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

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Answer 1 · 2018-07-18T17:38:53

Increasing

Given function:

#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#

Answer 2 · 2018-07-18T17:50:50

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

At first we not 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#