Mean Value Theorem

As a property of continuous function, we have Intermediate Value Theorem and Extreme Value Theorem. Then we ask what kind of properties differentiable functions have.

Mean Value Theorem

Theorem 2..7 Let $f(x)$

be continuous on $[a,b]$

and differentiable on $(a,b)$

. Then there exists at least one $\xi \in (a,b)$ satisfying

$\displaystyle \frac{f(b) - f(a)}{b - a} = f^{\prime}(\xi)$

**Figure 2.5:** Mean Value Theorem
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Understanding The Mean Value Theorem says that if you drive 60km in 1hr, then your average speed is 60km/hr and you must drive your car faster than 60km/hr at least once.

NOTE Note that $\displaystyle{\frac{f(b) - f(a)}{b - a}}$ can be thought of the slope of line passing through two points $(a,f(a)),(b,f(b))$ . Then $\displaystyle{\frac{f(b) - f(a)}{b - a} = f^{\prime}(\xi)}$ for $x_{1} < \xi < x_{2}$ can be thought of existence of tangent line with the slope is the same. Suppose that $f(x)$ is the position of a car and the interval $[a,b]$ represents time. Then, $\displaystyle{f(b) - f(a)}$ represents the distance moved during $b -a$ . In other words, $\frac{f(b) - f(a)}{x - a}$ represents the average speed. $f^{\prime}(\xi)$ represents the instantaneous speed.

Rolle's Theorem

Theorem 2..8 Let $f(x)$

be continuous on $[a,b]$

and differentiable on $(a,b)$

. If

, then there is at least one number $\xi$ in $(a,b)$

such that

$\displaystyle f^{\prime}(\xi) = 0$

Understanding If you shoot a ball upward and if the ball comes back to you, then there is a moment the ball stopped in the air.

Proof Since this function is continuous on $[a,b]$ , by Extreme Value Theorem, $f(x)$ attains the maimum value and the minimum value in the interval $[a,b]$ . Let $\xi$ be such that $f(\xi)$ is maximum. Then we have $f(\xi) \geq f(a) = f(b)$ . Thus

for $\displaystyle x > \xi \frac{f(x) - f(\xi)}{x - \xi} \leq 0,$

for $\displaystyle x < \xi \frac{f(x) - f(\xi)}{x - \xi} \geq 0.$

One Side $\footnotesize {f_{+}^{\prime}(\xi) = \displaystyle{\lim_{x \to \xi+}}\frac{f(x) - f(\xi)}{x- \xi}}$
$\footnotesize {f_{-}^{\prime}(\xi) = \displaystyle{\lim_{x \to \xi-}}\frac{f(x) - f(\xi)}{x- \xi}}$

Since $f(x)$ is differentiable, the left-hand side of the above inequalities is $f^{\prime}(\xi)$ and we have.

$\displaystyle f^{\prime}(\xi) \leq 0, f^{\prime}(\xi) \geq 0.$

Note that $f^{\prime}(\xi)$ exists. Thus

$\displaystyle f^{\prime}(\xi) = 0.$

Similarly for the minimum to have $f^{\prime}(\xi) = 0$ $\blacksquare$

Example 2..11 Find the admissible value of the following function.
$f(x) = x^3 - x^2 , [-1,1]$

**Figure 2.6:** Example2-11 Graph of
$% latex2html id marker 42761 \includegraphics[width=3.5cm]{SOFTFIG-2/reidai2-15_gr1.eps}$

SOLUTION $f(-1) = -1 - 1 = -2, f1. = 1 - 1 = 0$ and $f'(x) = 3x^2 - 2x$ . Then

$\displaystyle 3x^2 - 2x = \frac{f1. - f(-1)}{1-(-1)} = \frac{0 - (-2)}{2} = 1.$

Rewriting

$\displaystyle 3x^2 - 2x - 1 = (3x + 1)(x -1) = 0.$

Thus we have $x = -\frac{1}{3}$ , $x = 1$

. But $\xi$ must be in $(-1,1)$

. Therefore, $x = -\frac{1}{3}$ is the admissible value $\blacksquare$

Exercise 2..11 Find the admissible value of the following function.
$f(x) = \sin^{-1}{x} , [0,1]$

SOLUTION Note that $f(0) = \sin^{-1}{0} = 0, f1. = \sin^{-1}{1} = \frac{\pi}{2}$ and $f'(x) = \frac{1}{\sqrt{1 - x^2}}$ . Then we find $x$ satisfying

$\displaystyle \frac{1}{\sqrt{1 - x^2}} = \frac{f1. - f(0)}{1 - 0} = \frac{\pi}{2}$

$x^2 = 1 - \frac{4}{\pi^2}$ implies $x = \pm \sqrt{1 - \frac{4}{\pi^2}}.$ We note that $-\sqrt{1 - \frac{4}{\pi^2}}$ is not in $(0,1)$

$\displaystyle \xi = \sqrt{1 - \frac{4}{\pi^2}}\ensuremath{ \blacksquare}$

To find $x$ satisfying $\frac{1}{\sqrt{1-x^2}} = \frac{\pi}{2}$ , we squared both sides of the equation. Then $\frac{1}{1-x^2} = \frac{\pi^2}{4}$ . Now take the reciprocal. Then $1-x^2 = \frac{4}{\pi^2}$ which implies $x^2 = 1 - \frac{4}{\pi^2}$ .

