Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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

Can you answer this question?

1 Answer

0 votes
  • 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 $[-2,2]$
$f'(x)$ exists for all $x\in (-2,2)$
So $f(x)$ is differentiable on $(-2,2)$
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 $f'(c)=0$
$\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 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App