closed interval $[-1,2]\text{,}$ so we're able to use Theorem 5.54 to determine the absolute extrema. The derivative is $f'(x)=3x^2-2x\text{,}$ and the critical values are $x=0,2/3$ which are both in the interval $[-1,2]\text{.}$ In order to find the absolute extrema, we must consider all critical values that lie within the interval (that is, in $(-1,2)$) and the endpoints of the interval.

\begin{align*} f(-1)\amp =(-1)^3-(-1)^2+1=-1\\ f(0)\amp =(0)^3-(0)^2+1=1\\ f(2/3)\amp =(2/3)^3-(2/3)^2+1=23/27\\ f(2)\amp =(2)^3-(2)^2+1=5 \end{align*}

The absolute maximum is at (2,5) and the absolute minimum is at (-1,-1).

###### Example5.58. Determining Absolute Extrema.

Determine the absolute extrema of $f(x)=-9/x-x+10$ on the interval $[2,6]\text{.}$

Solution

First, notice $f$ is continuous on the closed interval $[2,6]\text{,}$ so we're able to use Theorem 5.54 to determine the absolute extrema. The function is not continuous at $x=0\text{,}$ but we can ignore this fact since 0 is not in $[2,6]\text{.}$ The derivative is $f'(x)=9/x^2-1\text{,}$ and the critical values are $x=\pm 3\text{,}$ but only $x=+3$ is in the interval. In order to find the absolute extrema, we must consider all critical values that lie within the interval and the endpoints of the interval.

\begin{align*} f(2)\amp =-9/(2)-(2)+10=7/2=3.5\\ f(3)\amp =-9/(3)-(3)+10=4\\ f(6)\amp =-9/(6)-(6)+10=5/2=2.5 \end{align*}

The absolute maximum is at (3,4) and the absolute minimum is at (6,2.5).

When we are trying to find the absolute extrema of a function on an open interval, we cannot use the Extreme Value Theorem. However, if the function is continuous on the interval, many of the same ideas apply. In particular, if an absolute extremum exists, it must also be a relative extremum. In addition to checking values at the relative extrema, we must check the behaviour of the function as it approaches the ends of the interval.

Some examples to illustrate this method.

###### Example5.59. Determining Absolute Extrema.

Find the extrema of $y=\sec(x)$ on $(-\pi/2,\pi/2)\text{.}$

Solution

Notice $\sec(x)$ is continuous on $(-\pi/2,\pi/2)$ and has one relative minimum at 0. Also

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

so $\sec(x)$ has no absolute maximum, but the point $(0,1)$ is the absolute minimum.

A similar approach can be used for infinite intervals.

###### Example5.60. Determining Absolute Extrema.

Find the extrema of $y=\ds\frac{x^2}{x^2+1}$ on $(-\infty,\infty)\text{.}$

Solution

Since $x^2+1\neq 0$ for all $x$ in $(-\infty,\infty)$ the function is continuous on this interval. This function has only one critical value at $x=0\text{,}$ which is the relative minimum and also the absolute minimum. Now, $\ds \lim_{x\to\pm\infty}\frac{x^2}{x^2+1}=1\text{,}$ so the function does not have an absolute maximum: It continues to increase towards 1, but does not attain this exact value.

##### Exercises for Section 5.5.2.

Find the absolute extrema for the following functions over the given interval.

1. $f(x)=-\frac{x+4}{x-4}$ on $[0,3]$

Absolute maximum $(3,7)\text{;}$ Absolute minimum $(0,1)$
Solution

We first look for any relative extrema of $f$ in the interval $[0,3]\text{.}$

\begin{equation*} f'(x)= \diff{}{x} \left(-\frac{x+4}{x-4}\right) = \frac{8}{(x-4)^2}\text{.} \end{equation*}

So $f'(x)=0$ has no solutions and $f$ is defined for all $x \in [0,3]\text{.}$ The absolute extrema must therefore occur at the endpoints. Since

\begin{equation*} f(0) = -\frac{4}{-4} = 1 \ \ \text{ and } \ \ \ f(3) = -\frac{7}{-1} = 7\text{,} \end{equation*}

we conclude that $(0,1)$ is the absolute minimum and $(3,7)$ is the absolute maximum.

