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Question
Verify Rolle's theorem for the following function on the indicated interval f (x) = \[\frac{\sin x}{e^x}\] on 0 ≤ x ≤ π ?
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Solution
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 \[\left( 0, \pi \right)\] .
\[f\left( x \right) = \frac{\sin x}{e^x}\]
\[ \Rightarrow f'\left( x \right) = \frac{\cos x - \sin x}{e^x}\]
\[\therefore f'\left( x \right) = 0\]
\[ \Rightarrow \frac{\cos x - \sin x}{e^x} = 0\]
\[ \Rightarrow \cos x - \sin x = 0\]
\[ \Rightarrow \tan x = 1\]
\[ \Rightarrow x = \frac{\pi}{4}\]
Thus,\[c = \frac{\pi}{4} \in \left( 0, \pi \right)\] such that\[f'\left( c \right) = 0\] .
Hence, Rolle's theorem is verified.
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