# nikki's question at Yahoo! Answers regarding the Mean Value Theorem

#### MarkFL

Staff member
Here is the question:

Calculus 1 Help on Mean Value Theorem?

I don't know how to do this.

Consider the function f(x)=7-8x^2 on the interval [-3,6]. Find the average or mean slope of the function on this interval, i.e. ((f(6)-f(-3))/(6-(-3)))=

By the Mean Value Theorem, we know there exists a c in the open interval (-3,6) such that f'(c) is equal to this mean slope. For this problem, there is only one c that works. Find it.

If someone could help, that would be great! Thanks!
Here is a link to the question:

Calculus 1 Help on Mean Value Theorem? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

#### MarkFL

Staff member
Re: nikki's question at Yahoo! Answers regarding the Mean value Theorem

Hello nikki,

First, let's compute the slope $m$ of the secant line:

$$\displaystyle m=\frac{f(6)-f(-3)}{6-(-3)}=\frac{\left(7-8(6)^2 \right)-\left(7-8(-3)^2 \right)}{6+3}=\frac{8\left(-6^2+3^2 \right)}{9}=\frac{8\cdot9\left(1-2^2 \right)}{9}=8(-3)=-24$$

Now, let's compute the derivative of the given function at $x=c$ and equate this to $m$:

$$\displaystyle f'(c)=m$$

$$\displaystyle -16c=-24$$

$$\displaystyle c=\frac{24}{16}=\frac{3\cdot8}{2\cdot8}=\frac{3}{2}$$

To nikki and any other guest viewing this topic, I invite and encourage you to register and post other calculus problems in our Calculus forum.

Best Regards,

Mark.