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

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

Figure 2.5: Mean Value Theorem

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 $${\frac{f(b) - f(a)}{b - a}}$$ can be thought of the slope of line passing through two points $(a,f(a)),(b,f(b))$. Then $${\frac{f(b) - f(a)}{b - a} = f^{\prime}(\xi)}$$ 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, $${f(b) - f(a)}$$ 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

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

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

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

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.

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

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

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

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 $y = x^3 - x^2$

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

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

Rewriting

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

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

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

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

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

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 $A$ and $B$ by straight line and think of this line as $x$-axis. Then the function takes 0 at $A$ and $B$ and we can use Rolle's Theorem.

Figure 2.7: Mean Value Theorem

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

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

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

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

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

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

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

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

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

Check

For $h < 0$, let $h = -k$. Then $k > 0$ and $f(a+h) < f(a)$ becomes $f(a - k) < f(a)$.

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

Thus $h > 0$ implies $f(a) < f(a+h)$ and $h < 0$ implies $f(a+h) < f(a)$. 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

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

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,

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

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

Same Derivatives

Theorem 2..11   Let $f(x)$ and $g(x)$ be continuous on $[a,b]$ and differentiable on $(a,b)$. If $f^{\prime}(x) = g^{\prime}(x)$ on $(a,b)$, then

$$f(x) = g(x) + c$$   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 $${f(x) = x - \sin{x}}$$ is strictly increasing on $${[-\frac{\pi}{2},\frac{\pi}{2}]}$$

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 $${(\frac{-\pi}{2},\frac{\pi}{2})}$$. Thus $f(x)$ is strictly increasing function on $${[-\frac{\pi}{2},\frac{\pi}{2}]}$$ $\blacksquare$

To compare, we can use strictly increasing.

Exercise 2..12   For $x > 0$, show the following inequality is true .

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

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

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

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

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

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

2.

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

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

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

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 $y = 4 - x^{2}$

(c) Find the largest possible area for a rectangle inscibed in a circle of radius 4

(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) $${f(x) = x^{3} - x^{2}, [-1,1]}$$ (b) $${f(x) = \sin^{-1}{x}, [0,1]}$$ (c) $${f(x) = \log{x}, [1,e]}$$

enshu:2-4-2

Show that $f(x) = x - \tan{x}$ is an strictly increasing function on the interval $${(-\frac{\pi}{2},\frac{\pi}{2})}$$

3.

Show the following inequalities are true

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