Functions: One-One, Inverses and Restricted Domains

Tell one-one from many-one, restrict a domain so an inverse exists, find inverses and composites with their domains and ranges, and sketch f with its inverse.

In the exam for IGCSE Additional Maths 0606 · 16 real past-paper questions below

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Function, domain, range

Press one button on a vending machine and you get one snack. Never two, never a surprise. A function works the same way: each input gives exactly one output.

Rule

A function gives exactly one output for each input. The domain is the set of inputs you are allowed to use. The range is the set of outputs you get.

You will see two ways to write the same function. Both say “double x and add 1, for x of 0 or more”:

Formula

f (x) = 2 x + 1 for x \geq 0

$f (x)$ — the output when the input is x.
$x \geq 0$ — the domain. The range here is $f (x) \geq 1$ .

Formula

f : x \mapsto 2 x + 1, x \geq 0

$\mapsto$ — read “maps to”: x goes to 2x + 1. It is the same function, written as an arrow.

Worked example

1
Explain why the graph of $x^{2} + y^{2} = 25$ (a circle) does not represent a function.
2
$x = 3 : 9 + y^{2} = 25$ — Pick one input and look for its outputs.
3
$y = 4 or y = - 4$ — One input gives two outputs.
4
$not a function$ — A vertical line at x = 3 meets the graph twice. A function allows at most once.

Rule

Write a range with the function's letter, not with x: $f (x) \geq 1$ , or $f \geq 1$ , or $y \geq 1$ . Writing $x \geq 1$ for a range loses the mark.

Key idea

The Swiss mathematician Leonhard Euler first wrote f(x) for a function in 1734. Almost three hundred years later, every exam paper still uses his notation.

Key idea

Domain is what goes in. Range is what comes out. A function never gives two outputs for one input.

One-one or many-one

Some functions send two different inputs to the same output. For $f (x) = x^{2}$ , both 3 and −3 give 9.

Rule

A one-one function: every output comes from exactly one input.

A many-one function: at least one output comes from two or more inputs.

To check from a graph, draw horizontal lines. If any horizontal line meets the graph more than once, the function is many-one.

Only a one-one function has an inverse. The inverse must take the output back to the input. If 9 came from both 3 and −3, the inverse would not know which one to give back.

Worked example

1
The function $f (x) = x^{2} - 6 x + 5$ has domain $0 \leq x \leq 4$ . Explain why $f^{- 1}$ does not exist.
2
$x^{2} - 6 x + 5 = (x - 3)^{2} - 4$ — Complete the square. The turning point is at x = 3.
3
$0 \leq 3 \leq 4$ — The turning point is inside the domain, so the curve goes down and then back up.
4
$f (2) = - 3, f (4) = - 3$ — Two inputs give the same output.
5
$f is many-one$ — So f has no inverse: $f^{- 1} (- 3)$ would need two answers.

Common mistake

Writing only “because it is a quadratic”. Say the function is many-one (not one-one), and back it up with two inputs that give the same output.

Key idea

No inverse means two inputs share an output. Name the function as many-one and show the pair.

Restricting the domain

A many-one function can be made one-one by cutting its domain. For a quadratic, cut at the turning point and keep one side.

Worked example

1
$f (x) = 2 x^{2} + 8 x + 3$ for $x \geq k$ . Find the least value of k for which $f^{- 1}$ exists, and the range of f.
2
$2 x^{2} + 8 x + 3 = 2 (x + 2)^{2} - 5$ — Complete the square. Take out the 2 first.
3
$turning point (- 2, - 5)$ — The bracket is 0 when x = −2, and then f is −5.
4
$k = - 2$ — From x = −2 onwards the curve only goes up, so f is one-one.
5
$f (x) \geq - 5$ — The range. The smallest output is at the turning point.

Notice: if k were smaller than −2, the domain would include both sides of the turning point and f would be many-one again.

Common mistake

Writing the range as $x \geq - 5$ . The range is about outputs, so write $f (x) \geq - 5$ .

Common mistake

Taking the x-coordinate of the turning point for the range. The range uses the y-coordinate, −5.

Key idea

To make a quadratic one-one, start the domain at the turning point. Its y-value starts the range.

Finding the inverse

To find an inverse, swap x and y, then make y the subject. Two facts tell you the domain and range at once.

Rule

Domain of $f^{- 1}$ = range of f.

Range of $f^{- 1}$ = domain of f.

Worked example

1
$f (x) = 2 (x + 2)^{2} - 5$ for $x \geq - 2$ . Find $f^{- 1} (x)$ and state its domain.
2
$x = 2 (y + 2)^{2} - 5$ — Write y = f(x), then swap x and y.
3
$(y + 2)^{2} = \frac{x + 5}{2}$ — Add 5, then divide by 2.
4
$y + 2 = \pm \frac{x + 5}{2}$ — Square root both sides. There are two signs for now.
5
$y \geq - 2 \Rightarrow take +$ — The range of the inverse is the domain of f, x ≥ −2. Only the plus sign gives values of −2 or more.
6
$f^{- 1} (x) = - 2 + \frac{x + 5}{2}, x \geq - 5$ — The domain of the inverse is the range of f.

Exponential functions come up often. You need one fact: ln undoes e. If $e^{k} = m$ , then $k = ln m$ .

