Question

The given below function f(x) satisfies the hypotheses of the mean value theorem
on the given interval (f(x) is continuous on the closed interval and f(x) is differentiable on the
open interval). Find all numbers c that satisfy the conclusion of that theorem.

f(x) = 3 - (1-x) ^ (2/3) on [1,2]


Answer 1

So you want to find \(c\) such that
\[f'(c)=\frac{f(2)-f(1)}{2-1}\]
What's your derivative?

Answer 2

2/3(1-x)^(-1/3)*[-1] yes?

Answer 3

Wait -2/3(1-x)^(-1/3)*[-1]

Answer 4

Yes, the negatives will cancel so you're left with \(f'(x)=\dfrac{2}{3}(1-x)^{-1/3}\). This means
\[f'(c)=\frac{2}{3}(1-c)^{-1/3}\]
By the MVT, we have
\[\frac{2}{3}(1-c)^{-1/3}=\frac{f(2)-f(1)}{2-1}=\frac{2-3}{1}=-1\]
All that's left is to solve for \(c\).

Answer 5

Dumb question, I'm out of brain power from finals... What do we do about the ^(-1/3) to get rid of it? Raise to ^-3?

Answer 6

Yes, that works. You could also multiply both sides by \((1-c)^{1/3}\) to get
\[\frac{2}{3}=-(1-c)^{1/3}\]
then raise both sides to the third power. Same result in the end.

Answer 7

Thank ya :-)

Answer 8

yw