Idea
Connect two points and by straight line and think of this line as -axis. Then the function takes 0 at and and we can use Rolle's Theorem.

**Figure 2.7:** Mean Value Theorem
$\includegraphics[width=3.5cm]{SOFTFIG-2/mvt.eps}$

Proof of Mean Value Theorem assuming Rolle's Theorem

The idea here is to create the function which satisfies the conditions of Rolle's theorem. The equation $y = P(x)$ of line passing through two points $(a,f(a))$ , $(b,f(b))$ is given by

$\displaystyle y = \frac{f(b) - f(a)}{b - a}(x - a) + f(a) = P(x).$

Now let

be the

. Then

$\displaystyle g(x) = f(x) - \big(\frac{f(b) - f(a)}{b - a}(x - a) + f(a)\big).$

Thus,

and

which satisfy the condition of Rolle's Theorem. Thus by Rolle's Theorem, there exists at least one $\xi \in (a,b)$ such that

$\displaystyle g^{\prime}(\xi) = f^{\prime}(\xi) - \frac{f(b) - f(a)}{b - a} = 0\ensuremath{ \blacksquare}$

Increasing/Decreasing The graph of a function $f(x)$ is said to be increasing .

Increasing/Decreasing Functions If $f(x)$ is defined in the neighborhood of $x=a$ and for $h(>0)$ , $f(x)$ satisfies $f(a-h) < f(a) < f(a+h)$ . Then $f(x)$ is increasing at $x=a$
If $f(x)$ is defined in the neighborhood of $x=a$ and for $h(>0)$ , $f(x)$ satisfies $f(a-h) > f(a) > f(a+h)$ . Then $f(x)$ is decreasing at $x=a$
The neighborhood of $x=a$ is the interval $(a-\delta, a + \delta)$ .

Increasing/Decreasing Functions

Theorem 2..9 Let $f(x)$

be differentiable function at $x=a$

. If $f^{\prime}(a) > 0$ , then $f(x)$

is increasing at $x=a$

. If $f^{\prime}(a) < 0$ , then $f(x)$

is decreasing at $x=a$

Understanding Consider the curve and a point whose slope of tangent line is positive. Then the neighborhood of this point, the curve is concave up

NOTE Consider $f^{\prime}(a) > 0$ .

$\displaystyle \lim_{h \rightarrow 0}\frac{f(a+h) - f(a )}{h} = f^{\prime}(a) > 0$

If $\vert h\vert$ is small enough, then

Check
For , let . Then and becomes . $\displaystyle \frac{f(a+h) - f(a)}{h} > 0$ Thus implies and implies . Therefore, is increasing at .

Check

For

, let

. Then

and

becomes

$\displaystyle \frac{f(a+h) - f(a)}{h} > 0$

Thus

implies

and

implies

. Therefore, $f(x)$

is increasing at $x=a$

Application of Mean Value Theorem

Properties of Differentiable Functions

Theorem 2..10 Let $f(x)$

be continuous on $[a,b]$

and differentiable on $(a,b)$

. Then

1. If $f^{\prime}(x) = 0$ for all $x$ in $(a,b)$ , then $f(x)$ is constant function on $[a,b]$ .

2. If $f^{\prime}(x) \geq 0$ for all $x \in (a,b)$ and there are only finite number of $x$ satisfying $f^{\prime}(x) \neq 0$ , then $f(x)$ is strictly increasing on $[a,b]$ .

Understanding The slope of tangent line becomes 0 only at finite point.

NOTE we need to show for any $x_{1}$ and $x_{2}$ satisfying $a \leq x_{1} < x_{2} \leq b$ , $f(x_1) = f(x_2)$ . Given a closed interval $[a,b]$ , we choose $x_1$ and $x_2$ so that $a \leq x_{1} < x_{2} \leq b$ . Then since $f'(x) = 0$ for all $x$ , for any $\xi$ satisfying $x_{1} < \xi < x_{2}$ , we have $f'(\xi) = 0$ . Thus

$\displaystyle f(x_{2}) - f(x_{1}) = (x_{2} - x_{1})f^{\prime}(\xi) = 0$

and $f(x_{1}) = f(x_{2})$ .

