Determine if $x=0$ is a point of relative extremum for $f(x)= \sin(x) + \frac{x^3}{6}$ 
Determine if $x=0$ is a point of relative extremum for $f(x)= \sin(x)
+ \frac{x^3}{6}$

I am trying to use this test 
Here,
$f(x)= \sin(x)
+ \frac{x^3}{6}$
$f'(x)=\cos(x) + \frac{x^2}{2} \Rightarrow f'(0)=1 \neq 0$
So I am unable to proceed further. 

Since the first derivative is positive, the function increases at that point.

