# Math 22 Increasing and Decreasing Functions

## Definitions of Increasing and Decreasing Functions

 A function  is increasing on an interval when, for any two numbers $x_{1}$ and
$x_{2}$ in the interval, $x_{2}>x_{1}$ implies $f(x_{2})>f(x_{1})$ A function  is decreasing on an interval when, for any two numbers $x_{1}$ and
$x_{2}$ in the interval, $x_{2}>x_{1}$ implies $f(x_{2}) ## Test for Increasing and Decreasing Functions

 Let $f(x)$ be differentiable on the interval $(a,b)$ .
1. If $f'(x)>0$ for all $x$ in $(a,b)$ , then $f$ is increasing on $(a,b)$ .
2. If $f'(x)<0$ for all $x$ in $(a,b)$ , then $f$ is decreasing on $(a,b)$ .
3. If $f'(x)=0$ for all $x$ in $(a,b)$ , then $f$ is constant on $(a,b)$ .


## Critical Numbers and Their Use

 If $f$ is defined at $c$ , then $c$ is a critical number of $f$ when $f'(c)=0$ or when $f'(c)$ is
undefined.


Exercises: Find critical numbers of

1) $f(x)=x^{2}+2x$ Solution:
$f'(x)=2x+2=2(x+1)=0$ So, $x=-1$ is critical number

2) $f(x)={\sqrt {x}}$ Solution:
$f(x)=x^{\frac {1}{2}}$ So, $f'(x)={\frac {1}{2}}x^{\frac {-1}{2}}={\frac {1}{2{\sqrt {x}}}}$ In this case, we have critical number when $f'(x)$ is undefined, which is when ${\sqrt {x}}=0$ . So critical number is $x=0$ ## Increasing and Decreasing Test

 1. Find the derivative of $f$ .
2. Locate the critical numbers of $f$ and use these numbers to determine test intervals.
That is, find all $f$ for which $f'(x)=0$ or $f'(x)$ is undefined.
3. Determine the sign of $f'(x)$ at one test value in each of the intervals.
4. Use the test for increasing and decreasing functions to decide
whether $f$ is increasing or decreasing on each interval.


Exercises: Find the intervals of increasing and decreasing of

1) $f(x)=x^{3}-3x^{2}$ Solution:
$f'(x)=3x^{2}-6x=3x(x-2)=0$ So, $x=0,2$ are critical numbers.
Hence, the test intervals are $(-\infty ,0),(0,2)$ and $(2,\infty )$ In each interval, choose a number and test on $f'(x)$ :
So, $f'(-1)=9>0$ , so $f(x)$ is increasing on $(-\infty ,0)$ $f'(1)=-3<0$ , so $f(x)$ is decreasing on $(0,2)$ $f'(3)=9>0$ , so $f(x)$ is increasing on $(2,\infty )$ Therefore, $f(x)$ is Increasing: $(-\infty ,0)\cup (2,\infty )$ Decreasing on $(0,2)$ 