Verify the Rolle’s theorem for each of the functions: f(x)= 4-x^2… - Vidyadip

Question

Verify the Rolle’s theorem for each of the functions:
$\text{f(x)}=\sqrt{4-\text{x}^2}\text{ in }[-2,2].$

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Answer

We have, $\text{f(x)}=\sqrt{4-\text{x}^2}=(4-\text{x}^2)^{\frac{1}{2}}$

$\text{f(x)}=\sqrt{4-\text{x}^2}$ is continuous function.

[since every polynomial function is a continuous function]

Hence, f(x) is continuous in [-2, 2]

$\text{f}'(\text{x})=\frac{1}{2}(4-\text{x}^2)^{\frac{-1}{2}}\cdot(-2\text{x})$

$=-\text{x}\cdot\frac{1}{\sqrt{4-\text{x}^2}},$ which exists everywhere except at $\text{x}=\pm2.$

Hence, f(x) is differentiable in (-2, 2).

$\text{f}(-2)=\sqrt{(4-4)}=0$ and $\text{f}(2)=\sqrt{(4-4)}=0$

$\Rightarrow\ \text{f}(-2)=\text{f}(2)$

Conditions of Rolle’s theorem are satisfied.

Hence, there exists a real number c such that f'(c) = 0

$\Rightarrow\ -\text{c}\frac{1}{\sqrt{4-\text{c}^2}}=0$

$\Rightarrow\ \text{c}=0\in(-2,2)$

Hence, Rolle’s theorem has been verified.

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