## Exercises3.4Mean Value Theorem

Use the Mean Value Theorem to solve the following problems.

###### 1.

Verify that the function

\begin{equation*} g(x)=\frac{3x}{x+7} \end{equation*}

satisfies the hypothesis of the Mean Value Theorem on the interval $[-1,2]\text{.}$ Then find all numbers $c$ that satisfy the conclusion of the Mean Value Theorem. Leave your final answer(s) exact.

Solution

Since $x+7\not= 0$ for all $x\in [-1,2]$ it follows that the function $g\text{,}$ as a rational function, is continuous on the closed interval $[-1,2]$ and differentiable on the open interval $(-1,2)\text{.}$ Therefore the function $g$ satisfies he hypothesis of the Mean Value Theorem on the interval $[-1,2]\text{.}$ By the Mean Value Theorem there is $c\in (-1,2)$ such that $\ds g'(c)=\frac{g(2)-g(-1)}{2-(-1)}\text{.}$ Thus the question is to solve $\ds \frac{21}{(c+7)^2}=\frac{7}{18}$ for $c\text{.}$ Hence $\ds c=-7\pm 3\sqrt{6}\text{.}$ Clearly $-7-3\sqrt{6}\lt -1$ and this value is rejected. From

\begin{equation*} -7+3\sqrt{6}>-1\Leftrightarrow 3\sqrt{6}>6 \mbox{ and } -7+3\sqrt{6}\lt 2\Leftrightarrow 3\sqrt{6}\lt 9 \end{equation*}

it follows that $c=-7+3\sqrt{6}\in(-1,2)$ and it is the only value that satisfies the conclusion of the Mean Value Theorem.

###### 2.

Use the Mean Value Theorem to show that $|\sin b-\sin a|\leq |b-a|\text{,}$ for all real numbers $a$ and $b\text{.}$

Solution

The inequality is obviously satisfied if $a=b\text{.}$ Let $a,b\in \mathbb{R}\text{,}$ $a\lt b\text{,}$ and let $f(x)=\sin x\text{,}$ $x\in [a,b]\text{.}$ Clearly the function $f$ is continuous on the closed interval $[a,b]$ and differentiable on $(a,b)\text{.}$ Thus, by the Mean value Theorem, there is $c\in (a,b)$ such that $\ds \cos c=\frac{\sin b-\sin a}{b-a}\text{.}$ Since $|\cos c|\leq 1$ for all real numbers $c$ it follows that $|\sin b-\sin a|\leq |b-a|\text{.}$

###### 3.

Two horses start a race at the same time and finish in a tie. Prove that at some time during the race they have the same speed.

Solution

Let $f(t)$ be the distance that the first horse covers from the start in time $t$ and let $g(t)$ be the distance that the second horse covers from the start in time $t\text{.}$ Let $T$ be time in which the two horses finish the race. It is given that $f(0)=g(0)$ and $f(T)=g(T)\text{.}$ Let $F(t)=f(t)-g(t)\text{,}$ $t\in [0,T]\text{.}$ As the difference of two position functions, the function $F$ is continuous on the closed interval $[0,T]$ and differentiable on the open interval $(0,T)\text{.}$ By the Mean value Theorem there is $c\in (0,T)$ such that $\ds F'(c)=\frac{F(T)-F(0)}{T-0}=0\text{.}$ It follows that $f'(c)=g'(c)$ which is the same as to say that at the instant $c$ the two horse have the same speed. (Note: It is also possible to use Rolle's theorem.)

###### 4.

Complete the following statement of the Mean Value Theorem precisely. Let $f$ be a function that is continuous on the interval . . . . . . . and differentiable on the interval . . . . . . . Then there is a number $c$ in $(a,b)$ such that $f(b)-f(a)=$ . . . . . . . . . . .

$[a,b]\text{;}$ $(a,b)\text{;}$ $f^\prime (c)(b-a)\text{.}$
Suppose that $-1\leq f^\prime (x)\leq 3$ for all $x\text{.}$ Find similar lower and upper bounds for the expression $f(5)-f(3)\text{.}$
Note that all conditions of the Mean Value Theorem are satisfied. To get the bounds use the fact that, for some $c\in (1,3)\text{,}$ $f(5)-f(3)=2f^\prime (c)\text{.}$
Suppose $g(x)$ is a function that is differentiable for all $x\text{.}$ Let $h(x)$ be a new function defined by $h(x)=g(x)+g(2-x)\text{.}$ Prove that $h^\prime (x)$ has a root in the interval $(0,2)\text{.}$
Note that $h(2)-h(0)=0$ and apply the Mean Value theorem for the function $h$ on the closed interval $[0,2]\text{.}$