Question

Given the function $f\left(x\right)=x^3-2x^2-x+1$. Then the value(s) of $c$ satisfying the conditions of the mean value theorem for the function on the interval $[-2,2]$, is

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

Answer 1

The correct option is C $-\frac{2}{3}$

The given function is a cubic polynomial and, hence, it is continuous and differentiable everywhere. Therefore the LMVT is applicable to this function.
$f(-2)=(-2{)}^3-2\cdot (-2{)}^2-(-2)+1$
$f(-2)=-8-8+2+1=-13$
$f\left(2\right)=2^3-2\cdot 2^2-2+1$
$f\left(2\right)=8-8-2+1=-1$
$f^{\prime }\left(x\right)=3x^2-4x-1$

According to LMVT, we have:
$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$
$\Rightarrow 3c^2-4c-1=\frac{-1-(-13)}{2-(-2)}$,
$\Rightarrow 3c^2-4c-1=3$,
$\Rightarrow 3c^2-4c-4=0$
$\Rightarrow (3c+2)(c-2)=0\Rightarrow c=-\frac{2}{3},2$
Only one root $c_1=-\frac{2}{3}$ belongs to the open interval $(-2,2)$