JC1 H2 Math Functions: Domain, Inverse & Composite

JC1 · H2 Mathematics · Study guide

Learn to reason from domains and ranges, choose the correct inverse branch, and check whether a composite exists before substituting.

Concept → worked examples → practice10 min read · practice linked below

Start with the allowed inputs

A function is more than a formula: it includes a domain. In H2 Mathematics, the same quadratic rule can have an inverse on one domain and fail to have one on another. Likewise, two perfectly valid functions may not form a composite in a particular order.

Before you begin

You should be comfortable rearranging equations, completing the square, solving simple inequalities, and sketching quadratics and square-root graphs. Interval notation matters: a square bracket includes an endpoint; a round bracket excludes it. Infinity always takes a round bracket.

By the end, you should be able to

Find a range using the stated domain and justify endpoint inclusion.
Test whether a function is one-to-one and find its inverse with the correct domain.
Use a domain restriction to select one quadratic branch.
Check composite existence before calculating its rule, domain and range.

1. Domain and range: inputs versus outputs

The domain $D_f$ is the set of permitted inputs. The range $R_f$ is the set of outputs actually attained. A function assigns exactly one output to each input; different inputs may still share an output.

When finding the largest possible real domain of a formula, exclude zero denominators, require square-root radicands to be non-negative, and require logarithm arguments to be positive. If a domain is already stated, use it: do not replace it with the largest possible domain.

Worked example 1 · A turning point and an excluded endpoint

Find the range of $f(x)=(x-1)^2-4$ , where $-1\leq x<4$ .

Locate the minimum. The vertex is $(1,-4)$ . Since $1$ belongs to the domain, the minimum output $-4$ is attained.
Find the upper limit. The domain gives $-2\leq x-1<3$ , so $|x-1|<3$ and $f(x)<5$ . At the included left endpoint, $f(-1)=0$ . As $x$ approaches $4$ from below, $f(x)$ approaches $5$ .
Check whether 5 occurs elsewhere. Solving $(x-1)^2-4=5$ gives $x=-2$ or $x=4$ . Neither is allowed, so $5$ is excluded.
State the range. The continuous curve takes every intermediate value, so $\boxed{R_f=[-4,5)}$ .

The endpoint values alone would miss the minimum. Always check whether a turning point lies inside the domain.

The filled points belong to the graph; the open point does not. The lowest output is

-4

, while

5

is approached but never reached.

2. One-to-one functions and inverses

A function is one-to-one if distinct allowed inputs have distinct outputs. Graphically, every horizontal line must intersect its graph at most once. Algebraically, show that $f(a)=f(b)$ forces $a=b$ for inputs in the domain.

An inverse reverses the mapping. A one-to-one function has an inverse defined on its range:

D_{f^{-1}}=R_f,\qquad R_{f^{-1}}=D_f.

To find it: write $y=f(x)$ , solve for $x$ , then exchange the variable names. State the inverse's domain. The notation $f^{-1}$ means inverse function, not reciprocal.

Worked example 2 · A rational inverse with a restricted domain

Let $h(x)=\dfrac{3x+2}{x-1}$ , with $x>1$ . Show that an inverse exists and find it.

Find the range. Rewrite $h(x)=3+\dfrac{5}{x-1}$ . For $x>1$ , the second term takes every positive value. Thus $R_h=(3,\infty)$ .
Justify one-to-one behaviour. If $1<a<b$ , then $\dfrac{5}{a-1}>\dfrac{5}{b-1}$ , so $h(a)>h(b)$ . The function is strictly decreasing and therefore one-to-one.
Rearrange to make the original input the subject. Start with $y=\dfrac{3x+2}{x-1}$ . Multiplying by $x-1$ is valid because $x>1$ : $\begin{aligned}y(x-1)&=3x+2\\yx-y&=3x+2\\x(y-3)&=y+2\\x&=\dfrac{y+2}{y-3}.\end{aligned}$ Division by $y-3$ is valid because the original range is $y>3$ .
Complete the definition. $\boxed{h^{-1}(x)=\dfrac{x+2}{x-3},\quad x>3}$ . Its domain is the original range, and its range is the original domain: $(1,\infty)$ .

A quick check: $h(2)=8$ and $h^{-1}(8)=2$ . This catches arithmetic slips; the preceding argument establishes the full result.

3. Restricting a quadratic to obtain an inverse

Worked example 3 · Choose the branch before the square root

Define $f(x)=(x-1)^2+2$ for $x\geq a$ . Find the least value of $a$ for which an inverse exists. Find the inverse for that least value of $a$ .

