Question

A function $f\left(x\right)=1-x^2+x^3$ is defined in the closed interval $[-1,1]$. The value of $x$, in the open interval $(-1,1)$ for which the mean value theorem is satifsied, is

• A. $-1/2$
• B. $-1/3$
• C. $1/3$
• D. $1/2$

Answer 1

The correct option is B $-1/3$

$f\left(x\right)=1-x^2+x^3,xϵ[-1,1]$
By L.M.V.T.
$f^{\prime }\left(x\right)=\frac{f\left(1\right)-f(-1)}{1-\left(1\right)}=\frac{2}{2}=1$
$-2x+3x^2=1$
$\therefore x=1,\frac{1}{3}$
$x=-\frac{1}{3}ϵ(-1,1)$