Question

Verify Rolle's theorem for the following function on the indicated interval f(x) = e^x cos x on [−π/2, π/2] ?

Answer 1

The given function is \[f\left( x \right) = e^x \cos x\] .

Since \[\cos x \text { and }  e^x\] are everywhere continuous and differentiable, \[f\left( x \right)\] being a product of these two is continuous on \[\left[ \frac{- \pi}{2}, \frac{\pi}{2} \right]\] and differentiable on \[\left( \frac{- \pi}{2}, \frac{\pi}{2} \right)\] .

Also, 

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

Thus, \[f\left( x \right)\] satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists \[c \in \left( \frac{- \pi}{2}, \frac{\pi}{2} \right)\] such that  \[f'\left( c \right) = 0\] .

We have

\[f\left( x \right) = e^x \cos x\]
\[ \Rightarrow f'\left( x \right) = e^x \left( \cos x - \sin x \right)\]

\[\therefore f'\left( x \right) = 0\]
\[ \Rightarrow e^x \left( \cos x - \sin x \right) = 0\]
\[ \Rightarrow \sin x - \cos x = 0\]
\[ \Rightarrow \tan x = 1\]
\[ \Rightarrow x = \frac{\pi}{4}\]

Since \[c = \frac{\pi}{4} \in \left( \frac{- \pi}{2}, \frac{\pi}{2} \right)\] such that \[f'\left( c \right) = 0\] .

​Hence, Rolle's theorem is verified.