Identify the symmetry. The vertex lies at $x=1$ . If $a<1$ , the distinct allowed inputs $a$ and $2-a$ have the same output, $(a-1)^2+2$ . Since both inputs lie in $[a,\infty)$ , the function is not one-to-one.
Keep one branch. For $a\geq1$ , every allowed $x$ has $x-1\geq0$ . Squaring is strictly increasing on non-negative inputs, so the retained branch is strictly increasing. Therefore the least value is $\boxed{a=1}$ , giving $R_f=[2,\infty)$ .
Select the sign. From $y-2=(x-1)^2$ , use $x\geq1$ to choose $x-1=+\sqrt{y-2}$ .
State the inverse. $\boxed{f^{-1}(x)=1+\sqrt{x-2},\quad x\geq2}$ , with range $[1,\infty)$ .

Keeping $x\leq1$ instead would give the minus branch. The original domain determines the sign; it is never an arbitrary choice.

An inverse swaps coordinates:

(3,6)

becomes

(6,3)

. With equal axis scales, the two graphs are reflections in

y=x

. The curves continue beyond the displayed window.

Explore: when does the inverse become possible?

Keep $f_a(x)=(x-1)^2+2$ , but vary its domain $x\geq a$ . Predict the threshold before moving the control.

Smallest allowed input,

a

Use the slider or arrow keys. The pale curve shows the unrestricted quadratic; the dark curve shows the allowed inputs.

One-to-one on this domain. The minimum output is 2. The inverse accepts inputs at least 2 and returns outputs at least 1.

For every $a\geq1$ , the inverse rule is $f_a^{-1}(x)=1+\sqrt{x-2}$ , but its domain starts at $(a-1)^2+2$ . Changing the original domain changes the inverse's domain too.

4. Composite functions: existence comes first

In this guide, $fg$ means $f\circ g$ , so $(fg)(x)=f(g(x))$ : apply $g$ first. The inner function's entire range must fit inside the outer function's domain.

fg\text{ exists on }D_g\quad\Longleftrightarrow\quad R_g\subseteq D_f.

When it exists, $D_{fg}=D_g$ . Find $R_{fg}$ by applying $f$ only to the outputs of $g$ , not automatically to all of $D_f$ . Reverse the roles to check $gf$ .

Worked example 4 · One order works; the other does not

Let $f(x)=\sqrt{x-2}$ , $x\geq2$ , and $g(x)=x^2+1$ , $x\in\mathbb R$ . Examine $fg$ and $gf$ , giving the rule, domain and range of each composite that exists on the stated domains. Here, $(fg)(x)=f(g(x))$ : apply $g$ first.

Record the sets. $R_f=[0,\infty)$ and $R_g=[1,\infty)$ .
Check $fg$ . Since $[1,\infty)\nsubseteq[2,\infty)$ , $fg$ does not exist on the given domain of $g$ . For example, $g(0)=1$ , but $f(1)$ is undefined.
Check $gf$ . Since $[0,\infty)\subseteq\mathbb R$ , $gf$ exists. Substitution gives $(gf)(x)=(\sqrt{x-2})^2+1=x-1$ .
State the domain and range. The inner function is $f$ , so the composite keeps $D_f=[2,\infty)$ . Thus $\boxed{(gf)(x)=x-1,\quad x\geq2}$ . This increasing rule has smallest output $2-1=1$ and no upper bound, giving $\boxed{R_{gf}=[1,\infty)}$ .

Although the simplified rule $x-1$ accepts all real numbers algebraically, this composite still accepts only the inputs allowed by $f$ .

Common mistakes to catch before submitting

Replace each shortcut with a justified statement
Mistake	Repair
Reading a range from endpoint values alone.	Check turning points and excluded endpoints.
Writing both square-root signs for an inverse.	Use the original domain to select one branch.
Checking $D_g\subseteq R_f$ for $fg$ .	Follow the input: $g$ acts first, so check $R_g\subseteq D_f$ .
Giving only a composite or inverse formula.	State its domain; simplification does not enlarge it.

Practise with our topic questions

Apply the methods in this guide using our JC1 H2 functions practice questions. The topic bank brings the questions and their existing worked solutions together, so you can attempt a question before checking the method.

State the domain and range, justify the inverse branch, and check whether a composite exists before calculating its rule.

Frequently asked questions

Is every function one-to-one?

No. The vertical line test checks whether each input has at most one output. The horizontal line test checks whether each output comes from at most one input. A full quadratic passes the first test and fails the second, so it is a function but has no inverse function on its full real domain.

Is the inverse the same as the reciprocal?

No. For $f(x)=2x+3$ , the inverse is $f^{-1}(x)=\dfrac{x-3}{2}$ , while the reciprocal is $\dfrac1{f(x)}=\dfrac1{2x+3}$ . The inverse undoes the mapping; the reciprocal divides one by its output. They have different rules and may have different domains.

Must the inner range equal the outer domain?

No. Containment is enough. Every possible inner output must be accepted by the outer function, but some outer inputs may remain unused. An overlap alone is insufficient because some allowed original inputs would still fail at the second step.

How is the range different from the codomain?

The codomain is a declared target set; the range consists of values actually reached. Here, an inverse is defined on the original range. If an inverse is instead required on an entire declared codomain, the original function must also reach every value in that codomain.