2. $f(x)= x^3+4x^2+4$ on $[-4,1]$

Absolute maximum $(0,4)\text{;}$ Absolute minimum $(-4,4)$
Solution

We compute:

\begin{equation*} f'(x) = 3x^2+8x \end{equation*}

This is another polynomial, and so $f'(x)$ is defined for all $x \in [-4,1]\text{.}$ Now solve:

\begin{equation*} f'(x) = x(3x+8) = 0 \implies x=0, -8/3\text{.} \end{equation*}

Both of these two points are in the domain of $f\text{,}$ and so $f$ has two critical points. We now compare the value of $f$ at both endpoints and at each critical point:

\begin{equation*} f(-4) = 4, \ \ f(-8/3) = \frac{367}{27}, \ \ f(0) = 4, \ \ f(1) = 9\text{.} \end{equation*}

Hence, $f(x)$ has an absolute maximum at $\left(-\frac{8}{3},\frac{367}{37}\right)$ and absolute minima at $(-4,4)$ and $(0,4)\text{.}$

3. $f(x)=\csc(x)$ on $[0,\pi]$

Absolute minimum $(\pi/2,1)\text{;}$ No absolute maximum
Solution

We first rewrite

\begin{equation*} f(x) = \csc(x) = \frac{1}{\sin (x)}\text{.} \end{equation*}

Now differentiate:

\begin{equation*} f'(x) = -\frac{\cos(x)}{\sin^2(x)}\text{.} \end{equation*}

Over the interval $[0,\pi]\text{,}$ $f'(x)$ is undefined at $x=0$ and $x=\pi\text{.}$ Since these are endpoints, we do not classify them as critical points. Furthermore,

\begin{equation*} f'(x) = 0 \implies \cos x = 0 \implies x = \frac{\pi}{2}+n\pi, n \in \mathbb{Z}\text{,} \end{equation*}

and so $f(x)$ has one critical point at $x=\frac{\pi}{2}\text{.}$ We now compare:

\begin{equation*} \lim_{x\to 0} f(x) = \infty, \ f(\pi/2) = 1, \ \lim_{x\to \pi} f(x) = \infty\text{.} \end{equation*}

Hence, $f(x)$ has an absolute minimum at $\left(\frac{\pi}{2}, 1\right)$ and no absolute maxima.

4. $f(x)=\ln(x)/x^2$ on $[1,4]$

Absolute maximum $(\sqrt{e}, 1/(2e))\text{;}$ Absolute minimum $(1,0) \text{.}$
Solution

We calculate

\begin{equation*} f'(x) = \diff{}{x} \left(\frac{\ln x}{x^2}\right) = \frac{1-2\ln x}{x^3}\text{,} \end{equation*}

and solve for $f'(x) = 0\text{.}$

\begin{equation*} \begin{split} f'(x) \amp = 0 \\ \frac{1-2\ln x}{x^3} \amp = 0 \\ \ln x^2 \amp = 1 \ \ (x > 0)\\ x \amp = \sqrt{e} \approx 1.65. \end{split} \end{equation*}

Since $f'$ is undefined only at $x=0\text{,}$ which is outside the interval $[1,4]\text{,}$ $x= \sqrt{e}$ gives our only interior critical point. Comparing

\begin{equation*} f(1)= 0, \ \ f(4) \approx 0.0867, \ \ f(\sqrt{e}) = 1/(2e) \approx 0.18 \end{equation*}

we find the absolute maximum of $f$ on $[1,4]$ occurs at $x=\sqrt{e}$ and the absolute minimum occurs at $x=1\text{.}$

5. $f(x)=x\sqrt{1-x^2}$ on $[-1,1]$

Absolute maximum$\left(\frac{1}{\sqrt{2}}, \frac{1}{2} \right)$ ; Absolute minimum $\left(-\frac{1}{\sqrt{2}}, -\frac{1}{2} \right)$
Solution

We look for relative extrema:

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

Hence, $f'(x)$ is undefined at $x=-1$ and $x=1\text{.}$ Since these points are endpoints, we do not consider them to be endpoints. Next, we look for roots of $f'(x)\text{:}$

\begin{equation*} f'(x) = \frac{1 - 2 x^2}{\sqrt{1 - x^2}}= 0 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}}\text{,} \end{equation*}

and so $f$ has two critical points. Now compare:

\begin{equation*} f(-1) = 0, \ f\left(-\frac{1}{\sqrt{2}}\right) = -\frac{1}{2}, \ f\left(\frac{1}{\sqrt{2}}\right) = \frac{1}{2}, \ f(1) = 0\text{.} \end{equation*}

