Skip to main content

# 1.1: Functions and Graphs

[ "article:topic", "authorname:green" ]

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

Definition: Functions

A function is a rule that assigns every element from a set (called the domain) to a unique element of a set (called the range).

Example 1:

Let the domain be US citizens and the range be the set of all fathers.  Let the rule of the function send each person to his or her fathers.

Example 2:

These functions have domain on the real numbers ($$\mathbb{R}$$):

\begin{align} f (x) &= x^2 \\ f (x) &= 3x + 1 \\ f (x) &= \dfrac{1}{x} \text{ (the domain is \mathbb{R}-\{0\})}\end{align}

Note that

$x^2+y^2=1$

is not a function since for $$x= 0$$, $$y$$ can be 1 or -1.

#### Vertical Line ﻿﻿Test

If every vertical line passes through the graph at most once, then the graph is the graph of a function.

### Finding the Domain

To find the domain of a function, we follow the three basic principals:

#### 1. Function Evaluation

If $$f$$ is an algebraic function we can write without the variables as in the following example:

If

$f(x) = x^2 - 2x$

we can write

$f = (x)^2 - 2(x).$

This more suggestively shows how to deal with non $$x$$ inputs.  For example

$f(x - 1) = (x - 1)^2 - 2(x - 1).$

#### 2. The Difference Quotient

We define the difference quotient for a function $$f$$ by

Definition: Difference Quotient

$\dfrac{f(x+h)-f(x)}{h}.$

Example 3:

Let

$f(x) = x^2 - 2x.$

Then

\begin{align} f(x + h) &= (x + h)^2 - 2(x +h) \\ &= x^2 + 2xh + h^2 -2x - 2h \end{align}

and

$f(x) = (x)^2 - 2(x)$

so that

\begin{align} \dfrac{f(x+h)-f(x)}{h}&=\dfrac{(x^2+2xh+h^2-2x-2h)-(x^2-2x)}{h} \\ &=\dfrac{2xh+h^2-2h}{h} \\&=\dfrac{h(2x+h-2)}{h} \\ &=2x+h-2.\end{align}

### Finding the domain and range of a function from its graph

We often use the graphing calculator to find the domain and range of functions.  In general, the domain will be the set of all $$x$$ values that has corresponding points on the graph.  We note that if there is an asymptote (shown as a vertical line on the TI 85) we do not include that x value in the domain.  To find the range, we seek the top and bottom of the graph.  The range will be all points from the top to the bottom (minus the breaks in the graph).

#### Zeros of Functions and the x-intercept Method

To find the x-intercepts of a complicated function, we can use a graphing calculator to view the graph, then use more math root to find the x-intercepts.

Exercise:

Find the roots of

$y = x^3 - 4x^2 + x + 2.$

If we want to find the intercept of two graphs, we can set them equal to each other and then subtract to make the left hand side zero. Then set the right hand side equal to $$y$$ and find the zeros.

Example 4:

Find the intercept of the graphs:

$y = 2x^3- 4$

and

$y = x^4 - x^2.$

Solution

We form the new function

$f(x) = (2x^3 - 4) - (x^4 - x^2)$

and use the root feature of our calculator to get

$x = 1.52 \text{ or } x = 2.$

#### Solving Inequalities Graphically

To solve an inequality graphically we first put 0 on the right hand side and $$f(x)$$ on the left hand side. Then we use the x-intercept method to find the zeros. If the inequality is a "<" we include the part of the graph below the x axis. If the inequality is a ">" we include the part of the graph above the x-axis.

Example 5:

Give and approximate solution of

$3x^5 - 14x^2 < x - 4.$

Solution

First, bring everything to the left hand side.

$3x^5 - 14x^2 - x + 4 < 0$

The graph shows that the roots lie at:

$x = -0.56, \;\;\; x = 0.51, \;\text{and} \; x = 1.64.$

The points lie below the x-axis to the left of -0.56 and between 0.51 and 1.64.  Hence the solution in interval notation is

$(-\infty,-0.56)\cup(0.51,1.64).$

Larry Green (Lake Tahoe Community College)

• Integrated by Justin Marshall.