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# 6.3: Mean Value Theorem

• • Dan Sloughter
• Professor (Mathematics) at Furman University
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##### Definition

We say $$f$$ is differentiable on an open interval $$I$$ if $$f$$ is differentiable at every point $$a \in I$$.

##### Definition

Suppose $$D \subset \mathbb{R}$$ and $$f: D \rightarrow \mathbb{R} .$$ We say $$f$$ has a local maximum at a point $$a \in D$$ if there exists $$\delta>0$$ such that $$f(a) \geq f(x)$$ for all $$x \in(a-\delta, a+\delta) \cap D .$$ We say $$f$$ has a local minimum at a point $$a \in D$$ if there exists $$\delta>0$$ such that $$f(a) \leq f(x)$$ for all $$x \in(a-\delta, a+\delta) \cap D .$$

##### Proposition $$\PageIndex{1}$$

Suppose $$D \subset \mathbb{R}, f: D \rightarrow \mathbb{R},$$ and $$a$$ is an interior point of $$D$$ at which $$f$$ has either a local maximum or a local minimum. If $$f$$ is differentiable at $$a,$$ then $$f^{\prime}(a)=0$$.

Proof

Suppose $$f$$ has a local maximum at $$a$$ (a similar argument works if $$f$$ has a local minimum at $$a$$ ). Choose $$\delta>0$$ so that $$(a-\delta, a+\delta) \subset D$$ and $$f(a) \geq f(x)$$ for all $$x \in(a-\delta, a+\delta) .$$ Then

$\frac{f(x)-f(a)}{x-a} \geq 0$

for all $$x \in(a-\delta, a)$$ and

$\frac{f(x)-f(a)}{x-a} \leq 0$

for all $$x \in(a, a+\delta) .$$ Hence

$\lim _{x \rightarrow a^{-}} \frac{f(x)-f(a)}{x-a} \geq 0$

and

$\lim _{x \rightarrow a^{+}} \frac{f(x)-f(a)}{x-a} \leq 0.$

Hence

$0 \leq \lim _{x \rightarrow a^{-}} \frac{f(x)-f(a)}{x-a}=f^{\prime}(a)=\lim _{x \rightarrow a^{+}} \frac{f(x)-f(a)}{x-a} \leq 0,$

so we must have $$f^{\prime}(a)=0$$. $$\quad$$ Q.E.D.

##### Theorem $$\PageIndex{2}$$

(Rolle's Theorem).

Let $$a, b \in \mathbb{R}$$ and suppose $$f$$ is continuous on $$[a, b]$$ and differentiable on $$(a, b) .$$ If $$f(a)=f(b),$$ then there exists a point $$c \in(a, b)$$ at which $$f^{\prime}(c)=0$$.

Proof

By the Extreme Value Theorem, we know $$f$$ attains a maximum and a minimum value on $$[a, b] .$$ Let $$m$$ be the minimum value and $$M$$ the maximum value of $$f$$ on $$[a, b] .$$ If $$m=M=f(a)=f(b),$$ then $$f(x)=m$$ for all $$x \in[a, b],$$ and so $$f^{\prime}(x)=0$$ for all $$x \in(a, b) .$$ Otherwise, one of $$m$$ or $$M$$ occurs at a point $$c$$ in $$(a, b) .$$ Hence $$f$$ has either a local maximum or a local minimum at $$c,$$ and so $$f^{\prime}(c)=0 .$$ $$\quad$$ Q.E.D.

##### Exercise $$\PageIndex{1}$$

Suppose $$f$$ is differentiable on $$(a, b)$$ and $$f^{\prime}(x) \neq 0$$ for all $$x \in(a, b) .$$ Show that for any $$x, y \in(a, b), f(x) \neq f(y)$$.

##### Exercise $$\PageIndex{2}$$

Explain why the equation $$x^{5}+10 x=5$$ has exactly one solution.

##### Exercise $$\PageIndex{3}$$

Let $$f(x)$$ be a third degree polynomial. Show that the equation $$f(x)=0$$ as at least one, but no more than three, solutions.

## 6.3.2 Mean Value Theorem

##### Theorem $$\PageIndex{3}$$

(Generalized Mean Value Theorem).

