Determine the turning points of the function $f(x)=3x^2-2x^3$.

First differentiate the function :

The turning points occur where the slopes of the tangents are flat, so the slope has to be $0$.

We found the $x$ coordinates of the turning points, to determine the $y$ coordinates, substitute back into the original function.

So, the turning points are :

In the previous example we've determined the turning points of the function, but this doesn't tell us which is the local minima and which one is the local maxima. To find this out, we need to do the second derivative test. Take the second derivative of the function $f$ :

If $x_1$ is a turning point, $x_1$ is a local minima if $f''(x_1)>0$.

If $x_1$ is a turning point, $x_1$ is a local maxima if $f''(x_2)<0$.

Let's test the point $x=0$ :

$6>0$, so the point $(0,0)$ is a local minima.

Now test the point $x=1$

$-6<0$, so the point $(1,1)$ is a local maxima.