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3.3 The mean-value theorem and applications

Consider the following graph:

Is there a point $x_{0}$ such that the slope of the curve is the same as the red (dashed) curve connecting $(a,f(a))$ with $(b,f(b)$ ? It seems so, and there is in fact a general theorem about this:

Theorem 3.3.1 (Mean-value theorem).

Let $f:[a,b]\rightarrow\mathbb{R}$ be a continuous function which is differentiable on $(a,b)$ . Then there is a point $x_{0}\in(a,b)$ such that

f^{\prime}(x_{0})=\frac{f(b)-f(a)}{b-a}.

Proof.

Let us first try to “normalise” the function in order to be in a more convenient form. We define

\phi:[a,b]\rightarrow\mathbb{R},\quad\phi(x)=f(x)-f(a)-\frac{f(b)-f(a)}{b-a}(x% -a).

It is continuous, and moreover differentiable on $(a,b)$ with $\phi(a)=\phi(b)=0$ . We have to show that there is $x_{0}\in(a,b)$ such that $\displaystyle f^{\prime}(x_{0})=\frac{f(b)-f(a)}{b-a}$ , which is equivalent to showing that

\phi^{\prime}(x_{0})=f^{\prime}(x_{0})-\frac{f(b)-f(a)}{b-a}=0.

Now if $\phi\equiv 0$ then $\phi^{\prime}\equiv 0$ and any number in $(a,b)$ may be chosen for $x_{0}$ . Let us therefore suppose that $\phi\not\equiv 0$ . We recall from MATH113 that a continuous function on a compact interval attains its infimum and supremum. Let $x_{0}\in[a,b]$ be a point such that

\phi(x_{0})=\sup_{x\in[a,b]}\phi(x)

if this number is positive. Otherwise, let $x_{0}$ be such that

\phi(x_{0})=\inf_{x\in[a,b]}\phi(x),

in which case $\phi(x_{0})<0$ . Notice that since $\phi\not\equiv 0$ , we cannot have both $\sup_{x\in[a,b]}\phi(x)$ and $\inf_{x\in[a,b]}\phi(x)$ equal to $0$ , so $\phi(x_{0})\not=0$ . Since $\phi(a)=\phi(b)=0$ we know that $x_{0}\not=a,b$ , so $x_{0}\in(a,b)$ . Then $x_{0}$ satisfies the condition of a local extremum of $\phi$ in Definition 3.2.3 (any $\delta>0$ does the job), namely a local maximum (or local minimum if we chose $x_{0}$ as the point where $\phi$ attains its infimum). Therefore $\phi^{\prime}(x_{0})=0$ by Proposition 3.2.4, which concludes the proof. ∎

Key idea of proof.

normalise $f$ and then study properties of local extrema, supremum and infimum on compact intervals

Corollary 3.3.2.

Suppose $f:[a,b]\rightarrow\mathbb{R}$ is a continuous function which is differentiable on $(a,b)$ .

$(i)$

If $f^{\prime}(x)\geq 0$ ( $f^{\prime}(x)>0$ ), for all $x\in(a,b)$ , then $f$ is increasing (strictly increasing) on $[a,b]$ .
$(ii)$

If $f^{\prime}(x)\leq 0$ ( $f^{\prime}(x)<0$ ), for all $x\in(a,b)$ , then $f$ is decreasing (strictly decreasing) on $[a,b]$ .
$(iii)$

If $f^{\prime}(x)=0$ , for all $x\in(a,b)$ , then $f$ is constant, i.e., there is $\lambda\in\mathbb{R}$ such that $f(x)=\lambda$ , for all $x\in I$ .
$(iv)$

If there is $M>0$ such that $|f^{\prime}(x)|\leqslant M$ , for all $x\in(a,b)$ , then $|f(x_{2})-f(x_{1})|\leqslant M|x_{2}-x_{1}|$ , for all $x_{1},x_{2}\in[a,b]$ .

Proof.

(i) and (ii), see Exercise A4.2.

