## Exercises3.8Miscellaneous

Solve the following problems.

###### 1.

For what values of the constant $c$ does $\ln x=cx^2$ have solutions. Assume that $c>0\text{.}$

$c\lt \frac{1}{2e}$
Solution

Let $f(x)=\ln x-cx^2\text{.}$ Note that the domain of $f$ is the interval $(0,\infty )$ and that $\ds \lim _{x\to \pm \infty }f(x)=-\infty\text{.}$ From $\ds f'(x)=\frac{1}{x}-2cx$ it follows that there is a local (and absolute) maximum at $\ds x=\frac{1}{\sqrt{2c}}\text{.}$ Since the function $f$ is continuous on its domain, by the Intermediate Value Theorem, it will have a root if and only if $\ds f\left( \frac{1}{\sqrt{2c}}\right) >0\text{.}$ Thus

\begin{equation*} \ln \frac{1}{\sqrt{2c}} -c\cdot\frac{1}{2c}>0\Leftrightarrow \ln \frac{1}{\sqrt{2c}}>\frac{1}{2}\Leftrightarrow c\lt \frac{1}{2e}\text{.} \end{equation*}
###### 2.

Show that $y=3x^3+2x+12$ has a unique root.

Solution

Note that the function $y=3x^3+2x+12\text{,}$ as a polynomial, is continuous on the set of real numbers. Also, $\ds \lim _{x\to -\infty }(3x^3+2x+12)=-\infty$ and $\ds \lim _{x\to \infty }(3x^3+2x+12)=\infty\text{.}$ By the Intermediate Value Theorem, the function has at least one root. Next, note that $y'=6x^2+2>0$ for all $x\in \mathbb{R}\text{.}$ This means that the function is increasing on its domain and therefore has only one root. (If there is another root, by Rolle's theorem there will be a critical number.)

###### 3.

Show that the equation $x^3+9x+5=0$ has exactly one real solution.

Hint
Take $y=x^3+9x+5\text{.}$
Solution

Let $y=x^3+9x+5\text{.}$ This function is a polynomial, and so is continuous. Since

\begin{equation*} y(0) = 5 > 0, \ \ y(-1) = -5, \end{equation*}

by the Intermediate Value Theorem that $y$ has at least one root in the interval $(-1,0) \text{.}$ Furthermore, since

\begin{equation*} y' = 3x^2+9 = 3(x^2+3), \end{equation*}

we see that $y$ is increasing for all $x \in \mathbb{R} \text{.}$ Therefore, $y$ can have no additional roots. Hence, $x^3+9x+5=0$ has exactly one real solution.

###### 4.

For which values of $a$ and $b$ is $(1,6)$ a point of inflection of the curve $y=x^3+ax^2+bx+1\text{?}$

Hint
Solve $f(10)=6$ and $f''(1)=0\text{.}$
$a=-3\text{,}$ $b=1\text{.}$
###### 5.

Prove that $\ds f(x)=\frac{1}{(x+1)^2}-2x+\sin x$ has exactly one positive root.

Solution

Note that the function $f$ is continuous on its domain $\mathbb{R}\backslash\{-1\}\text{.}$ Since $\ds \lim _{x\to \infty }\frac{1}{(x+1)^2}=0$ and $|\sin x|\leq 1\text{,}$ for all $x\in \mathbb{R}\text{,}$ it follows that $\ds \lim _{x\to \infty }f(x)=\lim _{x\to \infty }(-2x)=- \infty\text{.}$ Also, $\ds f(0)=1>0\text{.}$ By the Intermediate Value Theorem, the function has at least one root in the interval $(0,\infty )\text{.}$ Next, note that $\ds f'(x)=-\frac{2}{(x+1)^3} -2+\cos x\lt 0$ for all $x\in (0,\infty)\text{.}$ This means that the function is decreasing on $(0,\infty)$ and therefore has only one root.

###### 6.

A ball is thrown vertically upwards from a platform 12 feet above the ground so that after $t$ seconds have elapsed the height $s$ (in feet) of the ball above the ground is given by

\begin{equation*} s=12+96t-t^2\text{.} \end{equation*}

Compute the following quantities:

1. The initial velocity.

2. The time to highest point.

3. The maximum height attained.

1. From $s'(t)=v(t)=96-2t$ it follows that the initial velocity is $v(0)=96$ ft/sec.

2. The only critical number is $t=48\text{.}$ By the second derivative test there is a local (and absolute) maximum there.

3. $s(48)=2316$ feet.

