Answer
Relative maxima $x=\frac{3}{2}$
Relative minima $x=3, x=0$
Work Step by Step
When $x \in(0,3) ; f(x)=3 x-x^{2} $
When $x \in(-\infty, 0) \cup(3, \infty) ; f(x)=-3 x+x^{2}$
Let $f_{1}(x)=-x^{2}+3 x, f_{2}(x)=x^{2}-3 x$
$f_{1}^{\prime}(x)=3-2 x; f_{2}^{\prime}(x)=-3+2 x$
So in $(0,3), 3 x-x^{2}=0$ at $x=\frac{3}{2}$ $\operatorname{In}(-\infty, 0) \cup(3, \infty), 2 x-3 \neq 0$
When we combine these results and generalize for all the values of x , we have that the stationary points of the function are $\frac{3}{2}=x$
The critical points also include the value of $x$ where the function is not differentiable.
Hence, these values are $\mathrm{x}=0, \mathrm{x}=3$
Using the first derivative test, we know that:
(a) If $f(x)>0$ on an open interval extending left from $x_{0}$ and $f(x)<0$ on an open interval extending right from $x_{0},$ then $f$ has a relative maximum at
$x_{0}$
(b) If $f(x)<0$ on an open interval extending left from $x_{0}$ and $f(x)>0$ on an open interval extending right from $x_{0},$ then $f$ has a relative maximum at
$x_{0}$
So by this test, $x=3$ and $x=0$ are values of minima.