Extrema

The maxima and minima points of a function graph are called extrema.

Example

Find the extrema of the function $f(x)=x^3+2x^2-4x-8$.

1. Find derivatives

   $f'(x)=3x^2+4x-4$
   $f''(x)=6x+4$

2. Calculate the zero of the derivative

   There is a quadratic equation that can be solved e.g. with the PQ formula.
   $f'(x)=3x^2+4x-4$
   $x_E\Leftrightarrow f'(x_E)=0$
   
   $3x^2+4x-4=0\quad|:3$
   $x^2+\frac43x-\frac43=0$
   $x_{1.2} = -\frac{p}{2} \pm\sqrt{(\frac{p}{2})^2-q}$
   $x_{1.2} = -\frac{2}{3} \pm\sqrt{(\frac23)^2+\frac43}$
   $x_{1.2} = -\frac{2}{3} \pm\sqrt{\frac{16}{9}}$
   $x_{1.2} = -\frac{2}{3} \pm\frac43$
   $x_{E_{1}}=\color{blue}{-2} \quad x_{E_{2}}=\color{green}{\frac23}$

3. Insert zeros into the second derivative

   We use the just-determined places of extrema in the second derivation.
   $f''(x)=6x+4$
   
   $f''(\color{blue}{-2})=6\cdot\color{blue}{-2}+4=-8<0$
   => There is a maximum at the position $x=-2$
   $f''(\color{green}{\frac23})=6\cdot\color{green}{\frac23}+4=8>0$
   => There is a minimum point at the position $x=\frac23$
   
   Note: The calculated values $8$ and $-8$ were just to check if it's a maximum or minimum. They are no longer needed.

4. Specify extremas

   Maxima and minima points should be specified: Therefore calculate the y-coordinate with the original function.
   
   $f(\color{blue}{-2})$ $=(\color{blue}{-2})^3+2\cdot(\color{blue}{-2})^2-4\cdot(\color{blue}{-2})-8$ $=0$
   => Maximum point: $H(\color{blue}{-2}|0)$
   
   $f(\color{green}{\frac23})$ $=(\color{green}{\frac23})^3+2\cdot(\color{green}{\frac23})^2-4\cdot\color{green}{\frac23}-8$ $ \approx-9.48$
   => Minimum point: $T(\color{green}{\frac23}|-9.48)$