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  • Verify the Rolle’s Theorem for the function in <img alt="\begin{bmatrix} -\frac{\pi}{2} ,& \frac{\pi}{2} \end{bmatrix}" sr

Verify the Rolle’s Theorem for the function f(x) = e^ x \cos x in \begin{bmatrix} -\frac{\pi}{2} ,& \frac{\pi}{2} \end{bmatrix}

 

 
 
 
 
 

Answers (1)

Given:

f(x) = e^ x \cos x

(i) f(x) is continuous in \begin{bmatrix} -\frac{\pi}{2} ,& \frac{\pi}{2} \end{bmatrix}

(ii) f^{\prime}(x)=e^{x}(\cos x-\sin x) 

f(x) is differentiable in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

(iii) 

 f\left(\frac{-\pi}{2}\right)=f\left(\frac{\pi}{2}\right)=0

All three conditions of Rolle's Theoremare satisfied. So, there exist c\in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) such that f^{\prime}(c)=0.

f'(c) = e^{c}(\cos c-\sin c)=0

\tan c = 1

c=\frac{\pi}{4} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

Hence proved.

Posted by

Safeer PP

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