# Is f(x)= 2xsinx  increasing or decreasing at x=pi/3 ?

Dec 6, 2015

Increasing.

#### Explanation:

If $f ' \left(\frac{\pi}{3}\right) < 0$, the function is decreasing when $x = \frac{\pi}{3}$.

If $f ' \left(\frac{\pi}{3}\right) > 0$, the function is increasing when $x = \frac{\pi}{3}$.

Finding $f ' \left(x\right)$ will require the product rule:

$f ' \left(x\right) = \sin x \frac{d}{\mathrm{dx}} \left[2 x\right] + 2 x \frac{d}{\mathrm{dx}} \left[\sin x\right]$

Find each derivative separately.

$\frac{d}{\mathrm{dx}} \left[2 x\right] = 2$

$\frac{d}{\mathrm{dx}} \left[\sin x\right] = \cos x$

Plug them back in.

$f ' \left(x\right) = 2 \sin x + 2 x \cos x$

$f ' \left(\frac{\pi}{3}\right) = 2 \sin \left(\frac{\pi}{3}\right) + 2 \left(\frac{\pi}{3}\right) \cos \left(\frac{\pi}{3}\right)$

While we could continue to simplify this, it is clear that none of these values will be negative.

Thus $f ' \left(\frac{\pi}{3}\right) > 0$ and $f \left(x\right)$ is increasing when $x = \frac{\pi}{3}$.