$y=x^2+1\; for \;x \in [-2,2]$

Step 1:

y is a polynomial function it is continuous in $[-2,2]$ and differentiable is $(-2,2)$

$y=4+1=5$

when $x=2$

$y=4+1=5$

For $f(x)=y \qquad f(-2)=f(2)=5$

The conditions for Roller's theorem are satisfied.

Step 2:

There is atleast one value $c \;\in\; (-2,2)$

Where $\large\frac{dy}{dx}$$=0$ At this point the slope of the tangent is 0

$\therefore $ it will be $||$ to the x-axis

$\large\frac{dy}{dx}$$=0=>2x=0=>x=0$

When $x=0\qquad y=1$

The slope of the tangent at $(0,1)$ is parallel to the x-axis