Answer
$$
f(x)=-3 x^{3}+6 x^{2}-4 x-1
$$
Domain is $(-\infty, \infty)$
$$
\begin{aligned}
f(-x) & \ne f(x)
\end{aligned}
$$
No symmetry
The only critical number is: $x = \frac{2}{3}$.
Critical point:
$$ (\frac{2}{3}, -1.8) . $$
The function $f$ is decreasing on interval $ ( -\infty , \infty) . $
No relative extremum at $ (\frac{2}{3}, -1.8) . $
Point of inflection at $$ ( \frac{2}{3} , -1.8).$$
*
$f(x)$ is concave upward on $(-\infty, \frac{2}{3} )$.
**
$f(x)$ is concave upward on $(\frac{2}{3}, \infty )$.
y-intercept:
$$
f(0)=-1
$$
Work Step by Step
$$
f(x)=-3 x^{3}+6 x^{2}-4 x-1
$$
Domain is $(-\infty, \infty)$
$$
\begin{aligned}
f(-x) &=-3(-x)^{3}+6(-x)^{2}-4(-x)-1 \\
&=3 x^{3}+6 x^{2}+4 x-1\\
& \ne f(x)
\end{aligned}
$$
No symmetry
To find the intervals where the function is increasing and decreasing.
First, we find $f^{\prime}(x) $,
$$
\begin{aligned}
f^{\prime}(x) &=-9 x^{2}+12 x-4\\
\end{aligned}
$$
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
$$
\begin{aligned}
f^{\prime}(x) & =-9 x^{2}+12 x-4=0\\
& =-(3 x-2)^{2}=0 \\
&= (3 x-2)^{2} =0 \\
\Rightarrow\quad\quad\quad\quad\quad\\
x &=\frac{2}{3}\\
\end{aligned}
$$
So,
the only critical number is: $x = \frac{2}{3}$.
To find the critical points substituting in $f$ as follows:
$$
f(\frac{2}{3})=-3 (\frac{2}{3})^{3}+6(\frac{2}{3})^{2}-4 (\frac{2}{3})-1=-\frac{17}{9}\approx -1.8
$$
So, Critical point:
$$ (\frac{2}{3}, -1.8) . $$
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
$$
(-\infty, \frac{2}{3} ), \quad\quad (\frac{2}{3} , \infty ).
$$
(1)
Test a number in the interval $(-\infty, \frac{2}{3} )$ say $0$:
$$
\begin{aligned}
f^{\prime}(0) &=-9 (0)^{2}+12 (0)-4 \\
&=-4 \\
&\lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(-\infty, \frac{2}{3} )$.
(2)
Test a number in the interval $(\frac{2}{3}, \infty )$ say $1$:
$$
\begin{aligned}
f^{\prime}(1) &=-9 (1)^{2}+12 (1)-4 \\
&=-1 \\
& \lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(\frac{2}{3}, \infty )$.
So, the function $f$ is decreasing on interval $ ( -\infty , \infty) . $
No relative extremum at $ (\frac{2}{3}, -1.8) . $
Now, the second derivative is
$$
\begin{aligned}
f^{\prime \prime}(x) &=-18 x+12 \\
&=-6(3 x-2) \\
\end{aligned}
$$
Set $f^{\prime\prime}(x)$ equal to $0$ and solve for $x$.
$$
\begin{aligned}
f^{\prime \prime}(x) &=-18 x+12=0 \\
&=-6(3 x-2)=0 \\
& 3 x-2 =0 \\
\Rightarrow\quad\quad\quad\quad\quad\\
x &=\frac{2}{3}
\end{aligned}
$$
so $(\frac{2}{3} , f(\frac{2}{3}))$ is inflection point.
Now ,we find $f(\frac{2}{3})$ as follows:
$$
f(\frac{2}{3})=-3 (\frac{2}{3})^{3}+6 (\frac{2}{3})^{2}-4 (\frac{2}{3})-1 =-\frac{17}{9} \approx -1.8
$$
So, point of inflection at $$ ( \frac{2}{3} , -1.8).$$
*
$$
\begin{array}{l}
f^{\prime \prime}(0)=-18(0)+12=12>0 \\
f^{\prime \prime}(1)=-18(1)+12=-6<0
\end{array}
$$
Test a number in the interval $(-\infty, \frac{2}{3} )$ to see that $f^{\prime\prime}(x)$ is positive in that interval, so $f(x)$ is concave upward on $(-\infty, \frac{2}{3} )$.
**
Test a number in the interval $(\frac{2}{3}, \infty )$to find $f^{\prime\prime}(x)$ negative in that interval, so $f(x)$ is concave upward on $(\frac{2}{3}, \infty )$.
y-intercept:
$$
f(0)=-3 (0)^{3}+6 (0)^{2}-4 (0)-1=-1
$$
