# High School Calculus/Extrema and the Mean Value Theorem

## Extrema and the Mean Value Theorem

The extrema, or extreme values, of a function are the minimum and/or maximum of a function. They are also known as absolute maximums, or absolute minimums.

### Extreme Value Theorem

If a function $f$ is continuous on a closed interval $[a,b]$ then there exists both a maximum and minimum on the interval.

To find the relative extrema of a function, you first need to calculate the critical values of a function.

Do this by first finding the derivative of a function. Then set the derivative to equal 0, and solve for values of x.

Now, if the function is on a closed interval, $[a,b]$ , then evaluate all the critical points, and endpoints, in your original function.

For absolute minimums/maximums on a closed interval, the highest of your evaluated points is the maximum; the smallest is the minimum.

#### Example

What are the absolute maximums/minimums, of $f(x)=x^{3}$ on the interval $[-2,2]$ ?

To begin, differentiate the function,

$f'(x)=3x^{2}.$ Then set the derivative to 0 and solve for critical points.

$f'(x)=3x^{2}=0$ $x=0.$ Now calculate the original function at your critical points and endpoints,

$f(-2)=-8,f(2)=8,f(0)=0$ Since $f(-2)=-8$ is the lowest, we know that the point, $x=-2$ is the absolute minimum, and 8 is the highest, thus the point, $x=2$ is the absolute maximum.

### Rolle's Theorem

Let $f$ be a continuous function on a closed interval $[a,b]$ , and differentiable on the open interval $(a,b).$ If $f(a)=f(b)$ then there is at least one number $c$ in the interval $(a,b)$ such that $f'(c)=0.$ Essentially, Rolle's theorem tells us that if a function starts and ends at the same point in a closed interval, then there exists some point $c$ where the derivative $f'(c)=0.$ You can informally think of this as at some point, the function must switch from increasing to decreasing, or vice versa, to get back to the y-coordinate where it began.

### Mean Value Theorem

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

$f'(c)={\frac {f(b)-f(a)}{b-a}}$ *Note that the proof of the Mean Value Theorem uses Rolle's Theorem.

The Mean Value Theorem holds a couple different meanings. In Geometry, it tells us that if a secant line is drawn between our starting points, a and b, that there exists a tangent line parallel to the secant line, somewhere on the function.

The Mean Value Theorem also holds that somewhere on the open interval, $(a,b),$ that the instantaneous rate of change, the change at one point, is equal to the average rate of change over the whole interval.

When you evaluate with the Mean Value Theorem, it yields the average, or mean, rate of change over an interval.

#### Example

Determine if the Mean Value Theorem is applicable, and if it is, find all values of c in the open interval $(a,b)$ such that, $f'(c)={\frac {f(b)-f(a)}{b-a}}.$ Let $f(x)=x^{2},[-2,1].$ First, check if the function is continuous on the closed interval. Since we are dealing with $x^{2}$ , we already know it is continuous for all x.

Next, check if the function is differentiable on the open interval, $(-2,1).$ We know our derivative is $f'(x)=2x,$ which exists everywhere on the open interval.

Now that we know both conditions for the theorem have been met, we can apply it to the problem.

Begin by evaluating the function at your endpoints. $f(-2)=4,f(1)=1.$ Next, plug all the information into the theorem, $f'(c)={\frac {1-4}{1-(-2)}}$ $={\frac {-3}{3}}$ $=-1.$ This tells us that the average rate of change over the interval is $-1.$ To find all the values of $c$ , where $f'(c)=-1,$ replace $f'(c)$ with the derivative of $f(x).$ This looks like:

$2c=-1$ $c=-{\frac {1}{2}}$ The answer to the problem is $c=-{\frac {1}{2}}$ #### Practice Problems

Determine if the Mean Value Theorem is applicable, and if it is, find all values of c in the open interval $(a,b)$ such that, $f'(c)={\frac {f(b)-f(a)}{b-a}}.$ $1.f(x)=x^{3}-x^{2}-2x,[-1,1].$ $2.f(x)={\frac {x+1}{x}},[{\frac {1}{2}},2].$ $3.f(x)={\sqrt {2-x}},[-7,2].$ 