Question

Verify Rolle's theorem for the following function on the indicated interval f(x) = e^x sin x on [0, π] ?

Answer 1

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

Since\[\text { sin } x \text { and } e^{x} \] are everywhere continuous and differentiable.

Therefore, being a product of these two, \[f\left( x \right)\]is continuous on \[\left[ 0, \pi \right]\] and differentiable on \[\left( 0, \pi \right)\] .

Also,

\[f\left( \pi \right) = f\left( 0 \right) = 0\]

Thus, 

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

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

We have

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

\[\therefore f'\left( x \right) = 0\]
\[ \Rightarrow e^x \left( \sin x + \cos x \right) = 0\]
\[ \Rightarrow \sin x + \cos x = 0\]
\[ \Rightarrow \tan x = - 1\]
\[ \Rightarrow x = \pi - \frac{\pi}{4} = \frac{3\pi}{4}\]

Since

\[c = \frac{3\pi}{4} \in \left( 0, \pi \right)\] such that 

\[f'\left( c \right) = 0\] .

​Hence, Rolle's theorem is verified.