Question

Verify Rolle's Theorem for the function $f\left(x\right)=x^2+2x-8,x\in \left[-4,2\right]$

Answer 1

Given, $f\left(x\right)=x^2+2x-8,x\in \left[-4,2\right]$

Rolle's Theorem satisfied if

Condition $1$.

$f\left(x\right)=x^2+2x-8$ is continuous at $(-4,2)$

Since, $f\left(x\right)=x^2+2x-8$ is a polynomial

And every polynomial function is continuous for all $x\in R$

$\Rightarrow f\left(x\right)=x^2+2x-8$ is continuous at $x\in \left[-4,2\right]$

Condition $2$.

$f\left(x\right)=x^2+2x-8$ is differentiable at $(-4,2)$

$f\left(x\right)=x^2+2x-8$ is a polynomial

And every polynomial function is Differentiable for all $x\in R$

Therefore, $f\left(x\right)$ is differentiable at $(-4,2)$

Condition $3$.

$f\left(x\right)=x^2+2x-8$

$f\left(-4\right)={\left(-4\right)}^2+2\left(-4\right)-8=16-8-8=16-16=0$

And $f\left(2\right)={\left(2\right)}^2+2\left(2\right)-8=4+4-8=8-8=0$

Hence, $f(-4)=f\left(2\right)$

Now,

$\begin{array}{c}f\left(x\right)=x^2+2x-8\\ f^{\prime }\left(x\right)=2x+2-0\\ f^{\prime }\left(x\right)=2x+2\\ f^{\prime }\left(x\right)=2c+2\end{array}$

Since, all three conditions satisfied

$\begin{array}{c}{f}^{\prime }\left(c\right)=0\\ 2c+2=0\\ 2c=-2\\ c=\frac{-2}{2}\\ =-1\end{array}$

Value of $c$ $=-1\in \left(-4,2\right)$

Thus, Rolle's Theorem is satisfied.