Calculus/Mean Value Theorem for Functions

Mean Value Theorem

If $f(x) \$ is continuous on the closed interval $[a, b] \$ and differentiable on the open interval $(a,b) \$, there exists a number, $c \$, in the open interval $]a,b[ \$ such that

$f'(c) = \frac{d}{dx} \left[ f(c) \right] = \frac{f(b) - f(a)}{b - a}$.

Examples

What does this mean? As usual, let us utilize an example to grasp the concept. Visualize (or graph) the function $f(x) = x^3$. Choose an interval (anything will work), but for the sake of simplicity, [0,2]. Draw a line going from point (0,0) to (2,8). Between the points x = 0 and x = 2 exists a number x = c, where the derivative of $f$ at point c is equal to the slope of the line you drew.

Solution:

1:Using the definition of the mean value theorem

$f(b) - f(a) \over b - a$ insert values. Our chosen interval is [0,2]. So,

we have $f(2) - f(0) \over 2 -0$ = $8 \over 2$ $= 4$

2: By the definition of the mean value theorem, we know that somewhere in the interval exists a point that has the same slope as that point. Thus, let us take the derivative to find this point x = c.

$dy \over dx$$= 3x^2$. Now, we know that the slope of the point is 4. So, the derivative at this point c is 4. Thus, $4 = 3x^2$. The square root of 4/3 is the point.

Example 2: Find the point that satisifes the mean value theorem on the function$F(x) = sin(x)$ and the interval [0,$pi$]

Solution:

1: Always start with the defintion: $f(b) - f(a) \over b - a$.

so, $sin(\pi) - sin(0) \over \pi - 0$ = 0. (Remember, $sin(\pi)$ and $sin(0)$ are both 0).

2: Now that we have the slope of the line, we must find the point x=c that has the same slope. We must now get the derivative!

$dsin(x) \over dx$ $= cos(x)$ = 0. The Cosine function is 0 at $\pi/2 + \pi n$. Remember, we are bound by the interval [$0, \pi$], so $\pi/2$ is the point c that satisfies the Mean Value Theorem.

To expand on this, the "Mean" in Mean Value Theorem refers to the average rate of change along the interval. The line (it is a "secant line", as it touches in two places, the two endpoints) has a slope that is the average slope value along that curve bounded by the endpoints [a,b].

Proof that f'(c) is actually the average rate of change over the interval [a,b]:

Let f'(k) be the average rate of change over the interval [a,b] for n number of estimations. Then

$f'(k) \,$	$\approx$	$\frac{1}{n} (f'(x_1) + f'(x_2) + ... + f'(x_{n-1}) + f'(x_n))$
	$\approx$	$\frac{1}{n} \sum_{i=1}^n f'(x_i)$

Taking the limit of the expression above yields the precise value for f'(k)

$f'(k) = \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^n f'(x_i)$

Multiply the top and bottom of the summation argument by b-a

$f'(k) = \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^n f'(x_i) \left( \frac{b-a}{b-a} \right)$

Regroup the factors in a convenient manner

$f'(k) = \lim_{n \to \infty} \frac{1}{b-a} \sum_{i=1}^n f'(x_i) \left( \frac{b-a}{n} \right)$

Notice that $\frac{b-a}{n} = \Delta x.$ Then

$f'(k) = \lim_{n \to \infty} \frac{1}{b-a} \sum_{i=1}^n f'(x_i) \Delta x$

By definition of the definite integral

$f'(k) = \frac{1}{b-a} \int_a^b f'(x) dx$

And by the Fundamental Theorem of Calculus

$f'(k) = \frac{1}{b-a} (f(b) - f(a))$ and

$f'(c) = \frac{f(b)-f(a)}{b-a} = f'(k)$

Differentials

Assume a function $y = f(x)$ that is differentiable in the open interval (a,b) that contains x. $\Delta y =$$dy \over dx$ $\cdot \Delta x$

The "Differential of x" is the $\Delta x$. This is an approximate change in x and can be considered "equivalent" to $dx$. The same holds true for y. What is this saying? One can approximate a change in y by knowing a change in x and a change in x at a point very nearby. Let us view an example.

Example: A schoolteacher has asked her students to discover what $4.1^2$ is. The students, bereft of their calculators, are too lazy to multiply this out by hand or in their head and desire to utilize calculus. How can they approximate this?

1: Set up a function that mimics the procedure. What are they doing? They are taking a number (Call it x) and they are squaring it to get a new number (call it y). Thus, y = x^2 Write yourself a small chart. Make notes of values for x, y, $\Delta x$, $\Delta y$, and $dy \over dx$. We are seeking what y really is, but we need the change in y first.

2: Choose a number close by that is easy to work with. Four is very close to 4.1, so write that down as x. Your $\delta x$ is .1 (This is the "change" in x from the approximation point to the point you chose.)

3: Take the derivative of your function.

$dy \over dx$$= 2x$. Now, "split" this up (This is not really what is happening, but to keep things simple, assume you are "multiplying" $dx$ over.)

3b. Now you have $dy$$= 2x \cdot dx$. We are assuming $dy$ and $dx$ are approximately the same as the change in x, thus we can use $\Delta x$ and y.

3c. Insert values: $dy$$= 2 \cdot 4 \cdot .1$. Thus, $dy$ $= .8$.

4: To find $F(4.1)$, take $F(4) + dy$ to get an approximation. 16 + .8 = 16.8; This approximation is nearly exact (The real answer is 16.81. This is only one hundredth off!)

Definition of Derivative

The exact value of the derivative at a point is the rate of change over an infinitely small distance, approaching zero. Therefore, if h approaches 0 and the function is f(x):

$f'(x) = \frac{f(x+h)-f(x)}{(x+h)-x}$

If h approaches 0, then:

$f'(x) = lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$