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: 
$ x\_1=0 $ could be a maxima, but it could also be a saddle point. To verfiy if it's a saddle point, create the second derivative - if $ f''(x\_1)=0 $ then it's a saddle point, if $ f''(x\_1) \ne 0 $ then it's a maxima.
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.