Question: Use the First Derivative Test to find the points of local maxima and minima of the function 
$ƒ(x)=2x^3−x^4$.

To begin we have $f'(x)=6x^2-4x^3$

Then we can set $f'(x)=6x^2-4x^3=0$ and factor it as $2x^2(3-2x)=0$

The we get $\frac{3}{2}$ as a critical point and maxima. Is $x=0$ also a critical point?

Sorry if this is an easy question I just keep having problems with this in class.

Now I have this chart but it is saying I am wrong

Function $2x^3−x^4$ 

------------------Intervals

Properties: 

```
                  $(−∞,1.5)---- 1.5---(1.5,∞)

```

Signs of $f(x)$ >>>>>>>>> **+** >>>>>>>>>>>>>> **-**

Increase/decrease of $f(x)$ increasing >>local max>>decreasing

Answer: 
If you insist on using the First Derivative Test, what you need to do is check whether your function is increasing or decreasing at points close to your critical points, $\frac{3}{2}$ and $0$. That is, if $f'(1) > 0$ and $f'(2) < 0$, you can conclude that you have a local maximum at $x = \frac{3}{2}$. You're looking for a change from increasing to decreasing or decreasing to increasing to show something is a local maximum/local minimum. You do something similar for points around $x = 0$. 

Also, critical points come from three places:

1. Endpoints of an interval are always critical points (If you have a given interval).
2. Places for which your derivative function is undefined (Your derivative function is defined everywhere).
3. The $x$-values where $f'(x) = 0$ (Both $\frac{3}{2}$ and $0$ do this ).
The first derivative is positive before $x=3/2$ and negative after $x=3/2$. Since it vanishes at $x=3/2$, it must be a (local) maximum point.