Using rolle's theorem
f(x)=sin(5x) [
"\frac{\pi }{5}" alt="\frac{\pi }{5}" align="absmiddle" class="latex-formula"> ,

}
1 answer:
Step-by-step explanation:

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

Hence:
f(π/5)=f(2π/5)
<u>Rolle's</u><u> </u><u>Theorem</u><u>:</u>
There is at least one ξ in [π/5, 2π/5], such as: f'(ξ)=0
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