Is #f(x)=cot(2x)*tanx# increasing or decreasing at #x=pi/6#?

Question

1537 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-02T17:19:37

#f(x)# is decreasing at #x=pi/6#

Whether a function is increasing or decreasing at a point is decided by its first derivative at that point.

As #f(x)=cot2x*tanx=1/(tan2x) xxtanx#

#=(1-tan^2x)/(2tanx)xx2tanx=(1-tan^2x)/2=1/2-1/2tan^2x#

Hence, #f'(x)=(df)/(dx)=-1/2 xx 2tanx xx sec^2x=-tanxsec^2x#

and #f'(pi/6)=-tan(pi/6)sec^2(pi/6)=-1/sqrt3 xx (2/sqrt3)^2=-4/(3sqrt3)#

As #f'(pi/6) < 0#, #f(x)# is decreasing at #x=pi/6#

Observe that #x=pi/6=0.5236# the function is decreasing in the graph below.

graph{cot(2x)*tanx [-2.5, 2.5, -1.25, 1.25]}