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

Question

Christopher Allers · Accepted Answer

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 role is

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

Consequently,

#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#

Paisley Johnson · Answer

Increasing at #x=\pi/6#

Considering 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#

Scarlett Allison · Answer

undefined To determine if $ f(x) = \cos^2(x) + \sin(2x) $ is increasing or decreasing at $ x = \frac{\pi}{6} $, we need to evaluate the derivative of $ f(x) $ at that point. The derivative of $ f(x) $ is $ f'(x) = -2\sin(x)\cos(x) + 2\cos(2x) $. Evaluating this derivative at $ x = \frac{\pi}{6} $, we get $ f'(\frac{\pi}{6}) = -2\sin(\frac{\pi}{6})\cos(\frac{\pi}{6}) + 2\cos(\frac{\pi}{3}) $. Simplifying this expression, we find $ f'(\frac{\pi}{6}) = -\sqrt{3} + 1 $. Since $ f'(\frac{\pi}{6}) $ is positive, $ f(x) $ is increasing at $ x = \frac{\pi}{6} $.

Ezra Adams · Answer

undefined To determine if the function $ f(x) = \cos^2(x) + \sin(2x) $ is increasing or decreasing at $ x = \frac{\pi}{6} $, we need to examine the derivative of the function at that point.

1. Find the derivative of the function $ f(x) $.
2. Evaluate the derivative at $ x = \frac{\pi}{6} $.
3. Determine the sign of the derivative to ascertain if the function is increasing or decreasing at $ x = \frac{\pi}{6} $.

Let's calculate:

1. Derivative of $ f(x) $:
$$ f'(x) = -2\cos(x)\sin(x) + 2\cos(2x) $$

2. Evaluate $ f'(x) $ at $ x = \frac{\pi}{6} $:
$$ f'\left(\frac{\pi}{6}\right) = -2\cos\left(\frac{\pi}{6}\right)\sin\left(\frac{\pi}{6}\right) + 2\cos\left(\frac{\pi}{3}\right) $$
$$ = -2\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) + 2\left(\frac{1}{2}\right) $$
$$ = -\sqrt{3} + 1 $$

3. Since $ f'\left(\frac{\pi}{6}\right) = -\sqrt{3} + 1 $ is positive, the function $ f(x) $ is increasing at $ x = \frac{\pi}{6} $.