Let $$a, b \in \mathbb{R} .$$ If $$f$$ and $$g$$ are continuous on $$[a, b]$$ and differentiable on $$(a, b),$$ then there exists a point $$c \in(a, b)$$ at which

$(f(b)-f(a)) g^{\prime}(c)=(g(b)-g(a)) f^{\prime}(c).$

Proof

Let

$h(t)=(f(b)-f(a)) g(t)-(g(b)-g(a)) f(t).$

Then $$h$$ is continuous on $$[a, b]$$ and differentiable on $$(a, b) .$$ Moreover,

\begin{aligned} h(a) &=f(b) g(a)-f(a) g(a)-f(a) g(b)+f(a) g(a) \\ &=f(b) g(a)-f(a) g(b) \end{aligned}

and

\begin{aligned} h(b) &=f(b) g(b)-f(a) g(b)-f(b) g(b)+f(b) g(a) \\ &=f(b) g(a)-f(a) g(b). \end{aligned}

Hence, by Rolle's theorem, there exists a point $$c \in(a, b)$$ at which $$h^{\prime}(c)=0 .$$ But then

$0=h^{\prime}(c)=(f(b)-f(a)) g^{\prime}(c)-(g(b)-g(a)) f^{\prime}(c),$

which implies that

$(f(b)-f(a)) g^{\prime}(c)=(g(b)-g(a)) f^{\prime}(c).$

Q.E.D.

##### Theorem $$\PageIndex{4}$$

(Mean Value Theorem).

Let $$a, b \in \mathbb{R} .$$ If $$f$$ is continuous on $$[a, b]$$ and differentiable on $$(a, b),$$ then there exists a point $$c \in(a, b)$$ at which

$f(b)-f(a)=(b-a) f^{\prime}(c).$

Proof

Apply the previous result with $$g(x)=x$$. $$\quad$$ Q.E.D.

##### Exercise $$\PageIndex{4}$$

Prove the Mean Value Theorem using Rolle's theorem and the function

$k(t)=f(t)-\left(\left(\frac{f(b)-f(a)}{b-a}\right)(t-a)+f(a)\right).$

Give a geometric interpretation for $$k$$ and compare it with the function $$h$$ used in the proof of the generalized mean value theorem.

##### Exercise $$\PageIndex{5}$$

Let $$a, b \in \mathbb{R} .$$ Suppose $$f$$ is continuous on $$[a, b],$$ differentiable on $$(a, b),$$ and $$\left|f^{\prime}(x)\right| \leq M$$ for all $$x \in(a, b) .$$ Show that

$|f(b)-f(a)| \leq M|b-a|.$

##### Exercise $$\PageIndex{6}$$

Show that for all $$x>0$$,

$\sqrt{1+x}<1+\frac{x}{2}.$

##### Exercise $$\PageIndex{7}$$

Suppose $$I$$ is an open interval, $$f: I \rightarrow \mathbb{R},$$ and $$f^{\prime}(x)=0$$ for all $$x \in I .$$ Show that there exists $$\alpha \in \mathbb{R}$$ such that $$f(x)=\alpha$$ for all $$x \in I$$.

##### Exercise $$\PageIndex{8}$$

Suppose $$I$$ is an open interval, $$f: I \rightarrow \mathbb{R}, g: I \rightarrow \mathbb{R},$$ and $$f^{\prime}(x)=g^{\prime}(x)$$ for all $$x \in I .$$ Show that there exists $$\alpha \in \mathbb{R}$$ such that

$g(x)=f(x)+\alpha$

for all $$x \in I$$.

##### Exercise $$\PageIndex{9}$$

Let $$D=\mathbb{R} \backslash\{0\} .$$ Define $$f: D \rightarrow \mathbb{R}$$ and $$g: D \rightarrow \mathbb{R}$$ by $$f(x)=x^{2}$$ and

$g(x)=\left\{\begin{array}{ll}{x^{2},} & {\text { if } x<0,} \\ {x^{2}+1,} & {\text { if } x>0.}\end{array}\right.$

Show that $$f^{\prime}(x)=g^{\prime}(x)$$ for all $$x \in D,$$ but there does not exist $$\alpha \in \mathbb{R}$$ such that $$g(x)=f(x)+\alpha$$ for all $$x \in D .$$ Why does this not contradict the conclusion of the previous exercise?

##### Proposition $$\PageIndex{5}$$

If $$f$$ is differentiable on $$(a, b)$$ and $$f^{\prime}(x)>0$$ for all $$x \in(a, b)$$, then $$f$$ is increasing on $$(a, b)$$.

Proof

Let $$x, y \in(a, b)$$ with $$x<y .$$ By the Mean Value Theorem, there exists a point $$c \in(x, y)$$ such that

$f(y)-f(x)=(y-x) f^{\prime}(c).$

Since $$y-x>0$$ and $$f^{\prime}(c)>0,$$ we have $$f(y)>f(x),$$ and so $$f$$ is increasing on $$(a, b) .$$ $$\quad$$ Q.E.D.

##### Theorem $$\PageIndex{6}$$

If $$f$$ is differentiable on $$(a, b)$$ and $$f^{\prime}(x)<0$$ for all $$x \in(a, b)$$, then $$f$$ is decreasing on $$(a, b)$$.

##### Exercise $$\PageIndex{10}$$

State and prove similar conditions for nonincreasing and nondecreasing functions.