Answer
$$
f(x)=-2x^{3}+9x^{2}+168x-3
$$
$f(x)$ concave upward on
$$(-\infty,\frac{3}{2} )$$.
$f(x)$ concave downward on
$$(\frac{3}{2}, \infty )$$
Inflection point at
$$(\frac{3}{2}, \frac{525}{2})$$
Work Step by Step
$$
f(x)=-2x^{3}+9x^{2}+168x-3
$$
The first derivative is
$$
\begin{aligned}
f^{\prime}(x) &=-2 (3)x^{2}+9(2)x+168\\
&=-6x^{2}+18x+168,
\end{aligned}
$$
and the second derivative is
$$
\begin{aligned}
f^{\prime\prime}(x) &=-6(2)x+18+0\\
&=-12x+18.
\end{aligned}
$$
Set $f^{\prime\prime}(x)$ equal to $0$ and solve for $x$.
$$
\begin{aligned}
f^{\prime\prime}(x) =-12x+18 &=0\\
-12x &=-18\\
x &=\frac{-18}{-12}\\
x &=\frac{3}{2}.
\end{aligned}
$$
*
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 $(\frac{3}{2}, \infty )$.
***
Since the sign of $f^{\prime\prime}(x)$ changes from positive to negative at $\frac{3}{2}$, the graph changes from concave upward to concave downward at that point, so $(\frac{3}{2}, f(\frac{3}{2}))$ is inflection point.
Now ,we find $f(\frac{3}{2})$as follows:
$$
f(\frac{3}{2})=-2(\frac{3}{2})^{3}+9(\frac{3}{2})^{2}+168(\frac{3}{2})-3=\frac{525}{2}
$$
Finally, we have inflection point at $(\frac{3}{2}, f(\frac{3}{2}))=(\frac{3}{2}, \frac{525}{2})$.
