MATH 221

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 13 August 2014

Lecture 18

Rolle's Theorem and the Mean Value Theorem

Rolle's Theorem If $f\left(a\right)=f\left(b\right),$ and $f\left(x\right)$ is continuous between $a$ and $b,$ and
$f\left(x\right)$ is differentiable between $a$ and $b,$ then there is a point $c$ between $a$ and $b$ such that $\frac{df}{dx}{|}_{x=c}=0$ $$\begin{array}{c} a c c c c b x y y=f(x) \\ \text{4 possible choices for}\hspace{0.17em}c\text{!}\end{array}$$

Mean Value Theorem If $f\left(x\right)$ is continuous between $a$ and $b,$ and
$f\left(x\right)$ is differentiable between $a$ and $b,$ then there is a point $c$ between $a$ and $b$ such that $$\frac{df}{dx}{|}_{x=c}=\frac{f\left(b\right)-f\left(a\right)}{b-a}\text{.}$$ $$\begin{array}{c} a c c c c c b x f ( a ) f ( b ) y This line has slope f(b)-f(a)b-a \end{array}$$ Note: If $f\left(a\right)=f\left(b\right)$ then the line connecting $(a,f\left(a\right))$ and $(b,f\left(b\right))$ has slope $0$ and so Rolle's theorem is a special case of the mean value theorem.

Verify Rolle's theorem for the function $f\left(x\right)=e^{-x}\text{sin}\hspace{0.17em}x$ on the interval $[0,\pi ]\text{.}$ $$\begin{array}{c} π 4 π x y \end{array}$$ $$\begin{array}{ccc}{\displaystyle f\left(0\right)}& =& {\displaystyle e^0\text{sin}\hspace{0.17em}0=0,}\\ {\displaystyle f\left(\pi \right)}& =& {\displaystyle e^{\pi }\text{sin}\hspace{0.17em}\pi =0,}\\ {\displaystyle \frac{df}{dx}}& =& {\displaystyle e^{-x}\text{cos}\hspace{0.17em}x-e^{-x}\text{sin}\hspace{0.17em}x}\\ & =& {\displaystyle e^{-x}(\text{cos}\hspace{0.17em}x-\text{sin}\hspace{0.17em}x)\text{.}}\end{array}$$ So, if $\frac{df}{dx}=0$ then $\text{cos}\hspace{0.17em}x-\text{sin}\hspace{0.17em}x=0\text{.}$ So $\text{cos}\hspace{0.17em}x=\text{sin}\hspace{0.17em}x\text{.}$ So $x=\frac{\pi }{4}\text{.}$ So $c=\frac{\pi }{4}$ when $\frac{df}{dx}{|}_{x=c}=0\text{.}$

Verify the mean value theorem for $f\left(x\right)=e^x$ in the interval $[0,1]\text{.}$ $$\begin{array}{c} 1 x 1 3 y y=ex line with slope f(1)-f(0)1-0. \end{array}$$ $$\begin{array}{ccc}{\displaystyle f\left(0\right)}& =& {\displaystyle e^0=1,}\\ {\displaystyle f\left(1\right)}& =& {\displaystyle e^1=e,}\\ {\displaystyle \text{so}\hspace{0.17em}\frac{f\left(1\right)-f\left(0\right)}{1-0}}& =& {\displaystyle \frac{e-1}{1-0}=e-1\text{.}}\end{array}$$ dfdx=ex and we want $c$ so that $\frac{df}{dx}{|}_{x=c}=e-1\text{.}$ If $\frac{df}{dx}{|}_{x=c}=e^c=e-1\text{.}$ Then $c=\text{ln}(e-1)\approx \text{ln}\left(1.78\right)$ which is between $0$ and $1\text{.}$

