Using rolle's theorem f(x)=sin(5x) [(\pi )/(5) , (2\pi )/(5)}

Question

Using rolle's theorem
f(x)=sin(5x) [
`(\pi )/(5)` ,
`(2\pi )/(5)`}

Answer 1

Explanation:

`f(x) = sin(5x)`

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

`f( (\pi)/(5) ) = sin(5 * (\pi)/(5) ) = sin(\pi) = 0 \\ f( (2\pi)/(5) ) = sin(5 * (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