Composite functions: order, equations and when they exist

What you need to be able to do

What you should already know

fg means: do g first

\(f(x)=x^2+3\) and \(g(x)=2x+1\):

Forming and using a composite

1. Write the outer function with a bracket where its input goes: for \(fg(x)\), write \(f(\ \ )\) using \(f\)'s rule.
2. Put the whole inner function into that bracket, with brackets around it: \(f(2x+1)=(2x+1)^2+3\).
3. Simplify only as far as the question asks ("in the form …"). Leaving \((2x+1)^2+3\) unexpanded is often easier for the next part (an inverse).
4. To solve \(fg(x)=k\): set the composite equal to \(k\) and solve, with a method shown. Or peel the layers: \(f(\text{something})=k\Rightarrow\) something \(=f^{-1}(k)\).
5. Check every answer is in the domain of the inner function (and of the composite). Reject the rest and say so.

Example: solving with a composite. \(f(x)=x^2+3\), \(g(x)=2x+1\). Solve \(fg(x)=12\).

1. \(fg(x)=(2x+1)^2+3\), so \((2x+1)^2+3=12\)
2. Peel the outer layer: \((2x+1)^2=9\Rightarrow2x+1=\pm3\)
3. \(x=1\) or \(x=-2\). Check: \(g(1)=3,\ f(3)=12\) ✓

When can gf be formed?

Think of the machines again: \(f\)'s outputs are fed into \(g\). If \(f\) can produce a number that \(g\) refuses, the composite breaks.

The domain and range of a composite

When \(gf\) exists, its domain is the domain of \(f\) (the first machine decides which inputs are allowed). Its range comes from feeding the range of \(f\) through \(g\).

Example. \(f(x)=x^2\) for \(x\ge1\) (range \(f(x)\ge1\)); \(g(x)=3x-2\) for \(x\in\mathbb R\). \(gf(x)=3x^2-2\) for \(x\ge1\). The inputs to \(g\) are all \(\ge1\), and \(g\) is increasing, so \(gf(x)\ge3(1)-2=1\).

Key facts at a glance

• \(fg(x)=f(g(x))\): the function next to \(x\) acts first. \(fg\ne gf\) and \(fg\) is not a product.
• Form it by putting the whole inner function, in brackets, into the outer rule.
• Numbers: work inside out. Equations: set equal and solve, or peel the outer layer.
• \(gf\) exists only when the range of \(f\) lies within the domain of \(g\).
• Domain of \(gf\) = domain of \(f\); its range = the range of \(f\) fed through \(g\).

How this is examined

A composite is in almost every Functions question: form \(fg(x)\), evaluate it, or solve \(fg(x)=k\) (2-4 marks). The June 2021 report: some candidates "composed the functions in the wrong order or multiplied two or more functions together". For a numerical composite, the November 2021 report found that forming \(ff(x)\) first made the work "more complicated". Explaining why a composite cannot be formed was "the most challenging on the whole paper" (November 2022 report); "the range of \(g^{-1}\) is not within the domain of \(h\)" is the minimum answer schemes accept.

Mark-scheme focus

Explain why the composite function cannot be formed

Credited

• "The range of g does not lie in the domain of f" or "the domain of f does not include the whole of the range of g" B1
• an example outside the domain of f, e.g. \(g(-1)=-4\) but \(f\) needs \(x>0\)

Not credited

• "because g is not one-one" (irrelevant)
• "because you can't divide by zero" (only part of the story; \(g\) also gives negatives)
• mentioning only one function

Why

A composite fg exists when g's outputs can all be fed into f. The explanation must link the two sets in the right order.

From: 9709/12/O/N/22 Q9(e) and the November 2022 report.

Solve gf(x) = … (reject what is not in the domain)

Credited

• \(gf(x)=2((x+3)^2-12)-5\) *M1; \(2(x+3)^2-29=69\) A1
• solve (no method needed for this one) DM1
• \(x=4\) only A1

Not credited

• \(x=4\) and \(x=-10\): \(-10\) is not in the domain \(x\ge0\)
• \(f^{-1}\) then \(g^{-1}\) in the wrong order ("scores *M0")

Also accepted

peeling: \(g(\text{f(x)})=69\Rightarrow f(x)=g^{-1}(69)=37\Rightarrow(x+3)^2=49\Rightarrow x=4\).

