Answer
$$
f(x) =-\frac{2}{3} x^{3}-\frac{1}{2} x^{2}+3 x-4
$$
A relative minimum occurs at $x=\frac{-3}{2},$ and we find that:
$$
f(\frac{-3}{2}) =-\frac{59}{8} .
$$
A relative maximum occurs at $x=1,$ and we find that:
$$
f(1) =-\frac{13}{6}.
$$
Work Step by Step
$$
f(x) =-\frac{2}{3} x^{3}-\frac{1}{2} x^{2}+3 x-4
$$
To find the intervals where the function is increasing and decreasing.
First, we find $f^{\prime}(x) $,
$$
\begin{aligned}
f^{\prime}(x) &=-\frac{2}{3}(3)x^{2}-\frac{1}{2}(2) x+3 (1)\\
&=-2 x^{2}-x+3 \\
&=-\left(2 x^{2}+x-3\right) \\
&=-(2 x+3)(x-1)
\end{aligned}
$$
To find the critical numbers, we first find any values of $x$ that make $f^{\prime} =0 $. Now, solve the equation $f^{\prime} =0 $ to get
$$
\begin{aligned}
f^{\prime}(x)& =-(2 x+3)(x-1) =0\\
\Rightarrow\quad\quad\quad\quad\quad\\
2 x+3 =0 \quad & \text {or } \quad x-1 =0\\
\Rightarrow\quad\quad\quad\quad\quad\\
x =\frac{-3}{2} \quad & \text {or } \quad x =1\\
\end{aligned}
$$
So, the critical number are $x=1, \quad \frac{-3}{2}$
We can still apply the first derivative test, however, to find where $f$ is increasing and decreasing. Now, check the sign of $f^{\prime}(x)$ in the intervals
$$
(-\infty, \frac{-3}{2} ), \quad (\frac{-3}{2} , 1 ) \quad \text { and} \quad (1 , \infty ) .
$$
(1)
Test a number in the interval $(-\infty, \frac{-3}{2}) $ say $-2 $:
$$
\begin{aligned}
f^{\prime}(-2) &=-(2 (-2)+3)((-2)-1) \\
&=-3\\
& \lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(-\infty, \frac{-3}{2})$.
(2)
Test a number in the interval $(\frac{-3}{2} , 1) $ say $ 0$:
$$
\begin{aligned}
f^{\prime}(0) &=-(2 (0)+3)((0)-1) \\
&=3\\
& \gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(\frac{-3}{2} , 1) $.
(3)
Test a number in the interval $(1, \infty ) $ say $ 2 $:
$$
\begin{aligned}
f^{\prime}(2) &=-(2 (2)+3)((2)-1) \\
&=-7\\
& \lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(1 , \infty )$.
From (1), (2) and (3) we find that:
(*)
The function $f$ is decreasing on intervals $ (-\infty ,\frac{-3}{2} ), $ and is increasing on interval $(\frac{-3}{2}, 1). $ So, a relative minimum occurs at $x=\frac{-3}{2},$ and the value of the function $f$ at this value is given by:
$$
\begin{aligned}
f(\frac{-3}{2}) &=-\frac{2}{3} (\frac{-3}{2})^{3}-\frac{1}{2} (\frac{-3}{2})^{2}+3 (\frac{-3}{2})-4 \\
&=-\frac{59}{8} \\
\end{aligned}
$$
(**)
The function $f$ is increasing on interval $(\frac{-3}{2}, 1), $ and is decreasing on interval $(1, \infty). $ So, a relative maximum occurs at $x=1,$ and the value of the function $f$ at this value is given by:
$$
\begin{aligned}
f(1) &=-\frac{2}{3} (1)^{3}-\frac{1}{2} (1)^{2}+3 (1)-4 \\
&=-\frac{13}{6}\\
\end{aligned}
$$
