We determine the extrema of $y(x)=-2x^3+3x^2+12x+5$ for $-5 \leq x \leq 5$.
We use the following step-plan.
Step 1: Determine $y'(x)$
$y'(x)=-6x^2+6x+12$.
Step 2: Determine the 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: Make a sign chart of $y'(x)$
$y'(-3)=-60$, $y'(0)=12$ and $y'(3)=-24$.
Step 4: Determine the extremum locations
$x=-5$ is a maximum location
$x=-1$ is a minimum location
$x=2$ is a maximum location
$x=5$ is a minimum location
Step 5: Determine the extrema
$y(-5)=270$
$y(-1)=-3$
$y(2)=25$
$y(5)=-110$
Conclusion
$y(-5)=270$ is a boundary maximum
$y(-1)=-3$ is a minimum
$y(2)=25$ is a maximum
$y(5)=-110$ is a boundary minimum
