Question

The value of $c$ in $(0, 2)$ satisfying the mean value theorem for the function $f(x) = x(x - 1)^2, x \in [0, 2]$ is equal to

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $\frac{1}{3}$
• D. $\frac{2}{3}$

Answer 1

Answer: (B)

$f\left(x\right) =x\left(x-1\right)^{2} ; x\in\left[0,2\right] $
$f'\left(c\right) = \frac{f\left(b\right)-f\left(a\right)}{b-a} ; f\left(2\right) = 2, f\left(1\right) = 0 $
$f'\left(x\right) = 3x^{2} -4x+1 \Rightarrow f'\left(c\right) = 3c^{2} - 4c+1 $
Thus, $ 3c^{2} -4c+1 = \frac{f\left(2\right) -f\left(0\right)}{2-0} $
$ = \frac{2-0}{2-0} = 1 \Rightarrow c = \frac{4}{3} $