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

**bijosn** · March 28th 2011, 12:53 AM

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

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

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

**Drexel28** · March 28th 2011, 12:58 AM

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

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

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

I think you may be remembering the formula wrong. Remember that it's $\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}$ ...in our case $f'(c)=3(c-2)^2$

**bijosn** · March 28th 2011, 01:41 AM

thanks, I am using the mean value theorem formula as it is in the study guide

**HallsofIvy** · March 28th 2011, 04:57 AM

Then go back and read your study guide again! As Drexel28 said, it is $f'(x)= \frac{f(b)- f(a)}{b- a}$ , NOT " $f(x)= \frac{f(b)- f(a)}{b- a}$ " that you are using.

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