 |
Mean Value Theorem - Help Please?
f(x)= -2x^3+5x^2-4x+3 on closed interval [2,8]
Find the smallest value of c that satifies the conclusion of the Mean Value Theorem for this function defined on the given interval.
Can someone please help me on this problem. I've tried many times and still can't get the answer.
Thank you!
Your question is equivalent to the following: For f(x)=-2x^3+5x^2-4x+3 find the smallest value c in the closed interval [2,8] such that
f'(c)=(f(8)-f(2))/(8-2).
Since
f'(x)=-6x^2+10x-4
f(8)=-733
f(2)=-1
We need to solve the following equation:
-6x^2+10x-4=-122
6x^2-10x-118=0
This has solution(s)
x = (10 +/- sqrt(100+4(6)(118)))/12=(10+/-2sqrt(733))/12=(5+/-sqrt(733))/6
Since only (5+sqrt(733))/6 is in [2,8] we conclude that c=(5+sqrt(733))/6.
(Note that I am writing sqrt for the square root function).
