Is #f(x)= 4sin(4x-(3pi)/8) # increasing or decreasing at #x=pi/12 #?

Question

1519 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-12T20:02:16

The function is increasing

Calculate the first derivative and compute the sign at #x=pi/12#

The function is

#f(x)=4sin(4x-3/8pi)#

The derivative is

#f'(x)=16cos(4x-3/8pi)#

Plugging in the value of #x=pi/12#

#f'(pi/12)=16cos(4*pi/12-3/8pi)#

#=16cos(-9/24pi)#

#=6.12#

As #f'(pi/12) >0#, the function is increasing

graph{(y-4sin(4x-3/8pi))=0 [-4.385, 4.386, -2.19, 2.197]}

Answer 2 · 2018-07-12T20:06:40

#f'(pi/12)=4sqrt(8+2sqrt(6)+2sqrt(2))#

#f(x)=4sin(4x-3*pi/8)=-4cos(4x+1/8*pi)#

then

#f'(x)=16sin(4x+1/8*pi)#

so

#f'(pi/12)=16sin(11/24*pi)=4sqrt(8+2sqrt(6)+2sqrt(2))#
thus the function #f(x)=16sin(4x+1/8pi)# is increasing at #x=pi/12#