If $f^{\prime}(x) \geq 0$ , then $f(x_{1}) \leq f(x_{2})$ . If $x_{1} < x_{2}$ and $f(x_{1}) = f(x_{2})$ , then by 1), $f(x)$ is constant on $[x_{1},x_{2}]$ and $f^{\prime}(x) \equiv 0$ which violates the condition. Thus,

$\displaystyle x_{1} < x_{2} \Rightarrow f(x_{1}) < f(x_{2}).$

Therefore

is strictly increasing function on $[a,b]$

Same Derivatives

Theorem 2..11 Let $f(x)$

and

be continuous on $[a,b]$

and differentiable on $(a,b)$

. If $f^{\prime}(x) = g^{\prime}(x)$ on $(a,b)$

, then

$\displaystyle f(x) = g(x) + c$ where c is constant

Proof Let $F(x) = f(x) - g(x)$ . Then $F^{\prime}(x) = f^{\prime}(x) - g^{\prime}(x) =0$ implies $F(x)$ is constant. Thus $F(x) = f(x) - g(x) = c$ $\blacksquare$

Example 2..12 Show that the function $\displaystyle{f(x) = x - \sin{x}}$ is strictly increasing on $\displaystyle{[-\frac{\pi}{2},\frac{\pi}{2}]}$

Need to show $f'(x) = 0$ is satisfied by a finite number of $x$ .

SOLUTION Note that since $-1 \leq \cos{x} \leq 1$ , we have $f^{\prime}(x) = 1 - \cos{x} \geq 0$ . Now $f^{\prime}(x) = 0$ implies that $x = 0, \pm \pi, \pm 2\pi, \ldots$ . Then $x = 0$ is the only one which is in $\displaystyle{(\frac{-\pi}{2},\frac{\pi}{2})}$ . Thus $f(x)$ is strictly increasing function on $\displaystyle{[-\frac{\pi}{2},\frac{\pi}{2}]}$ $\blacksquare$

To compare, we can use strictly increasing.

Exercise 2..12 For $x > 0$

, show the following inequality is true .

$\displaystyle 1+x+\frac{x^{2}}{2} < e^{x}$

SOLUTION Let $\displaystyle{f(x) = e^{x} - (1+x+ \frac{x^{2}}{2})}$ . To show If we can show $f^{\prime}(x) > 0$ , then $f(x)$

is strictly increasing. So if $f(0) = 0$

, then for $x > 0$

. Then since $f(0) = 0$

, if we can show $f'(x) > 0$

, then we can show $f(x) > 0$

. So we find $f^{\prime}(x)$ . Since

$\displaystyle f^{\prime}(x) = e^{x} - 1 - x, f'(0) = 0$

we find $f^{\prime\prime}(x)$ . Then

$\displaystyle f^{\prime\prime}(x) = e^{x} - 1$

We know for $x > 0$

, we have $e^{x} > 1$ . Thus $f^{\prime\prime}(x) = e^{x} - 1 > 0$ which implies that $f^{\prime}(x) > 0$ . Therefore $f(x) > 0$

$\blacksquare$

Exercise A

1.

After varifying that the fuction safisfies the conditions of the mean-value theorem on the indicated interval, find the admissible values of $\xi$

(a) $\displaystyle{f(x) = x^{2}, [1,2]}$ (b) $\displaystyle{f(x) = x^{3}, [1,3]}$ (c) $\displaystyle{f(x) = \sqrt{1 - x^{2}}, [0,1]}$

2.

Find the intervals on which $f$

increases and the intervals on which $f$

decreases.

(a) $\displaystyle{f(x) = x^{3} - 3x + 2}$ (b) $\displaystyle{f(x) = x + \frac{1}{x}}$ (c) $\displaystyle{f(x) = x(x+1)(x+2)}$

(d) $\displaystyle{f(x) = \frac{x}{1+x^{2}}}$ (e) $\displaystyle{f(x) = \vert x-1\vert\vert x+2\vert}$

3.

Answer the following question

(a) Find the greatest possible value for $xy$ given that $x$ and $y$ are both positive and $x + y =40$

(b) Find the largest possible area for a rectangle with base on the $x$ -axis and upper vertices on the curve $y = 4 - x^{2}$

(d) Find the shortest distance between the ellipse $x^{2} + 2y^{2} = 2$ and a the line $x+y= 6$

Exercise B

1.: Find the admissible value for $\xi$
(a) $\displaystyle{f(x) = x^{3} - x^{2}, [-1,1]}$ (b) $\displaystyle{f(x) = \sin^{-1}{x}, [0,1]}$ (c) $\displaystyle{f(x) = \log{x}, [1,e]}$
enshu:2-4-2: Show that $f(x) = x - \tan{x}$ is an strictly increasing function on the interval $\displaystyle{(-\frac{\pi}{2},\frac{\pi}{2})}$
3.: Show the following inequalities are true
(a) For , $\displaystyle{\frac{x}{1+x} < \log{(1+x)}}$ (b) For , $\displaystyle{\frac{x}{1+x^{2}} < \tan^{-1}{x} < x}$ (c) $\displaystyle{e^{\pi} > \pi^{e}}$