Mean Value Theorem

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 $c\in (a,b)$ such that

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

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 definition:

{\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: $dy=2\cdot 4\cdot 0.1$ . 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\to 0}{\frac {f(x+h)-f(x)}{h}}

Cauchy's Mean Value Theorem

Cauchy's Mean Value Theorem

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

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