Increasing and decreasing

Differentiation: Applications of derivatives

Increasing and decreasing

A function $f$ is $\blue{\text{increasing}}$ in $x$ if $f(x)$ increases as $x$ increases.

A function $f$ is $\green{\text{decreasing}}\ (dashed)$ in $x$ if $f(x)$ decreases as $x$ increases.

In the example, we see that a function can also both increase and decrease. We say that the function increases on the interval $\ivoo{-\infty}{6}$ and decreases on the interval $\ivoo{6}{\infty}$ .

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We can check whether a function increases or decreases in a point $x$ by looking at the derivative in that point.

A function $f$ $\blue{\text{increases}}$ in a point $x$ if $f'(x)\gt 0$ .

A function $f$ $\green{\text{decreases}}$ in a point $x$ if $f'(x)\lt 0$ .

A function $f$ can transition from $\blue{\text{increasing}}$ to $\green{\text{decreasing}}$ (and the other way around) in a point $p$ if $f'(p)=0$ .

Example

$\begin{array}{rcll}f(x)&=&x^3-8x& \text{with } f'(x)=3x^2-8\\ f'(2)&=&4&\text{therefore }f\blue{\text{ increases }}\text{in }x=2\\f'(1)&=&-5&\text{therefore }f\green{\text{ decreases }}\text{in }x=1\end{array}$

Determine the interval at which the function increases / decreases

	Step-by-step	Example
	$f$ $\blue{\text{increases}}$ .	$f(x)=\left(x-4\right)^2+6$
Step 1	Determine the derivative of $f$ .	$f'(x)=2\left(x-4\right)$
Step 2	Determine the zeros of the derivative.	$x=4$
Step 3	For points to the left and right of the zeros, determine whether $f'$ is positive or negative.	$f'(0)=-8$ and $f'(6)=4$
Step 4	Now determine the interval / intervals on which $f$ increases. The function $f$ increases if $f'(x) \gt 0$ .	$f$ $\blue{\text{increases}}$ on $\ivoo{4}{\infty}$

This works the same way for intervals on which $f$ $\green{\text{decreases}}$ , only then $f'(x)\lt0$ applies.

The function $f(x)=-6x^2-9x+3$ is increasing on the interval $\ivoo{-\infty}{a}$ and decreasing on the interval $\ivoo{a}{\infty}$ . What is the value of $a$ ? Simplify your answer as much as possible.

$a=$ $-{{3}\over{4}}$

Step 1	We determine the derivative of $f$ using the power rule. This gives: $f'(x)=-12x-9$
Step 2	We solve the equation $-12x-9=0$ This goes as follows: $\begin{array}{rcl}-12x-9&=&0 \\ &&\phantom{xxx}\blue{\text{the equation we need to solve}} \\ -12x&=&9 \\ &&\phantom{xxx}\blue{\text{both sides minus }-9} \\ x&=&\displaystyle -{{3}\over{4}} \\ &&\phantom{xxx}\blue{\text{both sides divided by }-12} \\\end{array}$
Step 3	$f'(-3)=27$ $f'(1)=-21$
Step 4	Therefore, the function $f$ is increasing on the interval $\ivoo{-\infty}{-{{3}\over{4}}}$ and decreasing on the interval $\ivoo{-{{3}\over{4}}}{\infty}$ . Hence, $a=-{{3}\over{4}}$ .

The graph of $f$ is drawn below.

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