Is #f(x)= cot(-x+(5pi)/6) # increasing or decreasing at #x=pi/4 #?

Question

Is #f(x)= cot(-x+(5pi)/6)   # increasing or decreasing at #x=pi/4   #?

William Abdalla · Accepted Answer

Increasing

Assigned role: #f(x)=\cot(-x+{5\pi}/6)# Differentiating above equation w.r.t. #x# as follows #d/dxf(x)=d/dx\cot(-x+{5\pi}/6)# #f'(x)=-\cosec^2(-x+{5\pi}/6)d/dx(-x+{5\pi}/6)# #f'(x)=-\cosec^2(-x+{5\pi}/6)(-1)# #f'(x)=\cosec^2(-x+{5\pi}/6)# setting #x=\pi/4# in above equation, we get #f'(\pi/4)=\cosec^2(-\pi/4+{5\pi}/6)# #=\cosec^2({7\pi}/12)# #=\cosec^2 105^\circ# #=(\frac{2\sqrt2}{\sqrt3+1})^2# #=4(2-\sqrt3)>0# Since, #f'(\pi/4)>0# hence the function is increasing at #x=\pi/4#

Charles Arocha · Answer

since we get #f'(pi/4)=2(sqrt(3)-1)^2>0# so is #f(x)# increasing for this Point.

First, we weren't aware of that #cot(-x+5pi/6)=-cot(x+pi/6)# so #-cot(x+pi/6)=csc^2(x+pi/6)# and #f'(pi/4)=2(sqrt(3)-1)^2# Note that #pi/4+pi/6=(3pi+2pi)/12=(5pi)/12#

Samantha Adkison · Answer

undefined To determine whether $ f(x) = \cot\left(-x + \frac{5\pi}{6}\right) $ is increasing or decreasing at $ x = \frac{\pi}{4} $, we need to analyze the sign of its derivative $ f'(x) $ at that point.

First, find the derivative of $ f(x) $ using the chain rule:

$$ f'(x) = \frac{d}{dx} \cot\left(-x + \frac{5\pi}{6}\right) $$
$$ = -\csc^2\left(-x + \frac{5\pi}{6}\right) \cdot \frac{d}{dx} \left(-x + \frac{5\pi}{6}\right) $$
$$ = -\csc^2\left(-x + \frac{5\pi}{6}\right) \cdot (-1) $$

Now, evaluate $ f'(x) $ at $ x = \frac{\pi}{4} $:

$$ f'\left(\frac{\pi}{4}\right) = -\csc^2\left(-\frac{\pi}{4} + \frac{5\pi}{6}\right) \cdot (-1) $$

Since $ \csc^2\left(-\frac{\pi}{4} + \frac{5\pi}{6}\right) $ is positive (since $ \csc $ is positive in the second and third quadrants), $ f'\left(\frac{\pi}{4}\right) $ is negative.

Therefore, $ f(x) $ is decreasing at $ x = \frac{\pi}{4} $.