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Method · Functions

How to Find an Inverse Function

The inverse f1f^{-1} undoes ff: if f(a)=bf(a)=b then f1(b)=af^{-1}(b)=a. To find it, let y=f(x)y=f(x), make xx the subject, then write f1(x)f^{-1}(x) by replacing yy with xx.

What an inverse function is for

An inverse function reverses what the original function does. If ff takes an input and produces an output, then the inverse f1f^{-1} takes that output and returns the original input, so f(a)=bf(a)=b means exactly f1(b)=af^{-1}(b)=a.

Because of this undoing, composing a function with its inverse gives back the input unchanged: ff1(x)=xff^{-1}(x)=x and f1f(x)=xf^{-1}f(x)=x. This method answers questions such as "find f1(x)f^{-1}(x)", "evaluate f1(7)f^{-1}(7)", or "state the value of xx for which f1f^{-1} is undefined".

It is a core Form 4 Functions skill. A function only has an inverse if it is one-to-one, each output comes from exactly one input, which is why the topic is closely tied to domain and range.

Thinking of f1f^{-1} as a reverse machine keeps the algebra honest. Whatever ff does last, f1f^{-1} must undo first.

This picture also explains why the graph of f1f^{-1} is the reflection of the graph of ff in the line y=xy=x.

When to use this method

Reach for this method whenever a question uses the symbol f1f^{-1} and asks you to find it as an expression, to evaluate it at a number, or to state where it is defined. Typical wording is "find f1(x)f^{-1}(x)" or "hence find f1(5)f^{-1}(5)".

You also use it when a question asks you to show two functions are inverses of each other, which you confirm by checking that their composite is xx. Another cue is a graph question asking for a reflection in y=xy=x.

Because Functions opens the syllabus, inverse-function work appears in both Paper 1 and Paper 2, often linked to composite functions. If you see f1f^{-1} anywhere, this is the tool, but first make sure the function is one-to-one so the inverse actually exists.

The steps

Follow this order for a clean inverse every time:

  1. 1

    Write y=f(x)y=f(x)

    Replace the function name with yy, so a rule like f(x)=3x+1f(x)=3x+1 becomes y=3x+1y=3x+1.

  2. 2

    Make xx the subject

    Rearrange the equation to get xx alone on one side, undoing each operation in reverse order.

  3. 3

    Swap yy for xx

    Replace every yy with xx and write the result as f1(x)f^{-1}(x). This gives the inverse in standard form.

  4. 4

    State any restriction

    If the inverse involves a fraction or a square root, note the values of xx it excludes; the domain of f1f^{-1} equals the range of ff.

  5. 5

    Check with a composite

    Confirm your answer by showing ff1(x)=xff^{-1}(x)=x. If it simplifies to xx, the inverse is correct.

Undo in reverse order

If ff multiplies then adds, the inverse subtracts then divides. Reversing the order of operations is what makes the rearrangement reliable.

Worked example

Try this yourself first, then check each line against the solution.

Q1[4 marks]

The function ff is defined by f(x)=3x+1f(x)=3x+1. Find f1(x)f^{-1}(x), and hence evaluate f1(10)f^{-1}(10).

Show worked solution

Start by writing y=f(x)y=f(x):

y=3x+1y=3x+1

Make xx the subject. Subtract 1 from both sides, then divide by 3:

y1=3xx=y13y-1=3x \quad\Rightarrow\quad x=\frac{y-1}{3}

Now swap yy for xx to write the inverse in standard form:

f1(x)=x13f^{-1}(x)=\frac{x-1}{3}

Evaluate at x=10x=10:

f1(10)=1013=93=3f^{-1}(10)=\frac{10-1}{3}=\frac{9}{3}=3

Answer

f1(x)=x13f^{-1}(x)=\dfrac{x-1}{3} and f1(10)=3f^{-1}(10)=3. Check: ff1(x)=3(x13)+1=(x1)+1=xff^{-1}(x)=3\left(\dfrac{x-1}{3}\right)+1=(x-1)+1=x, so the inverse is correct.

Also f(3)=3(3)+1=10f(3)=3(3)+1=10, matching f1(10)=3f^{-1}(10)=3.

Notice the two independent checks in the note. Showing ff1(x)=xff^{-1}(x)=x confirms the general rule, while f(3)=10f(3)=10 confirms the single value.

When both agree, you can move on without doubt.

Common mistakes to avoid

  • Confusing f1(x)f^{-1}(x) with 1f(x)\frac{1}{f(x)}. The inverse is not a reciprocal; the 1-1 is a notation for "reverse", not a power.
  • Undoing operations in the wrong order. Reverse the order: if ff multiplies then adds, undo by subtracting then dividing.
  • Forgetting to swap yy and xx at the end, leaving the answer in terms of yy.
  • Missing the domain restriction when the inverse has a fraction or a root, so the excluded value is not stated.
  • Trying to invert a function that is not one-to-one; without a suitable domain restriction the inverse does not exist.

How a teacher helps you get it right

The single step that costs marks is the rearrangement, the moment you make xx the subject and one operation gets undone out of order. In a one-to-one lesson our teacher watches that exact line, sees whether you divided before subtracting or forgot to swap the variables, and has you redo just that step until it is automatic.

Because our teachers are experienced, you get someone who links the algebra back to the "reverse machine" idea so it makes sense, not just sticks. Lessons are online and taught in English, while SPM papers are set in both Malay and English.

We also build the habit of the composite check, ff1(x)=xff^{-1}(x)=x, so a wrong inverse is caught before it reaches the answer line.

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Frequently asked questions

Is f1(x)f^{-1}(x) the same as 1f(x)\frac{1}{f(x)}?

No. f1f^{-1} is the inverse function, which reverses what ff does; the 1-1 is notation, not a power.

The reciprocal 1f(x)\frac{1}{f(x)} is a completely different quantity.

How do I find f1(x)f^{-1}(x) step by step?

Write y=f(x)y=f(x), make xx the subject by undoing each operation in reverse order, then swap yy for xx and write the result as f1(x)f^{-1}(x). Finish by checking that ff1(x)=xff^{-1}(x)=x.

How can I check my inverse is correct?

Compose the function with your answer. If ff1(x)ff^{-1}(x) simplifies to xx (and f1f(x)=xf^{-1}f(x)=x), the inverse is right.

You can also test one value: if f(a)=bf(a)=b, then f1(b)f^{-1}(b) should give aa.

Why must a function be one-to-one to have an inverse?

If two different inputs gave the same output, the reverse process could not decide which input to return. Being one-to-one means every output comes from exactly one input, so the inverse is well defined.

What is the link between the graphs of ff and f1f^{-1}?

The graph of f1f^{-1} is the reflection of the graph of ff in the line y=xy=x. Points (a,b)(a,b) on ff become points (b,a)(b,a) on f1f^{-1}.

Source:SRC-DSKP-EN

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

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