Question
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\text{e}^{1-\text{x}^2}\text{ on }[-1,1]$

Answer

Here,

$\text{f}(\text{x})=\text{e}^{1-\text{x}^2}\text{ on }[-1,1]$

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.

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