Question

using rolle's theorem
f(x)=sin(5x) [$\frac{\pi }{5}$ , $\frac{2\pi }{5}$}

Answer 1

Step-by-step explanation:

$f(x) = sin(5x)$

• f is both continuous and differentiable in [π/5, 2π/5], since its a simple trigonometric function.

$f( \frac{\pi}{5} ) = sin(5 \times \frac{\pi}{5} ) = sin(\pi) = 0 \\ f( \frac{2\pi}{5} ) = sin(5 \times \frac{2\pi}{5} ) = sin(2\pi) = 0$

Hence:

f(π/5)=f(2π/5)

Rolle's Theorem:

There is at least one ξ in [π/5, 2π/5], such as: f'(ξ)=0