Question
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\text{e}^{1-\text{x}^2}\text{ on }[-1,1]$
$\text{f}(\text{x})=\text{e}^{1-\text{x}^2}\text{ on }[-1,1]$
$\text{f}(\text{x})=\text{e}^{1-\text{x}^2}\text{ on }[-1,1]$
We know that, exponantial function is continuous and differentiable everywhere. So, f(x) is continuous IS $[-1,1]$ and differentiable is $(-1,1).$ Now, $\text{f}(-1)=\text{e}^{1-1}=1$ $\text{f}(1)=\text{e}^{1-1}=1$ $\Rightarrow\text{f}(-1)=1$ So, Rolle's theorem is applicable, so there must exist a point $\text{c}\in(-1,1)$ such that f'(c) = 0. Now,$\text{f}(\text{x})=\text{e}^{1-\text{x}^2}$
$\text{f}'(\text{x})=\text{e}^{1-\text{x}^2}(-2\text{x})$ Now, $\text{f}'(\text{c})=0$ $-2\text{ce}^{1-\text{c}^2}=0$ $\Rightarrow\text{c}=0$ or $\text{e}^{1-\text{c}^2}=0$ $\Rightarrow\text{c}=0\in(-1,1)$ Hence, Rolle's theorem is verified.Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.