Lesson 6 ~45 min Functions · Path +90 XP

Domain and Range of Common Functions

Six graphs cover almost every domain and range question at this level. Learning them as shapes, together with the reason each has the domain and range it does, turns a calculation into a recognition.

Today's hook: Two of these six accept every real input and produce every real output. Two accept every input but produce only half the outputs. One rejects half the inputs. And one rejects a single point and misses a single output. Knowing which is which is worth more than any formula.

0/5QUESTS

Think First

warm-up

Sketch $y = x^2$ and $y = |x|$ on the same axes, roughly. They have the same domain and the same range. Say what is genuinely different about them, and then say what that tells you about how much domain and range describe a function.

Record your answer in your workbook.

The Big Idea

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Six shapes account for nearly every question: linear, quadratic, cubic, square root, reciprocal, absolute value. Each has a domain and a range that follow from one structural feature of its rule.

$$y = x, \quad x^2, \quad x^3, \quad \sqrt{x}, \quad \tfrac{1}{x}, \quad |x|$$

One question separates them all: does the rule destroy sign, and does it forbid anything? Squaring and absolute value destroy sign, so their ranges are one-sided. Roots forbid negatives and reciprocals forbid zero, so their domains are cut.

$\text{six shapes}$

Odd powers keep everything

$x$, $x^3$, $x^5$: every input allowed, every output produced.

Even powers halve the range

$x^2$, $x^4$, $|x|$: all inputs, but outputs only on one side.

Translations move both

A shift changes the domain or the range by the same amount. The shape is unchanged.

What You'll Master

objectives

Know

The domain and range of the six standard shapes
Which structural feature of each rule produces its domain and range
That domain and range do not determine a function

Understand

Why even powers give one-sided ranges and odd powers do not
Why a translation shifts a domain or range without changing its shape

Can Do

State the domain and range of any of the six shapes and of translated versions of them
Match a graph to its domain and range
Justify a domain or range from the structure of the rule

Words You Need

vocabulary

HyperbolaThe graph of a reciprocal function such as $y = \tfrac{1}{x}$, in two branches.

Absolute value$|x|$ is the size of $x$, ignoring sign. Its graph is a V.

BranchOne connected piece of a graph that comes in separate parts.

One-sided rangeA range bounded on one side only, such as $y \geq 0$.

Vertical asymptoteA vertical line the graph approaches but never meets, at an excluded input.

The Three That Accept Everything

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Linear, $y = x$. Domain all real numbers; range all real numbers. Nothing is forbidden going in, and the line reaches every height, so nothing is missed coming out. The same holds for any $y = mx + c$ with $m \neq 0$.

Cubic, $y = x^3$. Domain all real numbers; range all real numbers. Cubing accepts every input and, crucially, keeps the sign: a negative input gives a negative output. So negative outputs are reachable and nothing is missed.

Quadratic, $y = x^2$. Domain all real numbers; range $y \geq 0$. Squaring accepts every input but destroys the sign, so no negative output is ever produced.

The contrast between the last two is the single most useful idea in this lesson:

$$\text{odd power: sign kept, range all reals} \qquad \text{even power: sign destroyed, range one-sided}$$

The domain is identical in all three cases; only the ranges differ, and only because of what the power does to a negative.

The Two With Cut Domains

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Square root, $y = \sqrt{x}$. Domain $x \geq 0$; range $y \geq 0$.

The domain is cut because no real number squares to a negative, so a negative input has no output. The range is one-sided because the symbol $\sqrt{\ }$ denotes the non-negative root by convention. Both restrictions are half-lines, and both include their endpoint, since $\sqrt{0} = 0$ is perfectly well defined.

Reciprocal, $y = \dfrac{1}{x}$. Domain all real numbers except $0$; range all real numbers except $0$.

The domain excludes $0$ because division by zero is undefined. The range excludes $0$ for a different reason: $\dfrac{1}{x} = 0$ has no solution, since a fraction is zero only when its numerator is, and this numerator is $1$.

The graph comes in two separate branches, one in each of the regions where $x$ is positive and where it is negative, and it never crosses either axis. The two axes are asymptotes: lines the curve approaches without ever reaching. The vertical one sits at the excluded input; the horizontal one sits at the missing output.

Absolute Value

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$y = |x|$. Domain all real numbers; range $y \geq 0$.

The absolute value of a number is its size with the sign discarded: $|5| = 5$ and $|-5| = 5$. Every input is acceptable, and the output is never negative, so the domain and range match those of $y = x^2$ exactly.

