Question

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

Answer 1

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

Since \[\cos x \text { and } e^x\] are everywhere continuous and differentiable, being the quotient 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

\[f\left( x \right) = \frac{\sin x}{e^x}\]

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

\[\therefore f'\left( x \right) = 0\]

\[ \Rightarrow \frac{\cos x - \sin x}{e^x} = 0\]

\[ \Rightarrow \cos x - \sin x = 0\]

\[ \Rightarrow \tan x = 1\]

\[ \Rightarrow x = \frac{\pi}{4}\]

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

​Hence, Rolle's theorem is verified.