###### 7.

The table below gives the values of certain functions at certain points. Calculate each of the following or write “insufficient information” if there is no enough information to compute the value.

\begin{equation*} \begin{array}{|c|c|c|c|c|} \hline x\amp f(x)\amp f'(x)\amp g(x)\amp g'(x)\\ \hline \hline 1\amp 3\amp 3\amp 2\amp 2\\ \hline 2\amp 4\amp -1\amp 4\amp 0\\ \hline 3\amp 6\amp 1\amp 0\amp 4\\ \hline 4\amp -1\amp 0\amp 1\amp 1\\ \hline 5\amp 2\amp 4\amp 3\amp 3\\ \hline \end{array} \end{equation*}
1. What is $\ds \lim _{h\to 0}\frac{f(3+h)-f(3)}{h}\text{?}$

2. What is $\ds \lim _{h\to 0}\frac{f(2+h)g(2+h)-f(2)g(2)}{h}\text{?}$

3. Use differentials to find the approximate value of $f(0.98)\text{.}$

4. What are the coordinates of any point on the graph of $f$ at which there is a critical point?

1. $f'(3)=1\text{.}$

2. $(fg)'(2)=f'(2)g(2)+f(2)g'(2)=-4\text{.}$

3. $f(0.98)\approx f(1)+f'(1)(0.98-1)=3+3(-0.02)=2.94\text{.}$

4. There is a critical point at $(4,-1)\text{.}$

###### 8.
1. State domain and range of $f(x)=\arcsin x\text{.}$

2. Derive the differentiation formula $\ds \frac{d}{dx}[\arcsin x]=\frac{1}{\sqrt{1-x^2}}\text{.}$

3. Let $g(x)=\arcsin (\sin x)\text{.}$ Graph $g(x)$ and state its domain and range.

4. For the function $g(x)$ as defined in part (c) find $g'(x)$ using any method you like. Simplify your answer completely.

5. Explain carefully why the equation

\begin{equation*} 4x-2+\cos \left( \frac{\pi x}{2}\right) =0 \end{equation*}

has exactly one real root.

Solution
1. The domain of $f(x)=\arcsin x$ is the interval $[-1,1]$ and its range is $\ds \left[ -\frac{\pi }{2},\frac{\pi }{2}\right]\text{.}$

2. For $x\in (-1,1)$ let $y=\arcsin x\text{.}$ Then $\sin y=x$ and $y'\cos y=1\text{.}$ Since $\ds y \in \left( -\frac{\pi }{2},\frac{\pi }{2}\right)$ it follows that $\cos y>0\text{.}$ Thus $\ds \cos y =\sqrt{1-\sin ^2y}=\sqrt{1-x^2}\text{.}$

3. The domain of the function $g$ is the set of all real numbers and its range is the set $\ds \left[ -\frac{\pi }{2},\frac{\pi }{2}\right]\text{.}$

4. $\ds g'(x)=\left\{ \begin{array}{rl} 1& \mbox{if } x\in (4n-1,4n+1), \ n\in \mathbb{Z}\\ -1& \mbox{if } x\in (4n+1,4n+3), \ n\in \mathbb{Z}\\ \end{array} \right.\text{.}$ The derivative of $g$ is not defined if $x=2n+1\text{,}$ $n\in \mathbb{Z}\text{.}$

5. The function $F(x)=4x-2+\cos \left( \frac{\pi x}{2}\right)$ is continuous on the set of real numbers. From $\ds \lim _{x\to \pm \infty }F(x)=\pm \infty\text{,}$ by the Intermediate Value Theorem, the function $F$ has at least one root. From $\ds F'(x)=1-\frac{2}{\pi }\sin \left( \frac{\pi x}{2}\right)>0$ we conclude that $F$ is monotone.

###### 9.

Given that

\begin{equation*} \sinh x=\frac{e^x-e^{-x}}{2} \mbox{ and } \cosh x=\frac{e^x+e^{-x}}{2} \end{equation*}
1. Find $\ds \lim_{x\to \infty}\tanh x\text{.}$

2. Find the equation of the tangent line to the curve $y=\cosh(x)+3x+4$ at the point $(0,5)\text{.}$

1. $\displaystyle \ds \lim_{x\to \infty}\tanh x=1.$

2. $\displaystyle y_T = 3 x + 5.$

Solution
1. We have

\begin{equation*} \tanh x= \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x+e^{-x}} = \frac{1-e^{-2x}}{1+e^{-2x}}. \end{equation*}

Hence,

\begin{equation*} \lim_{x\to\infty}\tanh = \lim_{x\to\infty} \frac{1-e^{-2x}}{1+e^{-2x}} = 1. \end{equation*}
2. We differentiate:

\begin{equation*} \frac{dy}{dx}\big\vert_{x=0} = \left[\frac{d}{dx}\cosh x \right]_{x=0} +3 = 3. \end{equation*}

Therefore, the equation of the tangent line to $y=\cosh(x)+3x+4$ at the point $(0,5)$ is given by

\begin{equation*} y_T = 3 x + 5. \end{equation*}
###### 10.
1. Differentiate $\ds f(x)=\frac{\ln x}{x}\text{,}$ for $x>0\text{.}$

