Math 22 Functions

Basic Definitions

A function is a relationship between two variables such that to each value of the independent variable there corresponds exactly one value of the dependent variable.

The domain of the function is the set of all values of the independent variable for which the function is defined.

The range of the function is the set of all values taken on by the dependent variable.

Function notation: We usually denote a function f of x as $f(x)$. For example, function $y=2x^{2}+1$ can be written as $f(x)=2x^{2}+1$ in function notation.

Exercises Find the domain and range of the following functions:

1) $y={\sqrt {x+1}}$

Solution:
The domain is where the function defines (or all possible values of x). So, the radicand (everything under the square root) need to be non-negative.
So, $x+1\geq 0$
Answer: $x\geq -1$ or $[-1,\infty )$
The range is all of possible outcomes (values of y). Notice that ${\sqrt {x+1}}$ is never negative. So $y$ is never negative.
Answer: $y\geq 0$ or $[0,\infty )$

Evaluate a Function

To evaluate a function $f(x)$ at $x=a$. We just need to plug in $x=a$ to find $f(a)$.

Example: Find the value of the function $f(x)=4x^{2}+1$ at $x=1,2,3$

Answer:

$f(1)=4(1)^{2}+1=4+1=5$

$f(2)=4(2)^{2}+1=16+1=17$

$f(3)=4(3)^{2}+1=36+1=37$

Exercises Find the value of the function at the given values:

2) $y={\sqrt {x+1}}$ at $x=3,-3$

Solution:
$f(3)={\sqrt {3+1}}={\sqrt {4}}=2$
$x=-3$ isn't in the domain of $f(x)$. So, $f(-3)=$ undefined
OR
$f(-3)={\sqrt {-3+1}}={\sqrt {-2}}=undefined$

Composite Function

Let $f$ and $g$ be functions. The function given by $(f\circ g)(x)=f(g(x))$ is the composite function of $f$ and $g$.

Examples: Let $f(x)=2x+1$ and $g(x)=x^{2}+3$

So, $(f\circ g)(x)=f(g(x))=f(x^{2}+3)=2(x^{2}+3)+1=2x^{2}+7$

Notes:

This page were made by Tri Phan