#### Question

Verify Rolle's theorem for the following function on the indicated interval f(x) = 4^{sin}^{ x} on [0, π] ?

#### Solution

The given function is \[f\left( x \right) = 4^{ sin \ x}\].

Since sine function and exponential function are everywhere continuous and differentiable, \[f\left( x \right)\] is continuous on \[\left[ 0, \pi \right]\] and differentiable on \[\left( 0, \pi \right)\] .

Also,

\[f\left( x \right) = 4^{sin \ x } \]

\[ \Rightarrow f'\left( x \right) = 4^{sin x} \left( \cos x \right)\log4\]

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

\[ \Rightarrow 4^{sin x} \left( \cos x \right)\log4 = 0\]

\[ \Rightarrow 4^{ sin x } \cos x = 0\]

\[ \Rightarrow \cos x = 0\]

\[ \Rightarrow x = \frac{\pi}{2}\]

Thus,

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

Hence, Rolle's theorem is verified.