2. Sketch the graph of $\ds f(x)=\frac{\ln x}{x}\text{,}$ showing all extrema.

3. Use what you have done to decide which is larger, $\ds 99^{101}$ or $101^{99}\text{.}$

1. $\ds f'(x)=\frac{1-\ln x}{x^2}\text{.}$

2. Since $99 \gt e$ we have that $\ds \frac{\ln 99}{99}\lt \frac{\ln 101}{101}\text{.}$ This is the same as $101\ln 99\lt 99\ln 101\text{.}$ Hence $\displaystyle \ln 99^{101}\lt \ln 101^{99}$ and $\displaystyle 99^{101}\lt 101^{99}\text{.}$

###### 11.

Answer the following questions. No justification necessary.

1. What is the general antiderivative of $f(x)=6x^2+2x+5\text{?}$

2. What is the derivative of $g(x)=\sinh (x)$ with respect to $x\text{?}$

3. If $f^\prime (x)$ changes from negative to positive at $c$ the $f(x)$ has a (pick one)

1. local maximum at $c\text{.}$

2. local minimum at $c\text{.}$

3. global maximum at $c\text{.}$

4. global minimum at $c\text{.}$

4. If $x^5+y^5=1\text{,}$ what is $y^\prime$ in terms of $x$ and $y\text{?}$

5. If a point has polar coordinates $(r,\theta )=(3,3\pi )\text{,}$ what are its Cartesian coordinates?

1. $2x^3+x^2+5x+c, c\in \mathbb{R}\text{.}$

2. $\cosh (x)\text{.}$

3. local minimum at $c \text{.}$

4. $\displaystyle y^\prime =-x^4y^{-4}\text{.}$

5. $(-3,0)\text{.}$

###### 12.
1. State the definition of the derivative of a function $g$ at a number $x\text{.}$

2. State the Squeeze Theorem, clearly identifying any hypothesis and the conclusion.

3. State Fermat's Theorem, clearly identifying any hypothesis and the conclusion.

4. Give an example of a function with one critical point which is also an inflection point.

5. Give an example of a function that satisfies $f(-1)=0\text{,}$ $f(10)=0\text{,}$ and $f^\prime (x)>0$ for all $x$ in the domain of $f^\prime\text{.}$

1. $\displaystyle \displaystyle g'(x) = \lim_{h\to 0 } \frac{g(x+h)-g(x)}{h}.$

2. If $f(x)\leq g(x)\leq h(x)$ when $x$ is near $a$ (except possibly at $a$) and $\displaystyle \lim_{x\to a}f(x)=\lim _{x\to a}h(x)=L$ then $\displaystyle \lim_{x\to a}g(x)=L\text{.}$

3. If $f$ has a local maximum or minimum at $c\text{,}$ and $f'(c)$ exists, then $f'(c)=0\text{.}$

4. Take $f(x)=x^3$ and the point $(0,0)\text{.}$

5. For example, $\ds f(x)=x-\frac{1}{x}\text{.}$

###### 13.
1. State the definition of a critical number of a function $f\text{.}$

2. State the Mean Value Theorem, clearly identifying any hypothesis and the conclusion.

3. State the Extreme Value Theorem, clearly identifying any hypothesis and the conclusion.

4. State the definition of an inflection point of a function $f\text{.}$

5. Give an example of a function with a local maximum at which the second derivative is 0.

6. Give an example of a quadratic function of the form $f(x)=x^2+bx+c$ whose tangent line is $y=3x+1$ at the point $(0,1)\text{.}$

1. A critical number of a function $f$ is a number $c$ in the domain of $f$ such that either $f'(c)=0$ or $f'(c)$ does not exist.

2. Let $f$ be a function that satisfies the following hypotheses:

1. $f$ is continuous on the closed interval $[a,b]\text{.}$

2. $f$ is differentiable on the open interval $(a,b)\text{.}$

Then there is a number $c$ in $(a,b)$ such that $f'(c)=\ds \frac{f(b)-f(a)}{b-a}$ or, equivalently, $f(b)-f(a)=f'(c)(b-a) \text{.}$

3. If $f$ is continuous on a closed interval $[a,b]\text{,}$ then $f$ attains an absolute maximum value $f(c)$ and an absolute minimum value $f(d)$ at some numbers $c,d\in [a,b]\text{.}$

4. A point $P$ on a curve $y=f(x)$ is called an inflection point if $f$ is continuous there and the curve changes from concave upward to concave downward or from concave downward to concave upward at $P\text{.}$

5. Take $f(x)=-x^4$ and the point $(0,0)\text{.}$

6. $f(x)= x^2+3x+1\text{.}$

###### 14.
1. Give an example of a function that is strictly decreasing on its domain.

2. Give an example of a function $f$ and an interval $[a,b]$ such that the conclusion of the Mean Value Theorem is not satisfied for $f$ on this interval.

