We determine the extrema of $y(x)=-2x^3+3x^2+12x+5$ for $0 \leq x \leq 5$.
We use the following step-plan.
Step 1: Determine $y'(x)$
$y'(x)=-6x^2+6x+12$.
Step 2: Determine stationary points
$$\begin{align}
y'(x)=0 &\Leftrightarrow -6x^2+6x+12=0\\
&\Leftrightarrow x^2-x-2=0\\
&\Leftrightarrow (x-2)(x+1)=0\\
&\Leftrightarrow x=-1 \mbox{ or } x=2.
\end{align}$$
$x=-1$ is outside the domain of the function. Hence, $x=2$ is the unique stationary point.
Step 3: Determine $y(c)$
$y(2)=25$.
Step 4: Determine $y(a)$ for $a<c$
$y(0)=5$.
Step 5: Determine $y(b)$ for $b>c$
$y(5)=-110$.
Conclusion
Since $y(0)<y(2)$ and $y(5)<y(2)$:
$y(0)=5$ is a boundary minimum
$y(2)=25$ is a maximum
$y(5)=-110$ is a boundary minimum
