Math Curve sketching Monotony behavior

# Monotony behavior

Monotony behavior is the slope behavior of a function. This tells you when a function is going up or down.

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### General

A function is for two places $x_1$ and $x_2$ with $x_1<x_2$

• increasing monotonically if $f(x_1)\le f(x_2)$
• decreasing monotonically if $f(x_1)\ge f(x_2)$
• strictly (monotonically) increasing if $f(x_1)< f(x_2)$
• strictly (monotonically) decreasing if $f(x_1)> f(x_2)$
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### Remember

You can also determine the monotony behavior in an interval with the derivation.

The function is differentiable on the interval $I$ and

• increasing monotonically if for all $x\in I$ we have $f'(x)\ge0$
• decreasing monotonically if for all $x\in I$ we have $f'(x)\le0$
• strictly monotonically increasing if for all $x\in I$ we have $f'(x)>0$
• strictly monotonically decreasing if for all $x\in I$ we have $f'(x)<0$
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### Method

1. Find derivative
2. Calculate the zero (s) of the derivative
3. Determine intervals
4. Sample insertion into the derivative

### Example

Investigate the function $f(x)=x^3+2x^2-4x-8$ for monotory.

1. #### Find derivative

$f'(x)=3x^2+4x-4$
2. #### Calculate the zero of the derivative

There is a quadratic equation that can be solved, for example, with the PQ formula.
$f'(x)=3x^2+4x-4\quad|:3$
$f'(x)=x^2+\frac43x-\frac43$

$x_{1.2} = -\frac{p}{2} \pm\sqrt{(\frac{p}{2})^2-q}$
$x_{1.2} = -\frac{2}{3} \pm\sqrt{(\frac23)^2+\frac43}$
$x_{1.2} = -\frac{2}{3} \pm\sqrt{\frac{16}{9}}$
$x_{1.2} = -\frac{2}{3} \pm\frac43$
$x_1=\color{blue}{-2} \quad x_2=\color{green}{\frac23}$
3. #### Determine intervals

With the zeros of the derivative function you form the intervals at which the monotony behavior is investigated.
$x_1=\color{blue}{-2} \quad x_2=\color{green}{\frac23}$

$I_1(-\infty|\color{blue}{-2})$, $I_2(\color{blue}{-2}|\color{green}{\frac23})$, $I_3(\color{green}{\frac23}|\infty)$
4. #### Sample inserts in the derivative

Use any value from each interval in the derivative.

$I_1(-\infty|-2)$:
Sample insertion: $x=\color{red}{-3}$
$f'(\color{red}{-3})=3\cdot(\color{red}{-3})^2+4\cdot(\color{red}{-3})-4$ $=11 > 0$
=> The derivative is positive in interval $I_1$, i.e. the function is monotonically increasing in this interval.

$I_2(-2|\frac23)$:
Sample insertion: $x=\color{red}{0}$
$f'(\color{red}{0})=3\cdot\color{red}{0}^2+4\cdot\color{red}{0}-4$ $=-4 < 0$
=> The derivative is negative in the interval $I_2$, i.e. the function is monotonically decreasing in this interval.

$I_3(\frac23|\infty)$:
Sample insertion: $x=\color{red}{1}$
$f'(\color{red}{1})=3\cdot\color{red}{1}^2+4\cdot\color{red}{1}-4$ $=3 > 0$
=> The derivative is positive in the interval $I_3$, i.e. the function is monotonically increasing in this interval.