Example (film)

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).