Answer
$$
f(x)=-2 x^{3}-9 x^{2}+108 x-10
$$
Domain is $(-\infty, \infty)$
$$
\begin{aligned}
f(-x) &\ne f(x)
\end{aligned}
$$
No symmetry
the critical numbers are: $x =-6, \quad x =3$.
Critical points: (-6, -550) and (3, 179).
The function $f$ is decreasing on intervals $ ( -\infty , -6) $ and $ (3, \infty ) , $and increasing on interval $ ( -6 , 3) . $
Relative maximum at 3, relative minimum at 6
point of inflection at ( -1.5, 185.5).
$f(x)$ is concave upward on $(-\infty,-\frac{3}{2} )$.
$f(x)$ is concave downward on $(-\frac{3}{2}, \infty )$.
y-intercept:
$$
f(0)=-2 (0)^{3}-9 (0)^{2}+108 (0)-10 =-10
$$
Work Step by Step
$$
f(x)=-2 x^{3}-9 x^{2}+108 x-10
$$
Domain is $(-\infty, \infty)$
$$
\begin{aligned}
f(-x) &=-2(-x)^{3}-9(-x)^{2}+108(-x)-10 \\
&=2 x^{3}-9 x^{2}-108 x-10\\
&\ne f(x)
\end{aligned}
$$
No symmetry
To find the intervals where the function is increasing and decreasing.
First, we find $f^{\prime}(x) $,
$$
\begin{aligned}
f^{\prime}(x) &=-6 x^{2}-18 x+108 \\
&=-6\left(x^{2}+3 x-18\right) \\
&=-6(x+6)(x-3)
\end{aligned}
$$
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
$$
\begin{aligned}
f^{\prime}(x) =-6(x+6)(x-3) &=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
(x+6)(x-3) &=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
x+6 =0 & \quad \quad or \quad \quad x-3 =0 \\
x =-6 & \quad \quad or \quad \quad x =3 \\
\end{aligned}
$$
So,
the critical numbers are: $x =-6, \quad x =3$.
To find the critical points substituting in $f$ as follows:
$$
f(-6)=-2 (-6)^{3}-9 (-6)^{2}+108 (-6)-10=-550
$$
and
$$
f(3)=-2 (3)^{3}-9 (3)^{2}+108 (3)-10=179
$$
So, Critical points: (-6, -550) and (3, 179).
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
$$
(-\infty, -6 ), \quad\quad (-6 ,3 ), \quad\quad (3 , \infty) .
$$
(1)
Test a number in the interval $(-\infty, -6 )$ say $-7$:
$$
\begin{aligned}
f^{\prime}(-7) &=-6((-7)+6)((-7)-3)\\
&=-60 \\
&\lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(-\infty ,-6 )$.
(2)
Test a number in the interval $(-6, 3 )$ say $0$:
$$
\begin{aligned}
f^{\prime}(0) &=-6((0)+6)((0)-3)\\
&=108 \\
&\gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(-6, 3 )$.
(3)
Test a number in the interval $(3, \infty )$ say $4$:
$$
\begin{aligned}
f^{\prime}(4) &=-6((4)+6)((4)-3)\\
&=-60 \\
&\lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(3 , -\infty )$.
So, the function $f$ is decreasing on intervals $ ( -\infty , -6) $ and $ (3, \infty ) , $and increasing on interval $ ( -6 , 3) . $
Relative maximum at 3, relative minimum at 6
Now, the second derivative is
$$
\begin{aligned}
f^{\prime \prime}(x) &=-12 x-18 \\
\end{aligned}
$$
Set $f^{\prime\prime}(x)$ equal to $0$ and solve for $x$.
$$
\begin{aligned}
f^{\prime \prime}(x) &=-12 x-18 \\
&= -6(2 x+3)=0 \\
x &=-\frac{3}{2}
\end{aligned}
$$
so $(-\frac{3}{2} , f(-\frac{3}{2}))$ is inflection point.
Now ,we find $f(-\frac{3}{2})$as follows:
$$
f(-\frac{3}{2})=-2 (-\frac{3}{2})^{3}-9 (-\frac{3}{2})^{2}+108 (-\frac{3}{2})-10=-185.5
$$
So, point of inflection at ( -1.5, 185.5).
*
Test a number in the interval $(-\infty, -\frac{3}{2} )$ to see that $f^{\prime\prime}(x)$ is positive there, so $f(x)$ is concave upward on $(-\infty,-\frac{3}{2} )$.
**
Test a number in the interval $(-\frac{3}{2}, \infty )$to find $f^{\prime\prime}(x)$ negative in that interval, so $f(x)$ is concave downward on $(-\frac{3}{2}, \infty )$.
y-intercept:
$$
f(0)=-2 (0)^{3}-9 (0)^{2}+108 (0)-10 =-10
$$
