Is #f(x)=cos^2x+sin2x# increasing or decreasing at #x=pi/6#?

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Answer 1 · 2018-07-02T13:46:37

Increasing at #x=\pi/6#

Given that

#f(x)=\cos^2x+\sin2x#

#f'(x)=2\cosx(-\sin x)+2\cos2x#

#f'(x)=-\sin2x+2\cos2x#

Now,

#f'(\pi/6)=-\sin(2\cdot \pi/6)+2\cos(2\cdot \pi/6)#

#=-\sin(\pi/3)+2\cos(\pi/3)#

#=-\sqrt3/2+2\cdot1/2#

#=\frac{2-\sqrt3}{2}>0#

Since, #f'(\pi/6)>0# hence the function #f(x)# is increasing at #x=\pi/6#

Answer 2 · 2018-07-02T13:56:53

The function is increasing at #x=pi/6#

Calculate the first derivative #f'(x)# and look at the sign of #f'(pi/6)#.

If #f'(pi/6)>0#, the function is increasing

and

If #f'(pi/6)<0#, the function is decreasing

The function is

#f(x)=cos^2x+sin2x#

Therefore,

#f'(x)=-2cosxsinx+2cos2x=-sin2x+2cos2x#

And

#f'(pi/6)=-sin(2*pi/6)+2cos(2*pi/6)#

#=-sqrt3/2+2*1/2#

#=1-sqrt3/2#

#=0.13#

#f'(pi/6)>0#

Therefore, the function is increasing at #x=pi/6#