Answer
$$
f(x)=x^{3}-\frac{15}{2} x^{2}-18 x-1
$$
Domain is $(-\infty, \infty)$
$$
\begin{aligned}
f(-x) & \ne f(x)
\end{aligned}
$$
No symmetry
The critical numbers are: $x = 6, \quad x =-1$.
Critical points:
$$ (6, -163) \ \ \text{and } \ \ (-1, \frac{17}{2}). $$
The function $f$ is increasing on intervals $ ( -\infty , -1) $ and $ (6 , \infty ) , $and decreasing on interval $ ( -1 , 6) . $
Relative maximum at -1, relative minimum at 6
Point of inflection at $( \frac{5}{2} , -77.25).$
*
$f(x)$ is concave downward on $(-\infty, \frac{5}{2} )$.
**
$f(x)$ is concave upward on $(\frac{5}{2}, \infty )$.
y-intercept:
$$
f(0)=-1
$$
Work Step by Step
$$
f(x)=x^{3}-\frac{15}{2} x^{2}-18 x-1
$$
Domain is $(-\infty, \infty)$
$$
\begin{aligned}
f(-x) &=(-x)^{3}-\frac{15}{2}\left(-x\right)^{2}-18(-x)-1 \\
&=-x^{3}-\frac{15}{2} x^{2}+18 x-1\\
& \ne f(x)
\end{aligned}
$$
No symmetry
To find the intervals where the function is increasing and decreasing.
First, we find $f^{\prime}(x) $,
$$
\begin{aligned}
f^{\prime}(x) &=3 x^{2}-15 x-18\\
&=3\left(x^{2}-5 x-6\right)\\
&=3(x-6)(x+1)
\end{aligned}
$$
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
$$
\begin{aligned}
f^{\prime}(x) =3(x-6)(x+1) &=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
(x-6)(x+1) &=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
x-6 =0 & \quad \quad or \quad \quad x+1 =0 \\
x =6 & \quad \quad or \quad \quad x =-1 \\
\end{aligned}
$$
So,
the critical numbers are: $x = 6, \quad x =-1$.
To find the critical points substituting in $f$ as follows:
$$
f(6)=(6)^{3}-\frac{15}{2} (6)^{2}-18 (6)-1=-163
$$
and
$$
f(-1)=(-1)^{3}-\frac{15}{2} (-1)^{2}-18 (-1)-1=\frac{17}{2}
$$
So, Critical points:
$$ (6, -163) \ \ \text{and } \ \ (-1, \frac{17}{2}). $$
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
$$
(-\infty, -1 ), \quad\quad (-1 , 6 ), \quad\quad (6 , \infty) .
$$
(1)
Test a number in the interval $(-\infty, -1 )$ say $-2$:
$$
\begin{aligned}
f^{\prime}(-2) &=3((-2)-6)((-2)+1)\\
&=24 \\
&\gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(-\infty , -1 )$.
(2)
Test a number in the interval $(-1, 6 )$ say $0$:
$$
\begin{aligned}
f^{\prime}(0) &=3((0)-6)((0)+1)\\
&=-18 \\
& \lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(-1, 6 )$.
(3)
Test a number in the interval $(6, \infty )$ say $7$:
$$
\begin{aligned}
f^{\prime}(7) &=3((7)-6)((7)+1)\\
&=24 \\
&\gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(6 , \infty )$.
So, the function $f$ is increasing on intervals $ ( -\infty , -1) $ and $ (6 , \infty ) , $and decreasing on interval $ ( -1 , 6) . $
Relative maximum at -1, relative minimum at 6
Now, the second derivative is
$$
\begin{aligned}
f^{\prime \prime}(x) &=6 x-15 \\
\end{aligned}
$$
Set $f^{\prime\prime}(x)$ equal to $0$ and solve for $x$.
$$
\begin{aligned}
f^{\prime \prime}(x) &=6 x-15 \\
&= 3(2 x-5)=0 \\
x &=\frac{5}{2}
\end{aligned}
$$
so $(\frac{5}{2} , f(\frac{5}{2}))$ is inflection point.
Now ,we find $f(\frac{5}{2})$ as follows:
$$
f(\frac{5}{2})=(\frac{5}{2})^{3}-\frac{15}{2} (\frac{5}{2})^{2}-18 (\frac{5}{2})-1 =-\frac{309}{4}=-77.25
$$
So, point of inflection at $$ ( \frac{5}{2} , -77.25).$$
*
Test a number in the interval $(-\infty, \frac{5}{2} )$ to see that $f^{\prime\prime}(x)$ is negative in that interval, so $f(x)$ is concave downward on $(-\infty, \frac{5}{2} )$.
**
Test a number in the interval $(\frac{5}{2}, \infty )$to find $f^{\prime\prime}(x)$ positive in that interval, so $f(x)$ is concave upward on $(\frac{5}{2}, \infty )$.
y-intercept:
$$
f(0)=(0)^{3}-\frac{15}{2} (0)^{2}-18 (0)-1=-1
$$
