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

Jul 12, 2018

The function is increasing

#### Explanation:

Calculate the first derivative and compute the sign at $x = \frac{\pi}{12}$

The function is

$f \left(x\right) = 4 \sin \left(4 x - \frac{3}{8} \pi\right)$

The derivative is

$f ' \left(x\right) = 16 \cos \left(4 x - \frac{3}{8} \pi\right)$

Plugging in the value of $x = \frac{\pi}{12}$

$f ' \left(\frac{\pi}{12}\right) = 16 \cos \left(4 \cdot \frac{\pi}{12} - \frac{3}{8} \pi\right)$

$= 16 \cos \left(- \frac{9}{24} \pi\right)$

$= 6.12$

As $f ' \left(\frac{\pi}{12}\right) > 0$, the function is increasing

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

Jul 12, 2018

$f ' \left(\frac{\pi}{12}\right) = 4 \sqrt{8 + 2 \sqrt{6} + 2 \sqrt{2}}$

#### Explanation:

$f \left(x\right) = 4 \sin \left(4 x - 3 \cdot \frac{\pi}{8}\right) = - 4 \cos \left(4 x + \frac{1}{8} \cdot \pi\right)$

then

$f ' \left(x\right) = 16 \sin \left(4 x + \frac{1}{8} \cdot \pi\right)$

so

$f ' \left(\frac{\pi}{12}\right) = 16 \sin \left(\frac{11}{24} \cdot \pi\right) = 4 \sqrt{8 + 2 \sqrt{6} + 2 \sqrt{2}}$
thus the function $f \left(x\right) = 16 \sin \left(4 x + \frac{1}{8} \pi\right)$ is increasing at $x = \frac{\pi}{12}$