# Calc, Mean Value Theorem

posted by
**Michael**
.

Consider the function :

3x^3 - 2x^2 - 4x + 1

Find the average slope of this function on the interval. By the Mean Value Theorem, we know there exists a "c" in the open interval (-2,3) such that f'(c) is equal to this mean slope. Find the two values of "c" in the interval which work.

The average is 15, i know that's right.

f(b) - f(a)

-------------

b - a

not sure how to get the second part of the question tho.

You just need to calculate the derivative of the function. It is:

f'(x)= 6x^2 - 4x - 4

You then solve the equation

f'(c) = 15 --->

6c^2 - 4c - 4 = 15

wouldn't the derivative be:

f'(x)= 9x^2 - 4x - 4 ?

when i set that equal to 15, then i would get 7, and 19/3, both of which are not in the interval.

Yes, you are right!

Yes, you are right!

Let me see:

9x^2 - 4x - 4 = 15 -->

9x^2 - 4x - 19 = 0 --->

x = 4/18 +- 1/18*Sqrt[4^2 + 4*9*19]

So the solutions are x = 1.69 and

x = -1.247 which are inside the interval

Oh i c what i did wrong, i wasn't setting equation to zero. Thanks!