The graphs are not the same, though. $y = |x|$ is two straight half-lines meeting at a sharp corner at the origin; $y = x^2$ is a smooth curve that flattens as it approaches the origin. Near $x = 0.5$ they differ noticeably: $|0.5| = 0.5$ while $0.5^2 = 0.25$.

This is worth stating plainly because it is easy to slide past. Domain and range do not determine a function. They are a coarse description, useful for checking an answer and for knowing what a question can sensibly ask, but two very different functions can share both.

Reading a V

To sketch $y = |x - 3|$, find where the inside is zero: at $x = 3$. That is the corner. The two arms rise from it at the same steepness, so the range is $y \geq 0$ and the corner has moved rather than the shape changing.

The Whole Set, and What Moves Them

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The six, gathered:

$y = x$: domain all reals, range all reals.
$y = x^2$: domain all reals, range $y \geq 0$.
$y = x^3$: domain all reals, range all reals.
$y = \sqrt{x}$: domain $x \geq 0$, range $y \geq 0$.
$y = \dfrac{1}{x}$: domain $x \neq 0$, range $y \neq 0$.
$y = |x|$: domain all reals, range $y \geq 0$.

Translations move these without changing their shape, and the arithmetic is direct.

Adding a constant outside shifts the range by that amount: $y = x^2 + 5$ has range $y \geq 5$, and its domain is untouched.

Adding a constant inside shifts the domain: $y = \sqrt{x - 4}$ has domain $x \geq 4$, and its range stays $y \geq 0$.

A minus sign in front reflects the range about zero: $y = -\sqrt{x}$ has domain $x \geq 0$ and range $y \leq 0$, with the inequality reversed.

Combining them is just applying each step in turn: $y = -\sqrt{x-4} + 1$ has domain $x \geq 4$, and its range starts from $\sqrt{x-4} \geq 0$, becomes $-\sqrt{x-4} \leq 0$, then $\leq 1$ after the $+1$. So the range is $y \leq 1$.

Common Pitfalls

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Giving $y = x^3$ the range $y \geq 0$, by analogy with $y = x^2$.

Fix: cubing keeps the sign, so $(-2)^3 = -8$. Odd powers reach every real value; even powers do not.

Saying the range of $y = \dfrac{1}{x}$ is all real numbers.

Fix: the output $0$ is never produced. The excluded output mirrors the excluded input, and each corresponds to one asymptote.

Applying an inside shift to the range, writing the range of $y = \sqrt{x-4}$ as $y \geq 4$.

Fix: a change inside the root affects which inputs are allowed, so it shifts the domain. The range stays $y \geq 0$.

Forgetting to reverse the inequality after a reflection.

Fix: multiplying by $-1$ reverses an inequality, so $y \geq 0$ becomes $y \leq 0$. Check with one value: $-\sqrt{9} = -3$, which is indeed negative.

Watch Me Solve It · 3 examples

Watch Me Solve It · Reading a translated shape

+15 XP per step

PROBLEM

State the domain and range of $f(x) = \sqrt{x+5} - 2$.

1

Identify the base shape

It is a square-root function, so the base is $y = \sqrt{x}$ with domain $x \geq 0$ and range $y \geq 0$.
2

Handle the inside change for the domain

$x + 5 \geq 0 \ \Rightarrow \ x \geq -5$

The $+5$ is inside the root, so it changes which inputs are permitted.
3

Handle the outside change for the range

$\sqrt{x+5} \geq 0 \ \Rightarrow \ \sqrt{x+5} - 2 \geq -2$

The $-2$ is applied to the output, so it lowers the whole range by $2$.
4

Check the boundary is attained

$f(-5) = 0 - 2 = -2$

The value $-2$ is reached, so it is included. Domain $[-5,\infty)$, range $[-2,\infty)$.

AnswerDomain $x \geq -5$; range $y \geq -2$

Watch Me Solve It · A reflected reciprocal

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PROBLEM

State the domain and range of $g(x) = \dfrac{1}{x-3} + 4$, and name the two asymptotes.

1

Domain from the denominator

$x - 3 = 0 \ \Rightarrow \ x = 3 \ \text{excluded}$

Domain is all real numbers except $3$.
2

Range from the missing output

$\frac{1}{x-3} \neq 0$

The reciprocal part never equals zero, since its numerator is $1$.
3

Apply the outside shift

$\frac{1}{x-3} + 4 \neq 4$

Adding $4$ moves the missing output from $0$ to $4$. So the range is all real numbers except $4$.
4

Name the asymptotes

The vertical asymptote is $x = 3$, at the excluded input; the horizontal asymptote is $y = 4$, at the missing output. Each asymptote corresponds to one exclusion, which is a reliable way to check both answers at once.

