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!
1 answer