(iii) We prove this by contraposition. Suppose $f$ is not constant. Then there must be $x_{1},x_{2}\in I$ with $x_{1}<x_{2}$ and $f(x_{1})\not=f(x_{2})$ . By the mean-value theorem there is $x_{0}\in I$ such $x_{1}<x_{0}<x_{2}$ and $\displaystyle f^{\prime}(x_{0})=\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}$ , which is nonzero, so $f^{\prime}\not\equiv 0$ .

(iv) Given $x_{1}<x_{2}$ , the mean-value theorem yields $x_{0}\in(x_{1},x_{2})$ such that

|f(x_{2})-f(x_{1})|=|x_{2}-x_{1}|\cdot|f^{\prime}(x_{0})|.

But by assumption, $|f^{\prime}(x)|\leq M$ for all $x\in(a,b)$ , so we get

|f(x_{2})-f(x_{1})|=|x_{2}-x_{1}|\cdot|f^{\prime}(x_{0})|\leq|x_{2}-x_{1}|% \cdot M.

The constant $M$ is independent of the explicit choice of $x_{1},x_{2}\in(a,b)$ . ∎

Exercise L.

How does property (iv) relate to uniform continuity?

Example 3.3.3.

(1)

Let $f:[a,b]\rightarrow\mathbb{R}$ be continuous, and differentiable on $(a,b)$ , with $f^{\prime}(x)\not=1$ for all $x\in(a,b)$ . Let us show that there is at most one number $x\in[a,b]$ such that $f(x)=x$ .

To this end, suppose there were two such numbers $x_{1}$ and $x_{2}$ , with $a\leqslant x_{1}<x_{2}\leqslant b$ . Then, by the mean-value theorem, there exists a point $x_{0}\in(x_{1},x_{2})$ such that

$f^{\prime}(x_{0})=\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}=\frac{x_{2}-x_{1}}{x_{% 2}-x_{1}}=1,$

contradicting the fact that $f^{\prime}(x)\neq 1$ for all $x\in(a,b)$ .
(2)

Let $f:[a,b]\rightarrow\mathbb{R}$ be differentiable and such that $f^{\prime}(x)=\lambda f(x)$ for all $x\in[a,b]$ , where $\lambda\in\mathbb{R}$ is some constant. Let us show that $f(x)=c\operatorname{e}^{\lambda x}$ for some constant $c$ . Let $g:[a,b]\rightarrow\mathbb{R}$ be defined by $g(x)=\operatorname{e}^{-\lambda x}f(x)$ . By the product rule, $g$ is also differentiable on $(a,b)$ with derivative

$g^{\prime}(x)=-\lambda\operatorname{e}^{-\lambda x}f(x)+\operatorname{e}^{-% \lambda x}f^{\prime}(x)=\operatorname{e}^{-\lambda x}\big{(}-\lambda f(x)+% \lambda f(x)\big{)}=0,$

so $g(x)=c$ for all $x$ , where $c$ is a constant, by Corollary 3.3.2(iii). Hence $f(x)=c\operatorname{e}^{\lambda x}$ , for all $x\in[a,b]$ .
(3)

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be defined by $f(x)=2x^{3}-3x^{2}-12x$ , which is clearly differentiable. Let us find the greatest and least values of $f$ on $[-2,4]$ . How many solutions are there in this interval of the equation $f(x)=1$ ? Note first that $f^{\prime}(x)=6x^{2}-6x-12=6(x^{2}-x-2)=6(x+1)(x-2)$ . Hence $f$ has a local minimum at $2$ and a local maximum at $-1$ , and Corollary 3.3.2 shows that $f$ is increasing on $[2,\infty)$ and on $(-\infty,-1]$ , whereas $f$ is decreasing on $[-1,2]$ . In short:

$x$ -2 -1 2 4

$f(x)$ -4 7 -20 32

By the intermediate-value theorem, there is a solution of $f(x)=1$ in the interval $(-2,-1)$ , another in $(-1,2)$ and yet another in $(2,4)$ . In each interval there is only one solution, since $f^{\prime}(x)\neq 0$ on the open interval in question (so is strictly monotonic). Hence there are precisely three solutions in $[-2,4]$ .