AnswerDomain $x \neq 3$; range $y \neq 4$; asymptotes $x = 3$ and $y = 4$

Watch Me Solve It · Matching graphs to descriptions

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PROBLEM

Three functions have domain all real numbers. One has range all real numbers, one has range $y \geq 0$, and one has range $y \leq 3$. Suggest a function for each and justify the range.

1

Range all real numbers

$f(x) = x^3$

Cubing keeps the sign and grows without limit in both directions, so every real value is produced. Any linear function with non-zero gradient also works.
2

Range $y \geq 0$

$f(x) = |x|$

Absolute value accepts every input and never returns a negative, and $f(0) = 0$ so the boundary is attained. $x^2$ and $x^4$ also work.
3

Range $y \leq 3$

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

Since $x^2 \geq 0$, subtracting it from $3$ gives at most $3$, and $f(0) = 3$ is attained. The parabola opens downward with vertex at $(0,3)$.
4

Note that the answers are not unique

Each description is satisfied by infinitely many functions, since domain and range are a coarse description. $3 - |x|$ and $3 - x^4$ both also answer the third part.

AnswerFor example $x^3$, $|x|$ and $3 - x^2$; many other answers are valid

Brain Trainer · 5 problems

Brain Trainer · Name the domain and range

5 problems

Five items on the standard shapes. Work each one, then reveal the answer.

1 State the range of $y = x^4$.

An even power destroys the sign.$y \geq 0$
2 State the range of $y = x^5$.

An odd power keeps the sign.All real numbers
3 State the domain of $y = \sqrt{x-7}$.

Solve $x - 7 \geq 0$.$x \geq 7$
4 State the range of $y = |x| + 2$.

The absolute value is at least zero, so add $2$.$y \geq 2$
5 State the range of $y = \dfrac{1}{x} - 6$.

The reciprocal never gives zero, so the shifted version never gives $-6$.All real numbers except $-6$

Complete in your workbook.

Multiple Choice · 5 questions

MC1

Odd against even

+10 XP

Which of these has range all real numbers?

MC2

The reciprocal

+10 XP

The domain and range of $y = \dfrac{1}{x}$ are:

MC3

Inside or outside

+10 XP

The range of $y = \sqrt{x-6}$ is:

MC4

Reflection

+10 XP

The range of $y = -x^2 + 7$ is:

MC5

What they do not determine

+10 XP

$y = x^2$ and $y = |x|$ have:

Short Answer · 3 questions

The catalogue

+15 XP

SHORT ANSWER

State the domain and range of each, justifying the range in one sentence.
(a) $f(x) = x^3 - 8$
(b) $f(x) = (x-2)^2$
(c) $f(x) = \sqrt{9-x}$
(d) $f(x) = \dfrac{2}{x+1}$

Write your working in your book.

Shifts and reflections

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SHORT ANSWER

Let $f(x) = \sqrt{x}$, with domain $[0,\infty)$ and range $[0,\infty)$.
(a) State the domain and range of $y = f(x) + 3$.
(b) State the domain and range of $y = f(x-3)$.
(c) State the domain and range of $y = -f(x)$.
(d) State the domain and range of $y = 5 - \sqrt{x+2}$, showing your reasoning step by step.

Write your working in your book.

Reasoning about the six

+15 XP

SHORT ANSWER

(a) Explain why $y = x^2$ and $y = x^4$ have the same domain and range, but $y = x^2$ and $y = x^3$ do not.
(b) Explain why the graph of $y = \dfrac{1}{x}$ has two asymptotes and say what each one corresponds to.
(c) Give two functions with domain all real numbers and range $y \leq -1$, and say what makes them different.
(d) Explain why no function of the six standard shapes has domain all real numbers and range a single number, and give a function that does.

Write your working in your book.

(a) Both $x^2$ and $x^4$ are even powers, so both accept every real input, and both destroy the sign: a negative input raised to an even power gives a positive result. So both have domain all real numbers and range $y \geq 0$, with $0$ attained at $x = 0$. By contrast $x^3$ is an odd power, which keeps the sign, so $(-2)^3 = -8$ is negative and negative outputs are produced. Its range is therefore all real numbers, not $y \geq 0$, even though its domain matches.

