# Calculus/Mean Value Theorem

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Mean Value Theorem

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

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

## Examples

What does this mean? As usual, let us utilize an example to grasp the concept. Visualize (or graph) the function ${\displaystyle f(x)=x^{3}}$ . 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 ${\displaystyle x=0}$ and ${\displaystyle x=2}$ exists a number ${\displaystyle x=c}$ , where the derivative of ${\displaystyle f}$ at point ${\displaystyle c}$ is equal to the slope of the line you drew.

Solution

1: Using the definition of the mean value theorem

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

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

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

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 ${\displaystyle x=c}$ .

${\displaystyle {\frac {dy}{dx}}=3x^{2}}$

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

Example 2: Find the point that satisifes the mean value theorem on the function ${\displaystyle f(x)=\sin(x)}$ and the interval ${\displaystyle [0,\pi ]}$ .

Solution

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

so,

${\displaystyle {\frac {\sin(\pi )-\sin(0)}{\pi -0}}=0}$

(Remember, ${\displaystyle \sin(\pi )}$ and ${\displaystyle \sin(0)}$ are both 0.)

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

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

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

## Differentials

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

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

Example: A schoolteacher has asked her students to discover what ${\displaystyle 4.1^{2}}$ 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 ${\displaystyle x}$) and they are squaring it to get a new number (call it ${\displaystyle y}$). Thus, ${\displaystyle y=x^{2}}$ Write yourself a small chart. Make notes of values for ${\displaystyle x,y,\Delta x,\Delta y,{\frac {dy}{dx}}}$ . We are seeking what ${\displaystyle y}$ really is, but we need the change in ${\displaystyle y}$ 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 ${\displaystyle x}$ . Your ${\displaystyle \delta x}$ is .1 (This is the "change" in ${\displaystyle x}$ from the approximation point to the point you chose.)

3: Take the derivative of your function.

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

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

3c. Insert values: ${\displaystyle dy=2\cdot 4\cdot 0.1}$ . Thus, ${\displaystyle dy=0.8}$ .

4: To find ${\displaystyle F(4.1)}$ , take ${\displaystyle F(4)+dy}$ 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 ${\displaystyle f(x)}$ :

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

If h approaches 0, then:

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

## Cauchy's Mean Value Theorem

Cauchy's Mean Value Theorem

If ${\displaystyle f(x),g(x)}$ are continuous on the closed interval ${\displaystyle [a,b]}$ and differentiable on the open interval ${\displaystyle (a,b)}$ , ${\displaystyle g(a)\neq g(b)}$ and ${\displaystyle g'(x)\neq 0}$ , then there exists a number ${\displaystyle c\in (a,b)}$ such that

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