From: 9709/11/O/N/25 Q6(c).

Worked examples

Example 1 · fg, then an equation with both composites (Paper 1, 5 marks) ★★ Medium

1 Read it with me what is actually happening?

Two simple functions combined both ways. Part (a) puts \(g\) inside \(f\). Part (c) needs \(gf\) as well, then an ordinary quadratic. The domains decide which root survives.

• fg(x)g first, then f: f(2x + 1)
• gf(x)f first, then g: g(x² + 3)
• x > 0 / x > −½check the roots at the end

2 What is it asking for? the target, part by part

• (a)fg(x)1 mark
• (c)x4 marks

3 What do we know? everything the question gives us

• \(f(x)=x^2+3\)x > 0
• \(g(x)=2x+1\)x > −½

• \(gf(x)\) needs \(x>0\) (the domain of \(f\)); \(fg(x)\) needs \(x>-\tfrac12\)

4 Plan the route work backwards from what we want

FindMaths ideaAlready have itWork out firstTrap

• (a)fg(x)
  • Use\(f(2x+1)=(2x+1)^2+3\)
• (c)x
  • Also\(gf(x)=2(x^2+3)+1\)
  • Use\((2x+1)^2=2x^2+7\)
  • Trapkeeping x = −3

Why does the −3 make (c) nice?

\(fg(x)-3=(2x+1)^2\): the +3 cancels.

Which root is allowed?

Both composites need \(x>0\) for gf: \(x=1\) only.

5 Now solve it have a go on paper first, then open the worked solution

1. (a) \(fg(x)=(2x+1)^2+3\) B1
2. (c) \(gf(x)=2(x^2+3)+1\) B1; \((2x+1)^2+3-3=2x^2+7\) *M1
3. \(4x^2+4x+1=2x^2+7\Rightarrow2x^2+4x-6=0\Rightarrow(x+3)(x-1)=0\) DM1
4. \(x=1\) (\(x=-3\) is not in the domain) A1

The November 2020 report: "some omitted the '–3' on the left-hand side", and "some candidates did not show working for the solution of the quadratic, which resulted in the loss of a method mark."

6 Check it does the answer make sense?

\(fg(1)-3=9+3-3=9\); \(gf(1)=2(4)+1=9\) ✓.

Example 2 · Find two constants from composite values (Paper 1, 4 marks) ★★ Medium

1 Read it with me what is actually happening?

Two facts, two unknowns. \(f(7)=\tfrac52\) only involves \(a\): solve that first. Then \(gf(5)\): find \(f(5)\) with your \(a\), feed it into \(g\), set equal to 4.

• f(7) = 5/2an equation in a only
• gf(5)f(5) first (a number), then g of it

2 What is it asking for? the target, part by part

• (a)\(a\) and \(b\)4 marks

3 What do we know? everything the question gives us

• \(f(x)=1+\dfrac{2a}{x-a}\)
• \(g(x)=bx-2\)

4 Plan the route work backwards from what we want

FindMaths ideaAlready have itWork out firstTrap

• Finda
  • Use\(1+\tfrac{2a}{7-a}=\tfrac52\)
• Findb
  • Firstf(5) with a = 3
  • Use\(g(f(5))=4\)

Clear the fraction in 2a/(7 − a) = 3/2.

\(4a=3(7-a)\Rightarrow7a=21\).

What is f(5) when a = 3?

\(1+\tfrac{6}{2}=4\).

5 Now solve it have a go on paper first, then open the worked solution

1. \(1+\dfrac{2a}{7-a}=\dfrac52\Rightarrow\dfrac{2a}{7-a}=\dfrac32\Rightarrow4a=21-3a\Rightarrow a=3\) M1 A1
2. \(f(5)=1+\dfrac{6}{5-3}=4\); \(g(4)=4b-2=4\) M1
3. \(b=\tfrac32\) A1

The June 2023 report: "a common error was omitting to multiply the 1 when multiplying through by 7 – a", and for \(b\), "those who found f(5) first were generally more successful".

6 Check it does the answer make sense?

\(f(7)=1+\tfrac64=\tfrac52\) ✓; \(g(4)=6-2=4\) ✓.

