Critical points are where $\displaystyle f'(x)=0$ or where $\displaystyle f'(x)$ is undefined. Let's look at $\displaystyle f(x)=x^{2}$

$\displaystyle f'(x)=2x$ and $\displaystyle f'(x)=0$ when $\displaystyle x=0$, so that is a critical point. So now let's take a value to the "left" of 0 (any negative number). Let's just pick $\displaystyle x=-1$ for simplicity. $\displaystyle f'(-1)=-2$, which means that when $\displaystyle x<0$, $\displaystyle x^{2}$ is decrease (you can look at the graph to verify[/tex]. If $\displaystyle f'(x)<0$, that means $\displaystyle f(x)$ is decreasing. If we pick 1, we'll see the opposite.

Remember: If $\displaystyle f'(x)$ is negative on some interval, then $\displaystyle f(x)$ is decreasing on that interval, and the opposite is true. And if $\displaystyle f'(x)=0$, or if $\displaystyle f'(x)$ is undefined at some point, then that point is a critical point, and could be a local extreme.

A simple sign diagram for $\displaystyle f'(x)=2x$

A diagram that may be helpful: