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KBAT · Functions

KBAT: Reasoning About Functions

A functions KBAT question tests reasoning, not recall: you decide the order of a composite like gf(x)gf(x), find where it is undefined, invert a function, and judge whether an equation can be solved at all. The techniques are Form 4 functions; the higher-order part is choosing them and reading the result correctly.

What makes this a KBAT question

A routine functions question gives one clear instruction,'find fg(x)fg(x)' or 'find f1(x)f^{-1}(x)', and you carry out a single technique. A KBAT question strings several ideas together and asks you to reason about the result.

You might build a composite, decide where it is undefined, invert a function, and then judge whether an equation involving them can be solved at all. That is Kemahiran Berfikir Aras Tinggi, higher-order thinking: every individual step is familiar Form 4 work, but you meet them in an unfamiliar order and no one tells you which to use.

In Add Math this rewards students who understand what a function actually does, how inputs map to outputs, and why a value can be forbidden, rather than students who only memorise procedures.

One worked problem, in the style of Paper 2

This is an original question written in the style of SPM Paper 2. Try it yourself before reading the solution.

Q1[7 marks]

Two functions are defined by f(x)=2x+3f(x)=2x+3 and g(x)=xx1g(x)=\frac{x}{x-1}, x1x\neq 1. (a) Find the composite function gf(x)gf(x), and state the value of xx for which gf(x)gf(x) is undefined.

(b) Find the inverse function f1(x)f^{-1}(x). (c) A student claims that the equation gf(x)=1gf(x)=1 has a real solution.

By solving gf(x)=1gf(x)=1, determine whether the claim is correct and explain what your result means.

Show worked solution

Understand. We are given a linear function ff and a rational function gg that is undefined at x=1x=1.

Part (a) asks for gfgf, apply ff first, then gg, and the value where it breaks down. Part (b) is the inverse of the linear function.

Part (c) asks whether the composite can ever equal 11.

Plan. For gf(x)gf(x), substitute f(x)=2x+3f(x)=2x+3 into gg.

The composite is undefined where its denominator is zero, which is exactly where f(x)=1f(x)=1, the input gg cannot accept. For the inverse, let y=f(x)y=f(x) and make xx the subject.

For part (c), set gf(x)=1gf(x)=1 and solve honestly.

Execute and check. (a) Substitute f(x)=2x+3f(x)=2x+3 as the input of gg:

gf(x)=g(2x+3)=2x+3(2x+3)1=2x+32x+2gf(x)=g(2x+3)=\frac{2x+3}{(2x+3)-1}=\frac{2x+3}{2x+2}

The composite is undefined when the denominator is zero, 2x+2=02x+2=0, giving x=1x=-1. (This is precisely the value at which f(x)=1f(x)=1, the input gg is not allowed to take.)

(b) Let y=2x+3y=2x+3. Making xx the subject, x=y32x=\frac{y-3}{2}, so

f1(x)=x32f^{-1}(x)=\frac{x-3}{2}

(c) Solve gf(x)=1gf(x)=1:

2x+32x+2=1    2x+3=2x+2    3=2\frac{2x+3}{2x+2}=1 \;\Rightarrow\; 2x+3=2x+2 \;\Rightarrow\; 3=2

The statement 3=23=2 is impossible, so no value of xx satisfies the equation. The claim is therefore incorrect: gf(x)gf(x) is never equal to 11.

The line y=1y=1 is a horizontal asymptote of gfgf, the composite gets closer and closer to 11 as xx grows large, but never actually reaches it.

Check. Testing a large value, x=100x=100 gives gf(100)=2032021.005gf(100)=\frac{203}{202}\approx 1.005, close to 11 but not equal, while gf(0)=32gf(0)=\frac{3}{2}.

Both readings are consistent with a value that approaches 11 without attaining it, so the conclusion holds.

Finding a sensible first step

When a functions question looks tangled, the reliable first move is to read the notation slowly and settle the order before touching algebra. gf(x)gf(x) means 'do ff first, then gg', so the output of ff becomes the input of gg.

Write that substitution explicitly, g(f(x))g(f(x)), before simplifying, and you rarely reverse the order by accident. Next, note any restriction each function carries, because those restrictions travel into the composite.

Ask: what input is gg forbidden? Here gg cannot take 11, so the composite is undefined wherever f(x)=1f(x)=1.

Naming the order and the restriction at the start turns an unfamiliar question into a sequence of ordinary steps, and it protects the method marks that the later working depends on.

What markers reward

Marking is analytic, so method marks are awarded line by line. On a functions-reasoning question a marker looks for:

  • The correct order of composition, gfgf means substitute ff into gg, a mark frequently lost by reversing it.
  • The composite shown as a single simplified expression, such as 2x+32x+2\frac{2x+3}{2x+2}.
  • The excluded value stated with a reason, the denominator is zero, or gg cannot accept that input.
  • The inverse found by a valid method (let y=f(x)y=f(x), make xx the subject), with the answer written in terms of xx.
  • The reasoning part carried through to the contradiction 3=23=2, then a clear conclusion drawn from it.
  • An interpretation of the result,'no solution; y=1y=1 is never reached', not merely 'no answer'.

How a teacher helps

Reasoning with functions improves fastest when a student explains their thinking out loud and gets it checked. In a one-to-one lesson our teachers ask you to say the order before you write it, to name each restriction as it appears, and to state in words what an answer means.

We slow down the moment students usually rush, the jump from a contradiction like 3=23=2 to the conclusion 'therefore no solution', because that interpretation earns marks. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

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Frequently asked questions

What is the difference between fg(x)fg(x) and gf(x)gf(x)?

Order matters. fg(x)fg(x) means 'do gg first, then ff', while gf(x)gf(x) means 'do ff first, then gg'.

They usually give different expressions, so reversing the order is one of the most common lost marks, always write g(f(x))g(f(x)) explicitly before simplifying.

How do I state the restriction on a composite function?

A composite inherits every restriction of the functions inside it. Ask where the inner function's output is forbidden by the outer one.

Here gg cannot take the input 11, so gf(x)gf(x) is undefined where f(x)=1f(x)=1, i.e. x=1x=-1, the same value that makes the composite's denominator zero.

Can an equation involving functions really have no solution?

Yes. When solving gf(x)=1gf(x)=1 collapses to an impossible statement like 3=23=2, it means no value of xx works.

Geometrically the value 11 is a horizontal asymptote of gfgf, the graph approaches it but never touches it, so the equation has no real solution.

Can I earn marks if my final answer is wrong?

Yes. Add Math Paper 2 is 2 hours 30 minutes and 100 marks, and marking is analytic, method marks are awarded line by line.

A clear composite, a correct restriction and honest working can score well even if a later slip changes the final number.

Source:SRC-DSKP-ENSRC-FORMAT

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

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