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

$f^{\prime }(c)=\frac{d}{dx}[f(c)]=\frac{f(b)-f(a)}{b-a}$.

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

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

we have $\frac{f(2)-f(0)}{2-0}$ = $\frac{8}{2}$ $=4$

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.

$\frac{dy}{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: $\frac{f(b)-f(a)}{b-a}$.

so, $\frac{sin(\pi )-sin(0)}{\pi -0}$ = 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!

$\frac{dsin(x)}{dx}$ $=cos(x)$ = 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

$f^{\prime }(k)$	$\approx$	$\frac{1}{n}(f^{\prime }(x_1)+f^{\prime }(x_2)+...+f^{\prime }(x_{n-1})+f^{\prime }(x_n))$
	$\approx$	$\frac{1}{n}\sum _{i=1}^n{f}^{\prime }(x_i)$

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

$f^{\prime }(k)=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\sum _{i=1}^n{f}^{\prime }(x_i)$

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

$f^{\prime }(k)=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\sum _{i=1}^n{f}^{\prime }(x_i)\left(\frac{b-a}{b-a}\right)$

Regroup the factors in a convenient manner

$f^{\prime }(k)=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{b-a}\sum _{i=1}^n{f}^{\prime }(x_i)\left(\frac{b-a}{n}\right)$

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

$f^{\prime }(k)=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{b-a}\sum _{i=1}^n{f}^{\prime }(x_i)\mathrm{\Delta }x$

By definition of the definite integral

$f^{\prime }(k)=\frac{1}{b-a}\underset{a}{\overset{b}{\int }}{f}^{\prime }(x)dx$

And by the Fundamental Theorem of Calculus

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

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

Differentials

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

The "Differential of x" is the $\mathrm{\Delta }x$. 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, $\mathrm{\Delta }x$, $\mathrm{\Delta }y$, and $\frac{dy}{dx}$. 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.

$\frac{dy}{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 $\mathrm{\Delta }x$ 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):

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

If h approaches 0, then:

$f^{\prime }(x)=li{m}_{h\to 0}\frac{f(x+h)-f(x)}{h}$