Answer
$$
f(x)=-x^{3}-12x^{2}-45x+2
$$
$f(x)$ concave upward on
$$(-\infty,-4 )$$.
$f(x)$ concave downward on
$$(-4, \infty )$$
Inflection point at
$$(-4,54)$$
Work Step by Step
$$
f(x)=-x^{3}-12x^{2}-45x+2
$$
The first derivative is
$$
\begin{aligned}
f^{\prime}(x) &=-(3)x^{2}-12(2)x-45\\
&=-3x^{2}-24x-45,
\end{aligned}
$$
and the second derivative is
$$
\begin{aligned}
f^{\prime\prime}(x) &=-3(2)x-24-0\\
&=-6x-24.
\end{aligned}
$$
Set $f^{\prime\prime}(x)$ equal to $0$ and solve for $x$.
$$
\begin{aligned}
f^{\prime\prime}(x) =-6x-24 &=0\\
-6x &=24\\
x &=\frac{24}{-6}\\
x &=-4.
\end{aligned}
$$
*
Test a number in the interval $(-\infty,-4 )$ to see that $f^{\prime\prime}(x)$ is positive there, so $f(x)$ is concave upward on $(-\infty,-4 )$.
**
Test a number in the interval $(-4, \infty )$to find $f^{\prime\prime}(x)$ negative in that interval, so $f(x)$ is concave downward on $(-4, \infty )$.
***
Since the sign of $f^{\prime\prime}(x)$ changes from positive to negative at $-4$, the graph changes from concave upward to concave downward at that point, so $(-4 , f(-4))$ is inflection point.
Now ,we find $ f(-4)$as follows:
$$
f(-4)=-(-4)^{3}-12(-4)^{2}-45(-4)+2=54
$$
Finally, we have inflection point at $(-4, f(-4))=(-4,54)$.
