Functions Part 1 — Mathslive.ie

Section 1 of 9

What is a function?

A function is a rule that takes an input $x$ and gives back exactly one output $y$.

We write it as $f(x)$, read "f of x".

Given $f(x) = 5x + 1$, find $f(1)$, $f(2)$, $f(3)$.

$f(1)$ means: put $1$ in for $x$ and work out the answer.

$f(1) = 5(1) + 1 = 6$

$f(2) = 5(2) + 1 = 11$

$f(3) = 5(3) + 1 = 16$

Is it linear? Look at the outputs $6,\, 11,\, 16$ — they jump up by $5$ each time. The slope is constant:

$m = \dfrac{dy}{dx} = 5$

Yes — it is linear.

Key idea

$y = f(x)$

$x$ is the input (independent). $f(x) = y$ is the output (dependent).

For $y = mx + c$, the coefficient of $x$ is the slope: $\dfrac{dy}{dx} = m$.

You try

If $f(x) = 3x - 2$, find $f(4)$.

Put $4$ in for $x$ everywhere it appears.

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

$= 12 - 2$

$= 10$

$10$

You try

If $g(x) = x^{2} + 1$, find $g(-3)$.

Watch the sign: $(-3)^{2}$ is positive.

$g(-3) = (-3)^{2} + 1$

$= 9 + 1$

$= 10$

$10$

Section 2 of 9

The function machine

Think of a function as a machine. You feed in $x$, it applies a rule, and out comes $y$.

Domain = set of inputs · Range = set of outputs

An arrow diagram shows the same idea — each input is matched with its output:

$f(x) = 5x + 1$ · each input lands on exactly one output

Definition of a function · Must learn

A function is a rule where every element of the domain is used, and is used only once.

Section 3 of 9

Four ways to present a function

The same function can be shown in different ways. All four say the same thing:

(i) By a rule

$f(x) = 2x + 1$

(ii) On the Cartesian plane

A straight line through $(0,1)$ with slope $2$

(iii) By an arrow diagram

$1 \to 3,\; 2 \to 5,\; 3 \to 7$

(iv) By a story

"I pay you twice the hours you work, plus a $\,$€$1$ bonus." If $x$ is the hours and $y$ is the pay, then $y = 2x + 1$.

Section 4 of 9

Finding the rule from a diagram

Given an arrow diagram, your job is to spot the rule.

Find $f(x)$. Is this a function? What type?

Function? Yes — $\{1,2,3\}$ are all used and each is used only once.

Approach 1: Guess from the gap

Look at $7,\, 12,\, 17$ — the gap is $5$ each time. So the rule has $5x$ in it.

Try $f(x) = 6x + 1$: $f(2) = 6(2) + 1 = 13$ ✗ (should be $12$)

Try $f(x) = 5x + 2$: $f(1) = 7$ ✓ $f(2) = 12$ ✓ $f(3) = 17$ ✓

$f(x) = 5x + 2$

Approach 2: The build-up method

Write the couples as $(x,y)$ and use $y = ax + b$:

Couples: $(1,\,7), \;(2,\,12), \;(3,\,17)$

Sub $(1,7)$ into $y = ax + b$: $7 = a + b$

Sub $(2,12)$ into $y = ax + b$: $12 = 2a + b$

Subtract: $12 - 7 = 2a + b - (a + b)$

$5 = a$

Back-sub: $7 = 5 + b \;\Rightarrow\; b = 2$

$f(x) = 5x + 2$

You try

An arrow diagram sends $1 \to 4,\; 2 \to 7,\; 3 \to 10$. Find $f(x)$.

Gap is $3$. So the rule has $3x$ in it. Then adjust the constant.

Gap between outputs $= 3$, so $f(x) = 3x + b$

$f(1) = 3(1) + b = 4 \;\Rightarrow\; b = 1$

Check: $f(2) = 7$ ✓, $\; f(3) = 10$ ✓

$f(x) = 3x + 1$

You try

An arrow diagram sends $1 \to 1,\; 2 \to 5,\; 3 \to 9$. Use the build-up method to find $f(x)$.

Set up $y = ax + b$ using the first two couples.

$(1,1)$: $1 = a + b$

$(2,5)$: $5 = 2a + b$

Subtract: $4 = a$

Back-sub: $1 = 4 + b \;\Rightarrow\; b = -3$

$f(x) = 4x - 3$

Section 5 of 9

Injective (one-to-one)

A function is injective when every output comes from one and only one input. No two inputs share an output.

Think: "In · jective" — each one goes in to its own private spot.

Injective: each output has at most one arrow in. $7$ is not hit — it's in the co-domain but not in the range.

Three words you must know

Domain = the first set (inputs)

Co-domain = the second set (everything we could land on)

Range = the part of the co-domain that's actually used

Section 6 of 9

Surjective (onto)

A function is surjective when every element of the co-domain is hit by at least one arrow. Nothing on the right is left out.

Think: "Surge" = many. Several inputs may pile onto the same output, but every output gets used.

Surjective: every element on the right has at least one arrow in. Range = co-domain.

Section 7 of 9

Bijective (one-to-one and onto)

A function is bijective when it is both injective and surjective — a perfect pairing in both directions.

Bijective: one-to-one, both ways.

Must learn

Bijective is defined as being injective and surjective.

You try

A function sends $\{1, 2, 3\} \to \{4, 4, 9\}$ (so $1 \to 4,\; 2 \to 4,\; 3 \to 9$). Is it (a) injective, (b) surjective, (c) bijective?

Check: do any two inputs share an output? Is everything on the right used?

$1$ and $2$ both go to $4$ — so it is not injective.

Every output ($4$ and $9$) is used — so it is surjective.

Not injective, so it is not bijective.

Surjective only.

Surjective only (not injective, not bijective).

Section 8 of 9

The vertical line test

Given a graph, how do you tell if it's a function? Draw a vertical line anywhere on it.

Vertical line test · Must learn

Every $x$ must have one and only one $y$.

Any vertical line, going up or down, can hit the graph only once. If it does, the graph is a function.

Example 1: a straight line

This is a function.

Example 2: a circle

Not a function — one $x$ gives two $y$'s.

Bonus: horizontal line test

Pass the horizontal line test too (every horizontal line hits at most once) and the function is bijective.

Section 9 of 9

Restricting the domain

Look at the parabola $f(x) = (x-2)^{2} - 3$ with minimum at $(2,\,-3)$.

The horizontal line hits twice — parabola is not bijective on all of $\mathbb{R}$.

It passes the vertical line test — so it is a function. But it fails the horizontal line test, so it's not bijective.

We can fix this by restricting the domain.

For $y \ge -3$ (the full range): the function is surjective onto $[-3,\infty)$.

For $x \ge 2$ (just the right-hand half): the function is now bijective.

You try

For $f(x) = (x+1)^{2}$, what restriction on the domain would make $f$ bijective?

Find the vertex (minimum) first. Then take only one side.

Vertex at $x = -1$ (minimum of $(x+1)^{2}$).

Take either side of $x = -1$ so the horizontal line test passes.

$x \ge -1$ (or $x \le -1$)

End of Functions · Part 1.

You now know what a function is and how to spot one. Part 2 picks up with quadratic graphs, composition, and inverse functions.