Working with Functions

In short: A function is a rule that gives exactly one output for each input, written $f (x)$ . Its domain is the inputs it allows and its range is the outputs it can give. An even function is symmetric in the y-axis, and an odd function is symmetric about the origin.

Starting point: Years 9 and 10

Rules, tables and graphs

A rule like "double it and add 3" can be written as $y = 2 x + 3$ . A table of values lists inputs and outputs:

$x$	0	1	2	3
$y$	3	5	7	9

Plot the points and join them to see the graph. Any set of ordered pairs $(x, y)$ is called a relation.

What makes a function

A function is a relation where every $x$ gives exactly one $y$ .

Function notation

Instead of $y = 2 x + 3$ , write $f (x) = 2 x + 3$ . Read it as "f of x". To find $f (4)$ , put 4 in for $x$ :

$f (4) = 2 (4) + 3 = 11$

So $f (4) = 11$ means the point $(4, 11)$ is on the graph.

Functions

Idea

One input, one output
Vertical line test

Notation

f(x)
f(4) is the output at 4

Domain and range

Domain: allowed x
Range: possible y

Combining

Composite f(g(x))
Sum, product

Symmetry

Even: f(-x) = f(x)
Odd: f(-x) = -f(x)

Year 11Domain and range

The domain is the set of $x$ values you are allowed to put in. The range is the set of $y$ values that come out.

Two things shrink a domain:

You cannot divide by zero. $f (x) = \frac{1}{x - 2}$ has domain $x \neq = 2$ .
You cannot square root a negative. $g (x) = x + 1$ has domain $x \geq - 1$ and range $y \geq 0$ .

Interval notation uses square brackets to include an end and round brackets to leave it out. $[2, 5)$ means $2 \leq x < 5$ . $(- \infty, 3]$ means $x \leq 3$ .

$f (x) = x^{2}$ has domain all real numbers, $(- \infty, \infty)$ , and range $[0, \infty)$ .

Domains of combined functions. Every part must work. For $y = ln (x - 1) + 5 - x$ , the log needs $x > 1$ and the root needs $x \leq 5$ , so the domain is $1 < x \leq 5$ . For $y = ln x$ , the inside of the root must be at least 0, so $ln x \geq 0$ , which means $x \geq 1$ .

A root of a quadratic. For $y = 4 x - x^{2}$ , solve $4 x - x^{2} \geq 0$ : $x (4 - x) \geq 0$ , so $0 \leq x \leq 4$ . The inside is largest at the vertex $x = 2$ , where it equals 4, so the range is $0 \leq y \leq 2$ .

Range on a restricted domain. Find the values at both ends and at any turning point inside the interval. For $f (x) = x^{2} - 4 x$ on $1 \leq x \leq 5$ : $f (1) = - 3$ , the vertex gives $f (2) = - 4$ , and $f (5) = 5$ . So the range is $- 4 \leq y \leq 5$ .

Year 11Composite, odd and even functions

A composite function puts the output of one into another. $f (g (x))$ means "do $g$ first, then $f$ ."

If $f (x) = x^{2}$ and $g (x) = x + 3$ : $f (g (x)) = (x + 3)^{2}$ but $g (f (x)) = x^{2} + 3$ . Order matters.

For $f (g (x))$ to exist, the range of $g$ must fit inside the domain of $f$ .

Even functions satisfy $f (- x) = f (x)$ . Their graphs are mirror images in the y-axis, like $x^{2}$ and $x^{4}$ .

Odd functions satisfy $f (- x) = - f (x)$ . They look the same after a half turn about the origin, like $x^{3}$ and $\frac{1}{x}$ .

Test $f (x) = x^{3} - x$ : $f (- x) = - x^{3} + x = - (x^{3} - x) = - f (x)$ , so it is odd.

Domain and range of a composite. For $f (g (x))$ with $f (x) = x$ and $g (x) = x - 4$ : $f (g (x)) = x - 4$ , so the domain is $x \geq 4$ and the range is $y \geq 0$ . Work from the inside out: first which $x$ values $g$ accepts, then which of $g$ 's outputs $f$ accepts.

Using odd and even facts. If $f$ is odd, $f (- 2) = - f (2)$ . If $g$ is even, $g (- 2) = g (2)$ . Use these to evaluate expressions such as $f (g (- 2))$ step by step from the inside.

Quick summary

A function gives exactly one output for each input. Use the vertical line test.
$f (a)$ means substitute $a$ for $x$ .
Domain: allowed inputs. Range: possible outputs. Watch for dividing by zero and roots of negatives.
$f (g (x))$ : do $g$ first. Order matters.
Even: $f (- x) = f (x)$ , symmetric in the y-axis. Odd: $f (- x) = - f (x)$ , symmetric about the origin.

Quiz

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Try 3 questions now

If $f (x) = 3 x - 4$ , what is $f (5)$ ?
1. A 7
2. B 11
3. C 19
4. D $- 1$
Show answer
B: 11
$3 (5) - 4 = 11$ .
If $f (x) = x^{2} + 2 x$ , what is $f (- 2)$ ?
1. A 0
2. B $- 8$
3. C 8
4. D 4
Show answer
A: 0
$(- 2)^{2} + 2 (- 2) = 4 - 4 = 0$ .
Which of these is NOT a function?
1. A $y = x^{2}$
2. B $y = 2 x + 1$
3. C $x^{2} + y^{2} = 16$
4. D $y = x^{3}$
Show answer
C: $x^{2} + y^{2} = 16$
A circle fails the vertical line test.