Math Curve sketching Maxima and minima (Extrema)


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



Necessary criterion

The prerequisite for the existence of extrema is that the first derivative has a zero at this point:

Sufficient criterion

Whether it is a maximum, minimum, or neither is dependent on the second derivative
There is a

  • Maximum point, if $f''(x_E)<0$
  • Minimum point, if $f''(x_E)>0$
Maximum and minimum


  1. Find derivatives
  2. Calculate the zero(s) of the first derivative
  3. Insert zero(s) into the second derivative
  4. Specify extremas


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

  1. Find derivatives

  2. Calculate the zero of the derivative

    There is a quadratic equation that can be solved e.g. with the PQ formula.
    $x_E\Leftrightarrow f'(x_E)=0$

    $x_{1.2} = -\frac{p}{2} \pm\sqrt{(\frac{p}{2})^2-q}$
    $x_{1.2} = -\frac{2}{3} \pm\sqrt{(\frac23)^2+\frac43}$
    $x_{1.2} = -\frac{2}{3} \pm\sqrt{\frac{16}{9}}$
    $x_{1.2} = -\frac{2}{3} \pm\frac43$
    $x_{E_{1}}=\color{blue}{-2} \quad x_{E_{2}}=\color{green}{\frac23}$
  3. Insert zeros into the second derivative

    We use the just-determined places of extrema in the second derivation.

    => There is a maximum at the position $x=-2$
    => 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: $T(\color{green}{\frac23}|-9.48)$