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!

You're welcome! It's important to set the equation equal to zero when solving for the values of "c" using the mean value theorem. Remember, the mean slope is equal to the derivative of the function at a specific point, so setting the derivative equal to the mean slope will help you find the values of "c". If you have any more questions, feel free to ask!