Question

Verify Rolle's theorem for the function $f\left(x\right)=x^2+x-6$ in the interval $[-3,2]$.

Answer 1

$f\left(x\right)=x^2+x-6,x\in [-3,2]$

Since, $f\left(x\right)=x^2+x-6$ is a polynomial and every polynomial function is continuous for all $x\in R$

So, $f\left(x\right)$ is continuous at $x\in [-3,2]$

Also, every polynomial function is differentiable.

So, $f\left(x\right)$ is differentiable at $(-3,2)$

Now,

$f(-3)=9-3-6=0$

$f\left(2\right)=4+2-6=0$

Hence, $f(-3)=f\left(2\right)=0$

Now,

$f^{\prime }\left(x\right)=2x+1$

$f^{\prime }\left(c\right)=2c+1$

Since, all the conditions are satisfied,

$f^{\prime }\left(c\right)=0$

$2c+1=0$

$c=-\frac{1}{2}$

And, $-\frac{1}{2}\in [-3,2]$

Hence, Rolle's theorem is verified.