Example 3 · A composite equation with a rejected root (Paper 1, 4 marks) ★★ Medium

1 Read it with me what is actually happening?

\(fg(x)=\dfrac{1}{(2x+1)^2}-1\). Set it equal to 3, isolate the bracket, square-root (two signs), and keep only the root inside \(x<-\tfrac12\).

• fgg first: 1/(2x + 1) into f
• x < −½the domain of g: the final check

2 What is it asking for? the target, part by part

• (a)x4 marks

3 What do we know? everything the question gives us

• \(f(x)=x^2-1\)x < 0
• \(g(x)=\dfrac1{2x+1}\)x < −½

4 Plan the route work backwards from what we want

FindMaths ideaAlready have itWork out firstTrap

• Findx
  • Use\(\dfrac{1}{(2x+1)^2}-1=3\Rightarrow(2x+1)^2=\tfrac14\)
  • Trapkeeping \(x=-\tfrac14\)

Isolate the bracket.

\(\tfrac{1}{(2x+1)^2}=4\Rightarrow(2x+1)^2=\tfrac14\Rightarrow2x+1=\pm\tfrac12\).

Which root satisfies x < −½?

\(x=-\tfrac34\). The other, \(-\tfrac14\), is bigger than \(-\tfrac12\).

5 Now solve it have a go on paper first, then open the worked solution

1. \(fg(x)=\dfrac{1}{(2x+1)^2}-1\) B1
2. \(=3\Rightarrow4(2x+1)^2=1\) M1
3. \(2x+1=\pm\tfrac12\) A1
4. \(x=-\tfrac34\) only A1

Alternative (peeling): \(f(u)=3\Rightarrow u^2=4\Rightarrow u=-2\) (as \(u<0\)); \(g(x)=-2\Rightarrow2x+1=-\tfrac12\Rightarrow x=-\tfrac34\).

6 Check it does the answer make sense?

\(g(-\tfrac34)=\tfrac{1}{-1/2}=-2\); \(f(-2)=4-1=3\) ✓.

Example 4 · When can gf be formed? (Paper 1, 3 marks) ★★★ Hard

Why it is hard: it reverses the usual job: instead of a domain giving a range, a required range gives the domain.

1 Read it with me what is actually happening?

\(gf\) means \(f\) first, then \(g\). \(g\) only accepts inputs less than 8, so every output of \(f\) must be less than 8. The completed square makes that inequality easy.

• possible to form gfrange of f within the domain of g (x < 8)
• least p, greatest qthe widest interval where \(f(x)<8\)

2 What is it asking for? the target, part by part

• (b)\(p\) and \(q\)3 marks

3 What do we know? everything the question gives us

• \(f(x)=(2x-3)^2+4\)from (a)
• \(g\): \(x<8\)

4 Plan the route work backwards from what we want

FindMaths ideaAlready have itWork out firstTrap

• Findp, q
  • Use\(f(x)<8\)
    • Solve\((2x-3)^2<4\)

Solve (2x − 3)² < 4 without squaring out.

\(-2<2x-3<2\Rightarrow1<2x<5\Rightarrow\tfrac12<x<\tfrac52\).

Why < 8 and not ≤ 8?

\(g\)'s domain is \(x<8\): 8 itself is not allowed.

5 Now solve it have a go on paper first, then open the worked solution

1. (a) \((2x-3)^2+4\) B1 B1
2. (b) Need \((2x-3)^2+4<8\) *M1
3. \((2x-3)^2<4\Rightarrow\tfrac12<x<\tfrac52\) DM1
4. least \(p=\tfrac12\), greatest \(q=\tfrac52\) A1

6 Check it does the answer make sense?

At \(x=\tfrac12\) or \(\tfrac52\), \(f=8\): exactly the edge, excluded ✓.

Summary

• \(fg(x)=f(g(x))\): the function next to \(x\) acts first. Never multiply.
• Numbers: inside out, one step at a time.
• Solve composites by forming them, or by peeling one layer at a time; reject roots outside the domain.
• gf exists only if range f ⊆ domain g; explain with both sets named.

Can I do this?

Quiz

Concept checks

Answer each one in your head first, then open it.

Exam-style practice

New questions, not used above. Do them on paper, then mark your work against the official mark scheme.