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 $f(x_{2})<f(x_{1})$

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 $x=0$

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