Therefore, $f(x)$ has an absolute maximum at $\left(\frac{1}{\sqrt{2}}, \frac{1}{2} \right)$ and an absolute minimum at $\left(-\frac{1}{\sqrt{2}}, -\frac{1}{2} \right)\text{.}$

6. $f(x)=xe^{-x^2/32}$ on $[0,2]$

Absolute minimum $(0,0)\text{;}$ Absolute maximum $(2,2e^{1/8})$
Solution

We first look for any relative extrema:

\begin{equation*} \begin{split} f'(x) \amp = e^{-x^2/32} + x\left(\frac{-2x}{32}\right)e^{-x^2/32} \\ \amp = \left(1-16x^2\right)e^{-x^2/32} \end{split} \end{equation*}

So $f'(x)=0$ when

\begin{equation*} 0 = 1-16x^2 \implies x = \pm 4\text{.} \end{equation*}

Thus, $f$ has no relative extrema in the interval $[0,2]\text{.}$ Since

\begin{equation*} f(0) = 0, \text{ and } f(2) = 2e^{-1/8}\text{,} \end{equation*}

we see that $f$ has an absolute minimum at $0$ and an absolute maximum at $2$ over the interval $[0,2]\text{.}$

7. $f(x)=x-\tan^{-1}(2x)$ on $[0,2]$

Absolute minimum $(1/2,\frac{2-\pi}{4})\text{;}$ Absolute maximum $(2,2-\tan^{-1}(4))$
Solution

We first differentiate:

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

And so $f'(x)$ is defined over the domain of $f\text{.}$ We now compute:

\begin{equation*} f'(x) = 0 \implies \frac{2}{4 x^2 + 1} = 1 \implies x = \pm \frac{1}{2}\text{.} \end{equation*}

We reject the negative solution since the domain of $f$ is $[0,2]\text{.}$ Hence, $f(x)$ has once critical point at $x=\frac{1}{2}\text{.}$ Now compare:

\begin{equation*} f(0) = 0 \ f(1/2) \approx -0.285398 , \ f(2) \approx 0.674182\text{.} \end{equation*}

Therefore, $f(x)$ has an absolute minimum at $\left(\frac{1}{2}, -0.285398\right)$ and an absolute maximum at $\left(2,0.674182\right)\text{.}$

8. $f(x)=\frac{x}{x^2+1}$

Absolute maximum $(1,1/2)\text{;}$ Absolute minimum $(-1,-1/2)$
Solution

We are looking for absolute maxima and minima of $f$ over $(-\infty,\infty)\text{.}$ We look for any relative extrema:

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

Notice that $x^2+1=0$ has no real solutions, and so $f$ is defined over $\mathbb{R}\text{.}$ Hence, $f'(x)=0$ when

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

We now take the following limits:

\begin{equation*} \lim_{x\to \infty} \frac{x}{x^2+1} \Heq \lim_{x\to \infty} \frac{1}{2x} = 0 \end{equation*}
\begin{equation*} \lim_{x\to -\infty} \frac{x}{x^2+1} \Heq \lim_{x\to -\infty} \frac{1}{2x} = 0 \end{equation*}

Since $f(-1) = -1/2\text{,}$$f(1)=1/2$ and $f \to 0$ as $x \to \pm \infty\text{,}$ we deduce that $f$ has an absolute minimum at the point $(-1, -1/2)$ and an absolute maximum at the point $(1,1/2)\text{.}$

For each of the following, sketch a potential graph of a continuous function on the closed interval $[0,4]$ with the given properties.

1. Absolute minimum at 0, absolute maximum at 2, relative minimum at 3. Solution

We sketch a function on $[0,4]$ which has an absolute minimum at $x=0\text{,}$ an absolute maximum at $x=2\text{,}$ and a relative minimum at $x=3\text{:}$

2. Absolute maximum at 1, absolute minimum at 2, relative maximum at 3. Solution

We sketch a function on $[0,4]$ which has an absolute maximum at 1, an absolute minimum at 2, and a relative maximum at 3:

3. Absolute minimum at 4, absolute maximum at 1, relative minimum at 2, relative maxima at 1 and 3. Solution

We sketch a function on $[0,4]$ which has an absolute minimum at $x=4\text{,}$ an absolute maximum at $x=1\text{,}$ a relative minimum at $x=2$ and relative maxima at $x=1$ and $x=3\text{:}$