Question

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

Answer 1

Verify conditions of Rolle' s theorem
Given: $f\left(x\right)=x^2+2x-8,x\in [-4,2]$
Rolle's theorem is satisfied if
$\left(i\right)f\left(x\right)$ is continuous in $[a,b];f\left(x\right)=x^2+2x-8$ is a polynomial of degree 'two'. so, $f\left(x\right)$ is continuous in $[-4,2]$

$\left(ii\right)f\left(x\right)$ is differentiable in $(a,b);f\left(x\right)=x^2+2x-8$ is a polynomial of degree 'two'. so,$f\left(x\right)$ is differentiable in $(-4,2)$

$\left(iii\right)f\left(a\right)=f\left(b\right),f\left(x\right)=x^2+2x-8$
$f(-4)=(-4{)}^2+2(-4)-8$
$\Rightarrow f(-4)=16-8-8=16-16=0$
$f\left(2\right)=(2{)}^2+2(2)-8$
$\Rightarrow f\left(2\right)=4+4-8=8-8=0$
Here, $f(-4)=f\left(2\right)$
Hence, Function is satisfying the conditions of Rolle's theorem.

Finding ${}^{\prime }{c}^{\prime }$
Let ${}^{\prime }{c}^{\prime }$ be the any point in the interval $[-4,2],$ such that $f^{\prime }\left(c\right)=0$
$f\left(x\right)=x^2+2x-8$
$f^{\prime }\left(x\right)=2x+2-0$
Putting $x=c,f^{\prime }\left(c\right)=2c+2$
Since all $3$ conditions are satisfied so $f^{\prime }\left(c\right)=0$
$2c+2=0$
$2c=-2$
$c=-1$
Hence $c=-1\in (-4,2)$
Thus, Rolle's theorem is verified.