Question

Verify Rolle's Theorem for
$f(x) = \frac{ \sin(x) }{ {e}^{x} } \: in \: (0\pi)$
​

Answer 1

Given function,

$f(x) = \dfrac{\sin(x) }{ {e}^{x}} , \: x \in (0 , \pi)$

Firstly, let's verify if f(0) = f(π).

We know that, sin(0) = sin(π) = 0.

Thus, f(0) = f(π).

Also, both e^x and sin(x) are differentiable and continuous functions on (0,π).

Assume a point 'c' in the interval (0,π).

Here,

$f'(x) = {e}^{ - x} \cos(x) - {e}^{ - x} \sin(x)$

At point c,

$\implies f'(c) = {e}^{ - c} \cos(c) - {e}^{ - c} \sin(c) \\ \\ \implies {e}^{ - c} \cos(c) - {e}^{ - c} \sin(c) = 0 \\ \\ \implies {e}^{ - c} \cos(c) = {e}^{ - c} \sin(c) \\ \\ \implies \tan(c) = 1$

Generally, tangent function is 1 at nπ + π/4 but between (0,π), tangent gives one at π/4.

Thus,

$\implies \boxed{ \boxed{ c = \dfrac{\pi}{4} }}$