We determine the extrema of $y(x)=-2x^3+3x^2+12x+5$.
We use the following step-plan.
Step 1: Determine $y'(x)$
$y'(x)=-6x^2+6x+12$.
Step 2: Determine stationary points
$$\begin{align}
y'(x)=0 &\Leftrightarrow -6x^2+6x+12=0\\
&\Leftrightarrow x^2-x-2=0\\
&\Leftrightarrow (x-2)(x+1)=0\\
&\Leftrightarrow x=-1 \mbox{ or } x=2.
\end{align}$$
Step 3: Determine $y''(x)$
$y''(x)=-12x+6$.
Step 4: Determine extremum locations
$y''(-1)=18>0$: $x=-1$ is a minimum location,
$y''(2)=-18<0$: $x=2$ is a maximum location.
Step 5: Determine extrema
$y(-1)=-3$
$y(2)=25$
Conclusion
$y(-1)=-3$ is a minimum
$y(2)=25$ is a maximum
Remark: Compare this result with the result in Example (film) and Example (film).