Consider the mean value theorem for $f\left(x\right)=\frac{1}{x}$ in the interval $[-1,1]\text{.}$ $$f\left(1\right)=\frac{1}{1}=1\text{and}f(-1)=\frac{1}{-1}=-1\text{.}$$ So $$\frac{f\left(1\right)-f(-1)}{1-(-1)}=\frac{1-(-1)}{2}=\frac{2}{2}=1\text{.}$$ So we want $c$ so that $\frac{df}{dx}{|}_{x=c}=\frac{f\left(1\right)-f(-1)}{1-(-1)}=1\text{.}$ $$\frac{df}{dx}{|}_{x=c}=\frac{d\frac{1}{x}}{dx}{|}_{x=c}=-\frac{1}{x^2}{|}_{x=c}=-\frac{1}{c^2}\text{.}$$ Find $c$ so that $-\frac{1}{c^2}=1\text{.}$ (IMPOSSIBLE with real numbers.) What went wrong? $$\begin{array}{c} 1 -1 x 1 -1 y line with slope f(1)-f(-1)1-(-1)=1 \end{array}$$ $f\left(x\right)$ is not continuous or differentiable at $x=0\text{!!}$ So the mean value theorem does not apply.

Show that the equation $x^5+10x+3=0$ has exactly one real root.

Let $f\left(x\right)=x^5+10x+3\text{.}$ We have to show that there is only one real number that can be plugged into $f\left(x\right)$ to get $0\text{.}$

Notes:

(a)	As $x\to \infty ,$ $f\left(x\right)\to \infty \text{.}$
(b)	As $x\to -\infty ,$ $f\left(x\right)\to -\infty \text{.}$

$$\begin{array}{c} x y y=f(x) \end{array}$$ (a) and (b) tell us that $y=f\left(x\right)$ must cross the $x$ axis. Suppose it crosses twice, at $x=a$ and $x=b\text{.}$ $$\begin{array}{c} a b c x y \end{array}$$ Then $\frac{df}{dx}{|}_{x=c}=0$ for some $c$ between $a$ and $b\text{.}$ So $$0=\frac{df}{dx}{|}_{x=c}=\frac{d(x^5+10x+3)}{dx}{|}_{x=c}=5x^4+10{|}_{x=c}=5c^4+10\text{.}$$ But $5c^4+10$ is never $0,$ no matter what $c$ is. So $y=f\left(x\right)$ can't cross the $x$ axis twice. So it must cross it only once. So there is exactly one number (real number) that can be plugged into $f\left(x\right)$ to get $0\text{.}$

Discuss Rolle's theorem for $f\left(x\right)=(x-1)(2x-3)$ in the interval $1\le x\le 3\text{.}$ $$\begin{array}{c} 1 3 x 6 y \end{array}$$ $$\begin{array}{ccc}{\displaystyle f\left(1\right)}& =& {\displaystyle 0,}\\ {\displaystyle f\left(3\right)}& =& {\displaystyle (3-1)(6-3)=6\text{.}}\end{array}$$ Since $f\left(1\right)\ne f\left(3\right)$ we can't apply Rolle's theorem with $x=1$ and $x=3\text{.}$ Are there two points in the interval $[1,3]$ where $f\left(a\right)=f\left(b\right)\text{?}$ Yes, $f\left(1\right)=0$ and $f\left(\frac{3}{2}\right)=0\text{.}$ so we should be able to find $c$ between $1$ and $\frac{3}{2}$ so that $\frac{d{|}_{x=c}}{dx}=0\text{.}$ $$\frac{df}{dx}{|}_{x=c}=(x-1)·2+1·(2x-3){|}_{x=c}=4x-5{|}_{x=c}=4c-5\text{.}$$ so $\frac{df}{dx}{|}_{x=c}=0$ when $c=\frac{5}{4},$ and $\frac{5}{4}$ is between $1$ and $\frac{3}{2}\text{.}$

Notes and References

These are a typed copy of Lecture 18 from a series of handwritten lecture notes for the class MATH 221 given on October 18, 2000.

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