Is #f(x)= x/pi-2sin(3x-pi/4) # increasing or decreasing at #x=-pi/6 #?

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Answer 1 · 2016-08-01T13:46:22

#f# is #uarr# at #x=-pi/6#.

We recall that, #f'(x) >0 rArr f# is #uarr# at #x#.

#f(x)=x/pi-2sin(3x-pi/4)#

#rArr f'(x)=1/pi-2(cos(3x-pi/4))*3=1/pi-6cos(3x-pi/4)#

#rArr f'(-pi/6)=1/pi-6cos(3*(-pi/6)-pi/4))#

#=1/pi-6cos(-pi/2-pi/4)#

#=1/pi-6cos(pi/2+pi/4)............[as, cos(-theta)=costheta]#

#=1/pi-6(-sin(pi/4))...........[as, cos(pi/2+theta)=-sintheta]#

#=1/pi+6/sqrt2#, which is, #+ve#.

Hence, #f# is #uarr# at #x=-pi/6#.