# What value\s of c is (are) predicted by the Mean Value Theorem for f (x) = (x − 2)^3

• Mar 28th 2011, 12:53 AM
bijosn
What value\s of c is (are) predicted by the Mean Value Theorem for f (x) = (x − 2)^3
What value(s) of c is (are) predicted by the Mean Value Theorem for f (x) = (x − 2)^3 on the interval [0, 2]?

this is my attempt but its wrong I think

$\displaystyle \frac{(x-2)^4}{4}\big|_0^2$
$\displaystyle = 0 - \frac{16}{4}$
$\displaystyle = -4$

$\displaystyle f(c)(b-a) = -4$
$\displaystyle (c -2)^3(2 -0) = -4$
$\displaystyle (c -2)^3 = -2$
$\displaystyle (c -2) = (-2)^{\frac{1}{3}}$
$\displaystyle c = (-2)^{\frac{1}{3}} + 2$
• Mar 28th 2011, 12:58 AM
Drexel28
Quote:

Originally Posted by bijosn
What value(s) of c is (are) predicted by the Mean Value Theorem for f (x) = (x − 2)^3 on the interval [0, 2]?

this is my attempt but its wrong I think

$\displaystyle \frac{(x-2)^4}{4}\big|_0^2$
$\displaystyle = 0 - \frac{16}{4}$
$\displaystyle = -4$

$\displaystyle f(c)(b-a) = -4$
$\displaystyle (c -2)^3(2 -0) = -4$
$\displaystyle (c -2)^3 = -2$
$\displaystyle (c -2) = (-2)^{\frac{1}{3}}$
$\displaystyle c = (-2)^{\frac{1}{3}} + 2$

I think you may be remembering the formula wrong. Remember that it's $\displaystyle \displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}$...in our case $\displaystyle f'(c)=3(c-2)^2$
• Mar 28th 2011, 01:41 AM
bijosn
thanks, I am using the mean value theorem formula as it is in the study guide
• Mar 28th 2011, 04:57 AM
HallsofIvy
Then go back and read your study guide again! As Drexel28 said, it is $\displaystyle f'(x)= \frac{f(b)- f(a)}{b- a}$, NOT "$\displaystyle f(x)= \frac{f(b)- f(a)}{b- a}$" that you are using.