Respuesta :
Answer:
The number c=[tex]\frac{\pi}{2}[/tex] satisfies conclusion of Roller's theorem.
Step-by-step explanation:
Given function is,
[tex]f(x)=\cos 3x[/tex] which is,
(1) continuous on the closed interval [tex]\Big[\frac{\pi}{12},\frac{7\pi}{12}\Big][/tex]. Since,
[tex]\lim_{x\to \frac{\pi}{12}}f(x)=\frac{1}{\sqrt{2}}=f(\frac{\pi}{12})[/tex] and,
[tex]\lim_limits_{x\to \frac{7\pi}{12}}f(x)=\frac{1}{\sqrt{2}}=f(\frac{7\pi}{12})[/tex]
(2) derivable in the open interval [tex]\Big[\frac{\pi}{12},\frac{7\pi}{12}\Big][/tex] because of continuity.
(3) [tex]f(\frac{\pi}{12})=\frac{1}{\sqrt{2}}=f(\frac{7\pi}{12})[/tex]
Hence all conditions of Rollr's theorem satisfied, so there exist at least one value c, where [tex]\frac{\pi}{12}<c<\frac{7\pi}{12}[/tex] such that,
[tex]f'(c)=0\implies -3\sin 3c=0\implies 3c=n\pi\impliesc=\frac{\pi}{3}n[/tex] where n is an integer.
When,
n=1, c=[tex]\frac{\pi}{3}\in(\frac{\pi}{12},\frac{7\pi}{12})[/tex]
n=2, c=[tex]\frac{2\pi}{3}\notin(\frac{\pi}{12},\frac{7\pi}{12})[/tex]
Similarly for other values of n, c lies outside of the given interval.
Hence c=[tex]\frac{\pi}{2}[/tex] satisfies conclusion of Roller's theorem.
