# Verify the Rolle's theorem for the function $f(x)=\sqrt {4-x^2}\;in\;[-2,2]$

## 1 Answer

Toolbox:
• If $f[a,b]\rightarrow R$ is continuous on $[a,b]$ and differentiable on $(a,b)$ such that $f(a)=f(b)$,then there exists some c in $(a,b)$ such that $f'(c)=0$
Step 1:
$f(x)=\sqrt{4-x^2}$ in $[-2,2]$
Here $f(x)$ is defined for all $x\in [-2,2]$ and has a unique value for each $x\in [-2,2]$
Hence at every point of $[-2,2]$,the limit of $f(x)$ is equal to the value of the function.
Hence $f(x)$ is continuous on $[-2,2]$
$f'(x)=\large\frac{1}{2}$$(4-x^2)^{\Large\frac{-1}{2}}(-2x) \qquad=\large\frac{-x}{\sqrt{4-x^2}} f'(x) exists for all x\in (-2,2) So f(x) is differentiable on (-2,2) f(-2)=\sqrt{4-4}=0 f(2)=\sqrt{4-4}=0 Therefore f(-2)=f(2)=0 Thus all the three conditions of Rolle's theorem is satisfied. Step 2: Now we will have to see that there exists c\in (-2,2) such that f'(c)=0 f(x)=\sqrt{4-x^2} f'(x)=\large\frac{-x}{\sqrt{4-x^2}} f'(x)=0 \Rightarrow \large\frac{-x}{\sqrt{4-x^2}}$$=0$
$\Rightarrow x=0$
Since $c=0\in (-2,2)$ such that $f'(c)=0$.
Hence Rolle's theorem is verified.
answered Jul 3, 2013

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