Form and use composite functions, understanding that the order of composition is important

Functions – Composite Functions (IGCSE Additional Mathematics 0606)

Learning Objective

Students will be able to:

• Form and use composite functions.
• Explain why the order of composition matters.
• Determine the domain and range of a composite function.
• Restrict domains to obtain one‑to‑one functions, find inverses, and interpret graphs (including absolute‑value forms).

1. Key Definitions and Notation

2. Why the Order of Composition Matters

• In general \((f\circ g)(x)eq(g\circ f)(x)\). Swapping the order changes the algebraic expression, the domain, and often the range.
• Only when two functions *commute* (a rare situation) will the two compositions be identical.
• Reading the notation from right to left helps avoid mistakes: the right‑most function acts first.

3. Forming a Composite Function – Step‑by‑Step Procedure

1. Identify the inner function. This is the function that acts on the variable \(x\) first.
2. Replace every occurrence of \(x\) in the outer function with the whole inner expression. Keep the parentheses.
3. Simplify algebraically. Expand, factor, combine fractions, etc., only as far as needed.
4. Determine the domain.
   • Start with the domain of the inner function.
   • From that set keep only the values for which the inner output lies in the domain of the outer function (watch for denominators, square‑roots, logarithms, etc.).

4. Worked Example (Linear ↔ Quadratic)

Let \(f(x)=2x+3\) and \(g(x)=x^{2}-1\).

1. \((f\circ g)(x)\) – “\(g\) first, then \(f\)”

   (f∘g)(x)=f(g(x))=2\bigl(x^{2}-1\bigr)+3
   =2x^{2}-2+3=2x^{2}+1.

   Domain: \(g\) is defined for all real \(x\); \(f\) has no restriction ⇒ domain \(\mathbb R\).

2. \((g\circ f)(x)\) – “\(f\) first, then \(g\)”

   (g∘f)(x)=g(f(x))=(2x+3)^{2}-1
   =4x^{2}+12x+9-1=4x^{2}+12x+8.

   Domain: both \(f\) and \(g\) are defined for all real numbers ⇒ domain \(\mathbb R\).

Since the two results differ, the order of composition is essential.

5. Domain & Range of Composite Functions

For a composite \((f\circ g)\):

• Domain: \(\{x\in\text{Dom}(g)\mid g(x)\in\text{Dom}(f)\}\).
• Range: Apply the range of \(g\) to the function \(f\); often found by algebraic manipulation or by testing extreme values.

6. Inverses – Restricting Domains and Sketching

6.1 Making a function one‑to‑one

Many useful functions (e.g. \(x^{2},\sqrt{x},\sin x\)) are not one‑to‑one on their natural domains. To obtain an inverse we restrict the domain so that the function becomes injective.

• Example: \(f(x)=x^{2}\) on \([0,\infty)\) is one‑to‑one. Its inverse is \(f^{-1}(x)=\sqrt{x}\) with domain \(x\ge0\).
• Example: \(g(x)=\sqrt{x-2}\) (domain \(x\ge2\)) → \(g^{-1}(x)=x^{2}+2\) (domain \(x\ge0\)).

6.2 Finding the inverse (algebraic method)

1. Replace \(f(x)\) by \(y\). Write the equation \(y = \text{(expression in }x\text{)}\).
2. Swap \(x\) and \(y\) (solve for the new \(y\)).
3. Write the result as \(f^{-1}(x)\) and state its domain (the original range).

Example (rational function)

f(x)=\frac{1}{x-1},\;xeq1
y=\frac{1}{x-1}\;\Longrightarrow\;x=\frac{1}{y}+1
\Rightarrow f^{-1}(x)=\frac{1}{x}+1,\;xeq0

6.3 Sketching a function and its inverse

• Draw the graph of \(y=f(x)\) on a set of axes.
• Draw the line \(y=x\) (the line of symmetry for an inverse).
• Reflect each point of the original graph across the line \(y=x\); the reflected curve is the graph of \(y=f^{-1}(x)\).
• Only the portion of the original graph that lies on one side of the horizontal‑line test (i.e., is one‑to‑one) can be reflected.

6.4 Why a function may have **no** inverse

A function fails to have an inverse when it is not one‑to‑one. The horizontal‑line test provides a quick visual check: if any horizontal line cuts the graph in more than one point, the function is not invertible.

Illustration: The parabola \(y=x^{2}\) fails the test because the line \(y=4\) meets it at \(x=2\) and \(x=-2\). Hence no single‑valued inverse exists unless the domain is restricted (e.g., to \(x\ge0\)).

