Question

Let $f\left(x\right)=\frac{x-2}{x+2},$ then the value of $c$ that satisfy the mean value theorem for the function on the interval $[-1,2]$ is:

• A. $1.5$
• B. $0$
• C. $2$
• D. $1$

Answer 1

The correct option is B $0$

The function has a discontinuity at $x=-2$, but it is continuous on the interval $[-1,2]$ and differentiable on the interval $(-1,2)$. Hence, the LMVT is applicable on this function on the given interval
$f(-1)=\frac{-1-2}{-1+2}=-3$
$f\left(2\right)=\frac{2-2}{2+2}=0$
$f^{\prime }\left(x\right)={\left(\frac{x-2}{x+2}\right)}^{{}^{\prime }}\Rightarrow f^{\prime }\left(x\right)=\frac{x+2-x+2}{(x+2{)}^2}=\frac{4}{(x+2{)}^2}$

Using LMVT, we have:
$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\Rightarrow \frac{4}{(c+2{)}^2}=\frac{0-(-3)}{2-(-1)}=1$
$\Rightarrow (c+2{)}^2=4\Rightarrow c^2+4c=0$
$\Rightarrow c(c+4)=0$
As we can see only one root $c=0$ lies in the interval $(-1,2)$
$\therefore c=0$