Step-by-step explanation:
[tex]f(x) = sin(5x)[/tex]
- f is both continuous and differentiable in [π/5, 2π/5], since its a simple trigonometric function.
[tex]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[/tex]
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
