Math Curve sketching Maxima and minima (Extrema)

# Extrema

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

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### Remember

#### Necessary criterion

The prerequisite for the existence of extrema is that the first derivative has a zero at this point:
$f'(x_E)=0$

#### 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 $f''(x_E)>0$ i

### Method

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

### 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: $T(\color{green}{\frac23}|-9.48)$