Worked example

1
$g (x) = 4 e^{2 x} - 1$ for all real x. Find $g^{- 1} (x)$ and its domain.
2
$x = 4 e^{2 y} - 1$ — Swap x and y.
3
$e^{2 y} = \frac{x + 1}{4}$ — Add 1, then divide by 4.
4
$2 y = ln (\frac{x + 1}{4})$ — Take ln of both sides.
5
$g^{- 1} (x) = \frac{1}{2} ln (\frac{x + 1}{4}), x > - 1$ — $e^{2 x}$ is always positive, so the range of g is $g (x) > - 1$ . That is the domain of the inverse.

Common mistake

Thinking $f^{- 1} (x)$ means $\frac{1}{f ( x )}$ . The −1 is a label for the inverse, not a power.

Common mistake

Leaving the answer as “y = …”. Write it with the correct notation, $f^{- 1} (x) = \dots$ , and give its domain.

Key idea

Swap, solve, then swap the domain and range. They tell you which sign to keep.

Composite functions

A composite function feeds the output of one function into another. $gf (x)$ means $g (f (x))$ : do f first, then g. Read it from the inside out.

Worked example

1
$f (x) = x + 3$ and $g (x) = x^{2}$ . Find $gf (x)$ and $fg (x)$ .
2
$gf (x) = g (x + 3) = (x + 3)^{2}$ — f first: add 3. Then g squares the result.
3
$fg (x) = f (x^{2}) = x^{2} + 3$ — g first: square. Then f adds 3. A different answer, so order matters.

Worked example

1
$h (x) = 3 x - 2$ . Find $h^{2} (x)$ .
2
$h^{2} (x) = h (h (x)) = h (3 x - 2)$ — $h^{2}$ means h twice, not h squared.
3
$= 3 (3 x - 2) - 2 = 9 x - 8$ — Put 3x − 2 in place of x, then expand.

Rule

$gf$ exists only if every output of f is allowed into g: the range of f must fit inside the domain of g. Then the domain of gf is the domain of f.

Worked example

1
$f (x) = 2 x + 1$ for $x \geq 0$ and $g (x) = x - 1$ for $x \geq 1$ . Show gf exists, find its range, and solve $gf (x) = 4$ .
2
$f (x) \geq 1$ — The range of f. Every output is 1 or more.
3
$domain of g: x \geq 1$ — The outputs of f all fit, so gf exists. Its domain is x ≥ 0.
4
$gf (x) = (2 x + 1) - 1 = 2 x$ — f first, then g.
5
$gf (x) \geq 0$ — Feed the range of f, 1 or more, into g: the outputs start at $0 = 0$ .
6
$2 x = 4 \Rightarrow 2 x = 16 \Rightarrow x = 8$ — Square both sides. 8 is in the domain, so it is valid.

If the domain of f were $x \geq - 1$ instead, f could output −1. Then g would need $- 2$ , which is impossible, so gf would not exist.

Common mistake

Doing g first for gf. The function next to the x acts first.

Key idea

Inside out: the function nearest x acts first. Check that its range fits into the next function's domain.

Graphs of f and its inverse

An inverse swaps every input with its output. So the point (a, b) on $y = f (x)$ becomes (b, a) on $y = f^{- 1} (x)$ . Swapping x and y is a reflection in the line $y = x$ .

Rule

The graph of $y = f^{- 1} (x)$ is the reflection of $y = f (x)$ in the line $y = x$ .

Worked example

1
Sketch $y = f (x)$ and $y = f^{- 1} (x)$ for $f (x) = 2 (x + 2)^{2} - 5$ , $x \geq - 2$ .
2
$y = x$ — Draw this line first, as a dashed guide.
3
$(- 2, - 5) \to (- 5, - 2)$ — The end point of f swaps to the end point of the inverse.
4
$(0, 3) \to (3, 0)$ — The y-intercept of f becomes the x-intercept of the inverse.
5
$f^{- 1} (3) = - 2 + 4 = 0$ — Check with the formula from before: it agrees.

Asymptotes swap too. $y = 4 e^{2 x} - 1$ has asymptote $y = - 1$ , so its inverse has asymptote $x = - 1$ .

Some functions are their own inverse, such as $f (x) = 6 - x$ or $f (x) = \frac{1}{x}$ . Their graphs are symmetrical about $y = x$ .

Common mistake

Reflecting in the x-axis or the y-axis. The mirror line for an inverse is always $y = x$ .

Key idea

Draw $y = x$ first. Then swap the coordinates of each key point and mirror the curve across the line.

Which method? Read the question

Function questions come in parts that build on each other. Match the words to the method.

You see	You do
“Explain why this is not a function”	Find one input with two outputs
“Explain why $f^{- 1}$ does not exist”	Show f is many-one: two inputs, same output
“Least value of k so $f^{- 1}$ exists”	Complete the square; k is the x of the turning point
“Find the range of f”	Smallest (or largest) output; write $f (x) \geq \dots$
“Find $f^{- 1} (x)$ ”	Swap x and y, make y the subject, choose the sign
“Domain / range of $f^{- 1}$ ”	Swap the range and domain of f
“Explain why gf exists”	Range of f lies inside the domain of g
“Sketch f and $f^{- 1}$ ”	Reflect in $y = x$ ; swap the coordinates of key points

Common mistake

Losing the domain. An inverse or a composite is not finished until its domain is stated whenever the question asks for it.

Key idea

Find the range of f first. It becomes the domain of the inverse, and it must fit inside the domain of any function applied after f.

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