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

Aug 1, 2016

$f$ is $\uparrow$ at $x = - \frac{\pi}{6}$.

#### Explanation:

We recall that, $f ' \left(x\right) > 0 \Rightarrow f$ is $\uparrow$ at $x$.

$f \left(x\right) = \frac{x}{\pi} - 2 \sin \left(3 x - \frac{\pi}{4}\right)$

$\Rightarrow f ' \left(x\right) = \frac{1}{\pi} - 2 \left(\cos \left(3 x - \frac{\pi}{4}\right)\right) \cdot 3 = \frac{1}{\pi} - 6 \cos \left(3 x - \frac{\pi}{4}\right)$

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

$= \frac{1}{\pi} - 6 \cos \left(- \frac{\pi}{2} - \frac{\pi}{4}\right)$

$= \frac{1}{\pi} - 6 \cos \left(\frac{\pi}{2} + \frac{\pi}{4}\right) \ldots \ldots \ldots \ldots \left[a s , \cos \left(- \theta\right) = \cos \theta\right]$

$= \frac{1}{\pi} - 6 \left(- \sin \left(\frac{\pi}{4}\right)\right) \ldots \ldots \ldots . . \left[a s , \cos \left(\frac{\pi}{2} + \theta\right) = - \sin \theta\right]$

$= \frac{1}{\pi} + \frac{6}{\sqrt{2}}$, which is, $+ v e$.

Hence, $f$ is $\uparrow$ at $x = - \frac{\pi}{6}$.