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!
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