Calculus/Mean Value Theorem

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.

1: Using the definition of the mean value theorem
 * Solution


 * $$\frac{f(b)-f(a)}{b-a}$$

insert values. Our chosen interval is [0,2]. So, we have


 * $$\frac{f(2)-f(0)}{2-0}=\frac82=4$$

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 $$[0,\pi]$$. 1: Always start with the definition:
 * Solution


 * $$\frac{f(b)-f(a)}{b-a}$$

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!


 * $$\frac{d\sin(x)}{dx}=\cos(x)=0$$

The cosine function is 0 at $$\frac{\pi}{2}+k\pi $$, where $$k$$ is an integer. Remember, we are bound by the interval $$[0,\pi]$$, so $$\frac{\pi}{2}$$ is the point $$c$$ that satisfies the Mean Value Theorem.

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

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 $$x,y,\Delta x,\Delta y,\frac{dy}{dx}$$. 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 $$dy,dx$$ are approximately the same as the change in $$x$$, thus we can use $$\Delta x$$ and $$y$$.

3c. Insert values: $$dy=2\cdot4\cdot0.1$$. Thus, $$dy=0.8$$.

4: To find $$F(4.1)$$, take $$F(4)+dy$$ to get an approximation. 16 + 0.8 = 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 0. 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:
 * $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$