Question

Verify Rolle's theorem for the function $f\left(x\right)=x^2+2,xϵ[-2,2]$.

Answer 1

Given $f\left(x\right)=x^2+2$ is continuous in $[-2,2]$ and differentiable in $(-2,2)$
Also $f\left(a\right)=f(-2)=(-2{)}^2+2=6$
$f\left(b\right)=f\left(2\right)=2^2+2=6$
$\therefore f\left(a\right)=f\left(b\right)=6$
$\therefore$ Value of $f\left(x\right)$ at $x=-2$ and $2$ coincide
$\therefore$ According to Rolle's theorem $\exists cϵ(a,b)$ such that
$f^1\left(c\right)=0$
$\therefore f^1\left(x\right)=2x$
$\Rightarrow f^1\left(c\right)=2c=0$
$\therefore c=0ϵ(-2,2)$.