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Published July 19, 2026

Working with Composite and Inverse Functions

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Composite and Inverse Functions

Functions are rules that turn an input into an output. But what happens when you combine or reverse these rules? Composite and inverse functions are key higher-tier concepts for understanding multi-step processes — from calculating discounts to decrypting codes.

Combining and Reversing Functions

Composite Functions: Chaining Rules

A composite function applies one function to the result of another. The notation $fg(x)$ means apply $g$ first, then apply $f$ to the output.

Flowchart showing composite function f g of x. Input x enters the box labelled g, producing g of x. That output enters the box labelled f, producing f of g of x, which is the final output labelled f g of x. Four boxes connected by arrows. The first box is labelled Input x, the second g, the third f, and the fourth Output fg(x). Arrows flow left to right. Input x Apply g first g(x) Then apply f f(g(x)) Output fg(x)

Inverse Functions: Undoing the Rule

An inverse function, written $f^{-1}(x)$, reverses the original function. If $f(x)$ takes you from A to B, then $f^{-1}(x)$ takes you back from B to A.

Key facts:

  • $f^{-1}(f(x)) = x$.
  • The graph of $f^{-1}(x)$ is a reflection of $f(x)$ in the line $y = x$.
  • Important: $f^{-1}(x)$ does not mean $\dfrac{1}{f(x)}$.

Worked Examples

Example 1: Evaluating Composite Functions

Given $f(x) = 3x – 2$ and $g(x) = x^2 + 1$, find $fg(2)$.

  1. Work inside out — find $g(2)$ first.
    $g(2) = (2)^2 + 1 = 5$.
  2. Use this as the input for $f$: find $f(5)$.
    $f(5) = 3(5) – 2 = 13$.

Answer: $fg(2) = 13$

Example 2: Finding a Composite Expression

Given $f(x) = 2x + 5$ and $g(x) = x – 3$, find an expression for $fg(x)$.

  1. $fg(x) = f(g(x))$ — substitute $g(x)$ into $f$ in place of $x$.
  2. $f(g(x)) = 2(x – 3) + 5$.
  3. Expand and simplify: $2x – 6 + 5$.

Answer: $fg(x) = 2x – 1$

Example 3: Finding an Inverse Function

Find the inverse of $f(x) = 4x + 7$.

  1. Replace $f(x)$ with $y$: $y = 4x + 7$.
  2. Rearrange to make $x$ the subject:
    $y – 7 = 4x \implies x = \dfrac{y-7}{4}$.
  3. Swap $x$ and $y$: $y = \dfrac{x-7}{4}$.
  4. Replace $y$ with $f^{-1}(x)$.

Answer: $f^{-1}(x) = \dfrac{x-7}{4}$

Example 4: Inverse with a Square Root

Find the inverse of $g(x) = \sqrt{x – 3}$ for $x \geq 3$.

  1. Let $y = \sqrt{x – 3}$.
  2. Square both sides: $y^2 = x – 3$.
  3. Make $x$ the subject: $x = y^2 + 3$.
  4. Swap $x$ and $y$: $y = x^2 + 3$.

Answer: $g^{-1}(x) = x^2 + 3$

Tutor Insights

🤔 Common Misunderstandings

  • Order of operations: Mixing up $fg(x)$ and $gf(x)$. Work from right to left — apply $g$ first, then $f$.
  • Inverse notation: Thinking $f^{-1}(x)$ means $\frac{1}{f(x)}$. The $^{-1}$ means inverse, not a reciprocal.
  • Forgetting to swap: Missing the final step of swapping $x$ and $y$ when finding an inverse.

📝 Common Exam Mistakes

  • Algebraic errors when expanding brackets or rearranging during the inverse method.
  • Incorrect order: Calculating $gf(x)$ when $fg(x)$ was asked. Read carefully!
  • Not showing working: All steps of rearrangement must be visible to earn full method marks for inverse functions.

Practice Questions

  1. Given $f(x) = 2x – 3$ and $g(x) = x^2$, find $gf(4)$.
  2. Find the inverse function of $h(x) = 5x + 10$.
  3. Given $f(x) = 3x + 1$ and $g(x) = 2x – 7$, find an expression for $fg(x)$.
  4. Find the inverse function of $k(x) = \dfrac{x – 4}{3}$.
Show Answers
  1. Working: $f(4) = 2(4) – 3 = 5$. Then $g(5) = 5^2 = 25$.
    Answer: $gf(4) = 25$.
  2. Working: $y = 5x + 10 \implies y – 10 = 5x \implies x = \dfrac{y-10}{5}$. Swap $x, y$.
    Answer: $h^{-1}(x) = \dfrac{x-10}{5}$.
  3. Working: $fg(x) = f(2x-7) = 3(2x-7)+1 = 6x – 21 + 1$.
    Answer: $fg(x) = 6x – 20$.
  4. Working: $y = \dfrac{x-4}{3} \implies 3y = x – 4 \implies x = 3y + 4$. Swap $x, y$.
    Answer: $k^{-1}(x) = 3x + 4$.

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