Is #f(x)=cot(2x)*tanx^2# increasing or decreasing at #x=pi/3#?

Question

1434 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-14T17:13:46

#f(x)# is decreasing at #x=pi/3#.

#f(x) = cot(2x)*tan(x^2)#

#f'(x) = −2[csc^2(2x)tan(x^2)−xcot(2x)sec^2(x^2)]#

Calculating value of #f'(x)# at #x= pi/3#,

#f'(pi/3) = −2[csc^2((2pi)/3)tan((pi/3)^2)−(pi/3)cot((2pi)/3)sec^2((pi/3)^2)]#

#f'(pi/3) = −2[csc^2((2pi)/3)tan(pi^2/9)−(pi/3)cot((2pi)/3)sec^2(pi^2/9)]#

We know that,
#csc((2pi)/3) = 1.1547#, #tan((2pi)/3)=-1.7320#, #tan(pi^2/9) = 1.948453# and #sec(pi^2/9) = 2.19#

#=>f'(pi/3) = −2[1.1547^2*1.948453−(-pi/(3xx1.7320))*2.19^2]#

#=>f'(pi/3) = −2[1.3333*1.948453+(pi/5.196)*4.7961]#

#=>f'(pi/3) = −2[2.5979+(15.06739/5.196)]#

We can see that the slope #f'(x)# at #x=pi/3# is negative so the function #f(x)# is decreasing at #x=pi/3#.