Question

Verify Rolle’s Theorem for the function $f(x)=x^2+2x-8,x∈[-4,2]$

Answer 1

The given function,$f(x)=x^2+2x-8$,being a polynomial function, is continuous in $[−4,2]$ and is differentiable in $(−4,2).$
$f(−4)=(−4)^2+2(−4)-8=16-8-8=0$
$f(2)=(2)^2+2\times2-8=4+4-8=0$
$∴ f(−4)=f(2)=0$
⇒ The value of $f(x)$ at −4 and 2 coincides.
Rolle’s Theorem states that there is a point $c∈(−4,2)$ such that $f'(c)=0$
$f(x)=x^2+2x-8$
$⇒f'(x)=2x+2$
$∴f'(c)=0$
$⇒2c+2=0$
$⇒c=-1$,where $c=-1∈[-4,2]$
Hence, Rolle’s Theorem is verified for the given function.