3. Use a linear approximation to estimate $\sqrt{100.4}\text{.}$ Is your estimate larger or smaller than the actual value?

1. Take $y=-x$ for all $x \in \mathbb{R}.$

2. Take $y=\frac{1}{x^2}$ for $x \in [-1, 1] \text{.}$

3. $\sqrt{102} \approx 10.1\text{.}$ Overestimate.

Solution
1. Take $y=-x$ for all $x \in \mathbb{R}.$ Then $y'(x) = -1 \lt 0$ for all $x \in \mathbb{R}\text{,}$ and so the function is strictly decreasing on its entire domain.

2. Take $y=\frac{1}{x^2}$ for $x \in [-1, 1] \text{.}$ Then there is no number $c\in (-1,1)$ such that $f'(c) = \frac{1-1}{-2} = 0\text{.}$ The MVT fails because the function is not continuous on $[-1,1]\text{.}$

3. Let $f(x) = \sqrt{x}\text{.}$ Then we know that $f(100)=10$ and $f'(100)=1/20.$ Therefore, we compute:

\begin{equation*} \sqrt{102}=f(102) \approx f(100) + f'(100)(2) = 10 + \frac{1}{10} =10.1. \end{equation*}

Since $f(x)$ is concave down, this will be an overestimate. Indeed, using computer software, we find that $\sqrt{102} = 10.0995...\text{.}$

Short definitions, theorems and examples. No part marks given.

###### 15.

Suppose for a function $y=f(x)$ we have that if $x_1\not= x_2$ then $f(x_1)\not= f(x_2)$ for all $x_1,x_2$ in the domain of $f\text{.}$ What does this tell you about $f\text{?}$

Solution

From the statement, it follows that if $f(x_1) = f(x_2)\text{,}$ then $x_1=x_2$ for all $x_1,x_2$ in the domain of $f\text{.}$ Hence, we know that $f$ is a one-to-one function, and that it is invertible.

###### 16.

Suppose you need to show that $\displaystyle \lim_{x\to a}g(x)=L$ using the Squeeze Theorem with given functions $y=m(x)$ and $y=n(x)\text{.}$ What condition(s) must the functions $m$ and $n$ satisfy?

Solution

We must have that $\displaystyle \lim_{x\to a}m(x)=\lim_{x\to a}n(x)=L\text{,}$ and that for all $x$ in some neighbourhood of $a$ (but not necessarily at $a$), $g(x)$ is bounded above by one of $m(x), n(x),$ and bounded below by the other.

###### 17.

Identify the theorem that states the following: if any horizontal line $y=b$ is given between $y=f(a)$ and $y=f(c)\text{,}$ then the graph of $f$ cannot jump over the line; it must intersect $y=b$ somewhere provided that $f$ is continuous.

The Intermediate Value Theorem.
###### 18.

In the Figure below, which theorem is shown geometrically for a function $f$ on $[a,b]\text{?}$

The Mean Value Theorem.
###### 19.

Given a function $y=f(x)\text{.}$ Suppose that $f$ has a root in $[a,b]$ and you want to approximate the root using Newton's method with initial value $x=x_1\text{.}$ Show graphically, how Newton's method could fail.

###### 20.

The graph of the function $f$ is given in the Figure below. Clearly graph a possible inverse function for $f$ over its maximum domain directly on the given Cartesian coordinates.

###### 21.

Draw a graph of a function $f$ that has an inflection point at $x=2$ and $f^\prime(2)$ does not exist.

###### 22.

Draw a graph of a non-linear function $f$ that satisfies the conclusion of the Mean value Theorem at $c=-2$ in $[-4,2]\text{.}$

###### 23.

Let

\begin{equation*} f(x)=\left\{ \begin{array}{rl} x+4,\amp x\lt 1\\ 5,\amp x=1\\ 2x^2+3,\amp x>1. \end{array} \right. \end{equation*}

What can you say about differentiability of $f$ at $x=1\text{?}$

Not differentiable.
Solution
Since
\begin{equation*} \lim_{h\to 0+} \frac{f(1+h)-f(1)}{h} = 6, \end{equation*}
and
\begin{equation*} \lim_{h\to 0-} \frac{f(1+h)-f(1)}{h} = 1, \end{equation*}
it follows that $f$ is not differentiable at $x=1 \text{.}$
###### 24.

Show that $f(x)=\sinh x$ is an odd function.

Solution

For all $x \in \mathbb{R},$ we have that

\begin{equation*} f(-x) = \sinh(-x) = \frac{e^{-x}-e^{x}}{2} = -\frac{e^x-e^{-x}}{2} = -\sinh(x). \end{equation*}

Hence, $\sinh(x)$ is an odd function.