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

Mar 2, 2018

It is increasing.

#### Explanation:

First, we take $f ' \left(x\right)$ and input $\frac{\pi}{12}$. If the answer is more than zero, it is increasing, and if it is less than zero, it is decreasing.

We must find $\frac{d}{\mathrm{dx}} \left(4 \sin \left(4 x - \frac{3 \pi}{4}\right)\right)$

According to the product rule, $\left(f \cdot g\right) ' = f ' g + f g '$

Here, $f = 4$ and $g = \sin \left(4 x - \frac{3 \pi}{4}\right)$

But since $\frac{d}{\mathrm{dx}} 4 = 0$, the product rule reduces to:

$f g '$, or:

$4 \cdot \frac{d}{\mathrm{dx}} \sin \left(4 x - \frac{3 \pi}{4}\right)$

According to the chain rule, $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$, where $u$ is a function within $f$. Here:

$\frac{d}{\mathrm{dx}} \sin \left(4 x - \frac{3 \pi}{4}\right)$

$= \frac{d}{\mathrm{dx}} \sin \left(u\right) \cdot \frac{d}{\mathrm{dx}} \left(4 x - \frac{3 \pi}{4}\right)$

$= \cos \left(u\right) \cdot 4$

And as $u = \left(4 x - \frac{3 \pi}{4}\right)$, we have:

$4 \cos \left(4 x - \frac{3 \pi}{4}\right)$

Applying it into $f g '$:

$4 \cdot 4 \cos \left(4 x - \frac{3 \pi}{4}\right)$

$16 \cos \left(4 x - \frac{3 \pi}{4}\right)$ is our derivative. Inputting $\frac{\pi}{12}$ for $x$:

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

$16 \cos \left(\frac{\pi}{3} - \frac{3 \pi}{4}\right)$

$16 \cos \left(- \frac{5 \pi}{12}\right)$

Since $\cos \left(- x\right) = \cos \left(x\right)$, we have:

$16 \cos \left(\frac{5 \pi}{12}\right)$

$16 \cdot \frac{\sqrt{2 - \sqrt{3}}}{2}$

$8 \sqrt{2 - \sqrt{3}}$

$4.141$

As $4.141 > 0$, $4 \sin \left(4 x - \frac{3 \pi}{4}\right)$ is increasing at $x = \frac{\pi}{12}$

Graphing $4 \sin \left(4 x - \frac{3 \pi}{4}\right)$:

graph{4sin(4x-(3pi)/4) [-3.594, 3.334, -4.502, -1.04]}

We see that this is true.