Answer to Question #173899 in Calculus for Aarati

Question

Find the value of C in the rolle's theorem, where f(x) =cos x and (-pie/2)<C<(pie/2).

Answer 1

We have $f(x)=cosx$ .

Since cosine function is continuous and differentiable everywhere.

Therefore $f(x)$ is continuous on $[-\frac{\pi}{2},\frac{\pi}{2}]$ and differentiable on $(-\frac{\pi}{2},\frac{\pi}{2})$ .

Also , $f(-\frac{\pi}{2})=cos (-\frac{\pi}{2})=0$ and $f(\frac{\pi}{2})=cos (\frac{\pi}{2})=0$

$\therefore f(-\frac{\pi}{2})=f(\frac{\pi}{2})$ .

Thus all the conditions of Rolle's theorem are satisfied.

Then there exists at least one point $C\isin (-\frac{\pi}{2},\frac{\pi}{2})$ for which $f'(C)=0$ .

Now $f'(C)=0\implies-sinC=0$

$\implies C=0$

Hence the value of the $C$ is .