7. Absolute‑Value Forms

The absolute‑value transformation \(|f(x)|\) leaves positive outputs unchanged and reflects negative outputs above the \(x\)-axis.

• Algebraically: \(|f(x)|=\begin{cases}f(x),&f(x)\ge0\\-f(x),&f(x)<0\end{cases}\).
• Graphically: take the part of the graph of \(y=f(x)\) that lies below the axis, flip it upwards, and keep the part already above unchanged.

Example \(f(x)=x-2\):
\(|f(x)|=\begin{cases}x-2,&x\ge2\\-(x-2)=2-x,&x<2\end{cases}\). The vertex of the V‑shape occurs at \((2,0)\).

8. Common Pitfalls & How to Avoid Them

• Swapping the order of composition. Write the inner function first in your working notes.
• Dropping parentheses. When substituting, replace \(x\) with the **entire** inner expression, e.g. \(f(2x+3)\), not \(f(2)x+3\).
• Ignoring domain restrictions. After substitution, re‑check for zero denominators, negative radicands, or non‑positive logarithm arguments.
• Assuming a function has an inverse. Apply the horizontal‑line test; if it fails, restrict the domain or state that no inverse exists.
• Forgetting the absolute‑value effect. Remember that \(|f(x)|\) never yields negative values; adjust the graph accordingly.

9. Practice Questions

1. Given \(h(x)=3x-4\) and \(k(x)=\sqrt{x+5}\) (domain \(x\ge-5\)), find \((h\circ k)(x)\) and state its domain.
2. For \(p(x)=\dfrac{1}{x}\) and \(q(x)=x-2\), compute \((q\circ p)(x)\), simplify, and give the domain.
3. Determine whether \(f(x)=x^{3}\) and \(g(x)=x^{2}\) commute; i.e. check if \(f\circ g = g\circ f\).
4. Let \(r(x)=\dfrac{2x+1}{x-3}\) and \(s(x)=\dfrac{5}{x}\). Find \((r\circ s)(x)\) and list all values of \(x\) that must be excluded.
5. Find the domain of the composite \((\ln\circ\sqrt{\;})(x)=\ln\bigl(\sqrt{x}\bigr)\). Explain each restriction.
6. Given \(f(x)=\dfrac{1}{x-1}\), \(g(x)=\dfrac{x+2}{x}\), and \(h(x)=\sqrt{x-3}\), write \((f\circ g\circ h)(x)\) and state its domain.
7. Restrict the domain of \(u(x)=x^{2}\) so that an inverse exists, then find \(u^{-1}(x)\) and sketch both curves (use a simple hand‑drawn sketch or describe the shape).
8. Sketch the graph of \(v(x)=|x-2|\) and its inverse (if it exists). Explain why an inverse does or does not exist.
9. Find the inverse of the radical function \(w(x)=\sqrt{x-2}\) and state the domain of the inverse.
10. Consider the three‑function composition from the extension activity:
    • \(f(x)=\dfrac{1}{x}\), \(g(x)=x+2\), \(h(x)=\sqrt{x-1}\) (domain \(x\ge1\)).
    • Write \((f\circ g\circ h)(x)\) and \((h\circ g\circ f)(x)\); simplify and give the domains.

