Question

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

Answer 1

The function is increasing

Explanation:

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

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

Explanation:

`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`