Answer with Step-by-step explanation:
We are given that
[tex]f(x)=cos2x[/tex]
[[tex]\frac{\pi}{8},\frac{7\pi}{8}[/tex]]
1.Cos2x is continuous on given interval [[tex]\frac{\pi}{8},\frac{7\pi}{8}[/tex]]
2.Cos 2x is differentiable in ([tex]\frac{\pi}{8},\frac{7\pi}{8}[/tex])
3.[tex]f(\frac{\pi}{8})=Cos2(\frac{\pi}{8})=cos\frac{\pi}{4}=\frac{1}{\sqrt 2}[/tex]
[tex]f(\frac{7\pi}{8})=Cos(\frac{7\pi}{4}=Cos(2\pi-\frac{\pi}{4})=Cos\frac{\pi}{4}=\frac{1}{\sqrt 2}[/tex]
Using the formula
[tex]Cos(2\pi -x)=Cos x[/tex]
Therefore, f(a)=f(b)
Hence,Cos 2x satisfies the three hypothesis of Roll's theorem on the given interval.
[tex]f'(x)=-2Sin2 x[/tex]
Substitute x=c
[tex]f'(c)=-2Sin2x[/tex]
[tex]f'(c)=0[/tex]
[tex]-2Sin2x=0[/tex]
[tex]2x=n\pi[/tex]
[tex]x=\frac{n\pi}{2}[/tex]
Where [tex]n\in Z[/tex]
Substitute n=1
[tex]x=\frac{\pi}{2}[/tex] lies in the given interval.
Hence, the value of c=[tex]\frac{\pi}{2}[/tex]
