Question

Verify Rolle's theorem for the function $y=x^2+2,a=-2$ and $b=2$.

Answer 1

Verify conditions of Rolle's theorem
$y=x^2+2,a=-2$ and $b=2$
Let $f\left(x\right)=x^2+2.$ Rolle's theorem is satisfied if
$\left(i\right)f\left(x\right)$ is continuous in $[a,b]$
As, $f\left(x\right)=x^2+2$ is a polynomial of degree 'two', $f\left(x\right)$ is continuous in $[-2,2]$

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

$\left(iii\right)f\left(a\right)=f\left(b\right)$,
$\Rightarrow f(-2)=(-2{)}^2+2=6$
$\Rightarrow f\left(2\right)=(2{)}^2+2=6$
Hence, $f\left(2\right)=f(-2)$
Hence, Function is satisfying the conditions of Rolle's theorem.

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