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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 2.2: Mean Value Theorem

\includegraphics[width=6cm]{SOFTFIG-2/Fig2-4-1.eps}
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 $f(a) = f(b)$, then there is at least one number $\xi$ in $(a,b)$ such that

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

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. $

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.
$f(x) = x^3 - x^2 ,\ [-1,1]$

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 $(0,1)$

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

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 $g(x)$ be the $f(x) - P(x)$. Then

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

Thus, $g(a) = f(a) - P(a) = f(a) - f(a) = 0$ and $g(b) = f(b) - P(b) = f(b) - f(b) = 0$ 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 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 $x = a$
2.2

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 $x = a$.

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

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

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

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

Thus implies and implies . Therefore, is increasing at . $h > 0$$f(a) < f(a+h)$$h < 0$$f(a+h) < f(a)$$f(x)$$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 $[a,b]$.

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

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

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

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

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

Therefore is strictly increasing function on . $f(x)$$[a,b]$

Same Derivatives

  Let and be continuous on and differentiable on . If on , then
Theorem ..211$f(x)$$g(x)$$[a,b]$$(a,b)$$f^{\prime}(x) = g^{\prime}(x)$$(a,b)$

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

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

Example ..212   Show that the function is strictly increasing on $\displaystyle{f(x) = x - \sin{x}}$$\displaystyle{[-\frac{\pi}{2},\frac{\pi}{2}]}$

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

Exercise ..212   For , show the following inequality is true . $x > 0$

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

Let . Then since , if we can show , then we can show . So we find . Since SOLUTION$\displaystyle{f(x) = e^{x} - (1+x+ \frac{x^{2}}{2})}$$f(0) = 0$$f'(x) > 0$$f(x) > 0$$f^{\prime}(x)$

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

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

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

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

(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 increases and the intervals on which decreases.
$f$$f$

(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?D

Find the greatest possible value for given that and are both positive and ?D

(a)$xy$$x$$y$$x + y =40$

=2.6zw =1 Find the largest possible area for a rectangle with base on the -axis and upper vertices on the curve

(b)$x$$y = 4 - x^{2}$

Find the largest possible area for a rectangle inscibed in a circle of radius 4?D

(c)

Find the shortest distance between the ellipse and a the line ?D

(d)$x^{2} + 2y^{2} = 2$$x+y= 6$

Exercise B


1.
Find the admissible value for ?D
$\xi$

(a)$\displaystyle{f(x) = x^{3} - x^{2}, \ [-1,1]}$(b)$\displaystyle{f(x) = \sin^{-1}{x}, \ [0,1]}$(c)$\displaystyle{f(x) = \log{x}, \ [1,e]}$
2.
Show that is an strictly increasing function on the interval ?D
$f(x) = x - \tan{x}$$\displaystyle{(-\frac{\pi}{2},\frac{\pi}{2})}$
3.
Show the following inequalities are true?D

For , For ,

(a)$x > 0$$\displaystyle{\frac{x}{1+x} < \log{(1+x)}}$(b)$x > 0$$\displaystyle{\frac{x}{1+x^{2}} < \tan^{-1}{x} < x}$(c)$\displaystyle{e^{\pi} > \pi^{e}}$



   
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