10. Answers & Explanations

1. \((h\circ k)(x)=h\bigl(k(x)\bigr)=3\sqrt{x+5}-4\).
   Domain: \(x\ge-5\) (required by the square‑root). No further restriction from the linear function.
2. \((q\circ p)(x)=q\bigl(p(x)\bigr)=\dfrac{1}{x}-2=\dfrac{1-2x}{x}\).
   Domain: \(xeq0\) (denominator of \(p\)). No extra restriction from \(q\).
3. \(f\circ g(x)=f(g(x))=(x^{2})^{3}=x^{6}\).
   \(g\circ f(x)=g(f(x))=(x^{3})^{2}=x^{6}\).
   The two expressions are identical ⇒ the functions commute.
4. \[ (r\circ s)(x)=r\!\left(\frac{5}{x}\right)=\frac{2\left(\frac{5}{x}\right)+1}{\frac{5}{x}-3} =\frac{\frac{10}{x}+1}{\frac{5-3x}{x}} =\frac{10+x}{5-3x}. \] Exclusions: \(x=0\) (from \(s\)) and \(5-3x=0\Rightarrow x=\frac{5}{3}\) (final denominator).
   Domain: \(\mathbb R\setminus\{0,\frac{5}{3}\}\).
5. \((\ln\circ\sqrt{\;})(x)=\ln(\sqrt{x})=\tfrac12\ln x\).
   Restrictions: \(\sqrt{x}\) needs \(x\ge0\); \(\ln\) needs its argument \(>0\). Hence \(x>0\) (zero is excluded because \(\ln0\) is undefined).
   Domain: \(x>0\).
6. First \(h\): \(h(x)=\sqrt{x-3}\) ⇒ \(x\ge3\).
   Next \(g\): \(g\bigl(h(x)\bigr)=\dfrac{\sqrt{x-3}+2}{\sqrt{x-3}}\) – denominator \(eq0\) ⇒ \(\sqrt{x-3}eq0\) ⇒ \(x>3\).
   Finally \(f\): \(f\bigl(g(h(x))\bigr)=\dfrac{1}{\dfrac{\sqrt{x-3}+2}{\sqrt{x-3}}-1} =\dfrac{1}{\dfrac{2}{\sqrt{x-3}}}= \dfrac{\sqrt{x-3}}{2}\).
   Domain: \(x>3\).
7. Restrict \(u(x)=x^{2}\) to \(x\ge0\). Then \(u\) is one‑to‑one.
   Inverse: \(y=x^{2}\Rightarrow x=\sqrt{y}\) ⇒ \(u^{-1}(x)=\sqrt{x}\) with domain \(x\ge0\).
   Sketch: a right‑hand half of the parabola for \(u\); the inverse is the same curve reflected across \(y=x\) (still a right‑hand half‑parabola). Both meet at \((0,0)\).
8. \(v(x)=|x-2|\) is V‑shaped, vertex \((2,0)\). Because the graph fails the horizontal‑line test (e.g., the line \(y=1\) meets it at two points), no single‑valued inverse exists. (If the domain were restricted to \(x\ge2\) or \(x\le2\), an inverse could be defined.)
9. Let \(y=\sqrt{x-2}\) ⇒ \(y^{2}=x-2\) ⇒ \(x=y^{2}+2\).
   Hence \(w^{-1}(x)=x^{2}+2\) with domain \(x\ge0\) (the range of the original square‑root).
10. (i) \((f\circ g\circ h)(x)\): \[ h(x)=\sqrt{x-1},\; g(h)=\sqrt{x-1}+2,\; f(g(h))=\frac{1}{\sqrt{x-1}+2}. \] Domain: \(x\ge1\) (from \(h\)).
    
    (ii) \((h\circ g\circ f)(x)\): \[ f(x)=\frac{1}{x},\; g(f)=\frac{1}{x}+2,\; h(g(f))=\sqrt{\frac{1}{x}+2-1} =\sqrt{\frac{1}{x}+1}. \] Domain: \(\frac{1}{x}+1\ge0\) and denominator \(eq0\) ⇒ \(x>0\).
    The two results differ, confirming that order matters.

11. Quick Checklist for Composite Functions

• Write the inner (right‑most) function first.
• Substitute the whole inner expression for every \(x\) in the outer function; keep parentheses.
• Simplify step‑by‑step; expand only when it helps.
• Determine the domain:
  1. Start with the inner function’s domain.
  2. Apply any extra restrictions that arise when the inner output is used in the outer function (denominator ≠ 0, radicand ≥ 0, log argument > 0, etc.).
• Check whether the two functions commute only after you have both explicit expressions.
• When an inverse is required, first verify the function is one‑to‑one (or restrict the domain appropriately).
• Remember: \(|f(x)|\) never produces negative values; adjust the graph accordingly.

12. Summary Table

13. Extension Activity – Three‑Function Composition

Let \(f(x)=\dfrac{1}{x}\), \(g(x)=x+2\), \(h(x)=\sqrt{x-1}\) (domain \(x\ge1\)).

1. Write and simplify \((f\circ g\circ h)(x)=f\bigl(g(h(x))\bigr)\).
   Answer: \(\displaystyle \frac{1}{\sqrt{x-1}+2},\; \text{domain }x\ge1.\)
2. Write and simplify \((h\circ g\circ f)(x)=h\bigl(g(f(x))\bigr)\).
   Answer: \(\displaystyle \sqrt{\frac{1}{x}+1},\; \text{domain }x>0.\)
3. Compare the two results. Discuss how the order changes both the algebraic form and the domain, reinforcing the central idea that composition is not commutative.