(b) The vertical asymptote is the line $x = 0$. It corresponds to the excluded input: as $x$ approaches $0$ the outputs grow without bound in size, so the curve runs alongside that vertical line without ever meeting it. The horizontal asymptote is the line $y = 0$. It corresponds to the missing output: as $|x|$ grows the outputs get arbitrarily close to $0$ but never equal it, since $\tfrac{1}{x} = 0$ has no solution. So the two exclusions from the domain and the range appear on the graph as the two asymptotes.

(c) For example $f(x) = -x^2 - 1$ and $g(x) = -|x| - 1$. Both accept every input; in both, the first part is at most $0$ and subtracting $1$ makes the outputs at most $-1$, attained at $x = 0$. They differ in shape: $f$ is a smooth downward parabola while $g$ is an inverted V with a sharp corner at $(0,-1)$. This is the same point as in the lesson: sharing a domain and a range does not make two functions alike.

(d) A range consisting of a single number means the function returns the same output for every input. None of the six does this: the linear and cubic reach every real value, the quadratic and absolute value take every value from $0$ upwards, the square root every value from $0$ upwards on its own domain, and the reciprocal every value but $0$. Each produces infinitely many different outputs. A function that does have a single-number range is a constant function such as $f(x) = 7$, whose domain is all real numbers and whose range is the one-element set $\{7\}$. Its graph is a horizontal line, which passes the vertical line test and so is a genuine function.

Marking guidance: 2 marks for (a). 3 marks for (b): one per asymptote, one for linking each to an exclusion. 2 marks for (c). 2 marks for (d).

Stretch Challenge · Reading the domain and range off any rule

+25 XP

CHALLENGE

(a) Describe a general procedure for finding the domain of any function built from the six shapes, and apply it to $f(x) = \dfrac{\sqrt{x+1}}{x^2-4}$.
(b) Explain why finding a range has no comparable general procedure, and describe the two techniques that usually work.
(c) Find the range of $f(x) = \dfrac{1}{x^2+1}$, showing your reasoning.

(a) The procedure: start from all real numbers, then work through the rule and impose one condition for each denominator (it must not be zero) and one for each even root (what is inside must be at least zero). Take all the conditions together, since every one of them must hold at once. Applying it to $f(x) = \dfrac{\sqrt{x+1}}{x^2-4}$: the root requires $x + 1 \geq 0$, so $x \geq -1$; the denominator requires $x^2 - 4 \neq 0$, so $x \neq 2$ and $x \neq -2$. Combining, and noting that $-2$ is already excluded by $x \geq -1$, the domain is $x \geq -1$ with $2$ removed: $[-1,2) \cup (2,\infty)$.

(b) There is no comparable procedure because the domain depends only on which operations appear, while the range depends on what the function actually does with its inputs, which the surface of the rule does not reveal: $x^2 - 4$ and $x^3 - 4$ are written almost identically and have completely different ranges. Two techniques usually work. The first is to reason outward from the innermost part, using known facts such as "a square is at least zero" or "a root is at least zero" and applying each outer operation in turn, remembering to reverse an inequality at a minus sign. The second is to set $f(x) = c$ and ask for which values of $c$ that equation has a solution inside the domain; the set of such $c$ is exactly the range. The second is more reliable and is the one to use when the first stalls.

(c) Reason outward. For every real $x$, $x^2 \geq 0$, so $x^2 + 1 \geq 1$; in particular it is never zero, so the domain is all real numbers. Taking reciprocals of positive numbers reverses order, so from $x^2 + 1 \geq 1$ we get $\dfrac{1}{x^2+1} \leq 1$. Also the denominator is always positive, so the output is always positive: $f(x) > 0$. So the range is contained in $(0,1]$. Both ends behave as that suggests: $f(0) = 1$, so the value $1$ is attained; and as $|x|$ grows the denominator grows without bound, so the outputs get arbitrarily close to $0$ without reaching it. Every value strictly between is attained too, which the second technique confirms: solving $\dfrac{1}{x^2+1} = c$ gives $x^2 = \dfrac{1}{c} - 1$, which has a real solution exactly when $\dfrac{1}{c} - 1 \geq 0$, that is when $0 < c \leq 1$. The range is $(0,1]$.

Quick Review

recap

Odd powers

All inputs, all outputs

Even powers

All inputs, outputs one-sided

Roots and reciprocals

Cut domains, and matching cut ranges

Inside or outside

Inside moves the domain, outside moves the range

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