# Solving function equations and inequalities

> IB Mathematics: Analysis and Approaches HL · Functions
> Source: https://www.owlsprep.com/study/ib-math-aa-hl-u2-solving-function-equations-and-inequalities/

This module covers algebraic and graphical methods to solve equations and inequalities involving functions. You will learn to identify domain restrictions, handle extraneous solutions, and approach common IB exam question types for this topic.

**Prerequisites:** [Domain and range of functions](https://www.owlsprep.com/study/ib-math-aa-hl-u2-domain-and-range-of-functions/); [Solving quadratic equations and inequalities](https://www.owlsprep.com/study/ib-math-aa-hl-u1-quadratic-equations-inequalities/)

## Learning objectives

- Solve common types of function equations algebraically
- Identify and reject extraneous solutions caused by domain restrictions
- Solve function inequalities using algebraic and graphical methods
- Avoid common exam traps when working with function relations

## Solving Function Equations Algebraically

**Function equation** — An equation that involves one or more functions, requiring either the form of an unknown function or input values that satisfy the relation

*Example:* $f(x+1) = 2x + 3$ or $f(2x) = f(x)^2$

Most IB exam questions ask you to find input values that satisfy an equation involving a known function. Common strategies include substitution of composite inputs, equating coefficients for unknown functions, and using function properties like odd/even symmetry.

**Worked example:** Given $f(x) = x^2 - 4x$, solve $f(x+1) = 5$

1. Substitute $(x+1)$ into the function definition and expand:
2. $$f(x+1) = (x+1)^2 - 4(x+1) = x^2 - 2x - 3$$
3. Set equal to 5 and rearrange to standard quadratic form:
4. $$x^2 - 2x - 3 = 5 \implies x^2 - 2x - 8 = 0$$
5. Factor and solve for $x$:
6. $$(x-4)(x+2) = 0 \implies x = 4, \; x = -2$$
7. $f(x)$ is a polynomial with domain all real numbers, so both solutions are valid.

**Check your understanding**

1. Which of the following is a function equation?

   - $3x + 5 = 0$
   - $f(x) = 3x + 5$
   - $f(x+2) = 4x - 1$
   - $y = 2x^2$

   *Answer:* $f(x+2) = 4x - 1$

   *Why:* A is a linear equation in $x$, B is a function definition, C is a function equation that asks for either $x$ or the form of $f$.

## Domain Restrictions and Extraneous Solutions

When solving function equations, the original function has a fixed domain. Any solution that lies outside this domain is extraneous and must be rejected. Common sources of restrictions include square roots, denominators, and logarithms.

> **warning**
>
> You will lose 1-2 marks in IB exams if you do not reject extraneous solutions. Always check all solutions against the original equation's domain.

**Worked example:** Solve $\sqrt{2x + 5} = x + 1$ for real $x$

1. First identify domain restrictions: the expression under the square root must be non-negative, and the left side (a square root) is non-negative, so the right side must also be non-negative:
2. $$2x + 5 \geq 0 \implies x \geq -\frac{5}{2}, \quad x + 1 \geq 0 \implies x \geq -1$$
3. Square both sides to eliminate the square root:
4. $$(\sqrt{2x + 5})^2 = (x+1)^2 \implies 2x + 5 = x^2 + 2x + 1$$
5. Simplify and solve:
6. $$x^2 - 4 = 0 \implies x = 2, \; x = -2$$
7. Check against the domain: $x = -2 < -1$, so it is extraneous. Only $x=2$ is valid. Verify in the original equation to confirm.

## Solving Function Inequalities

**Function inequality** — An inequality that requires finding all input values that satisfy a relation between two function expressions

*Example:* $\frac{1}{x} > x$

Two methods are common: algebraic interval testing and graphical intersection. The key to avoiding mistakes is to never multiply both sides by a variable term that can be negative, as this flips the inequality sign unpredictably.

> **tip**
>
> Always rearrange all terms to one side of the inequality, leaving zero on the other side. This avoids sign errors entirely.

**Worked example:** Solve $\frac{1}{x} > x$ for real $x \neq 0$

1. Rearrange all terms to the left side:
2. $$\frac{1}{x} - x > 0 \implies \frac{1 - x^2}{x} > 0 \implies \frac{(1-x)(1+x)}{x} > 0$$
3. Identify critical points (where numerator/denominator is zero): $x = -1, 0, 1$. These split the number line into 4 intervals.
4. Test the sign of the expression in each interval: 1. $x < -1$: negative → no solution; 2. $-1 < x < 0$: positive → solution; 3. $0 < x <1$: negative → no solution; 4. $x>1$: positive → solution.
5. The final solution set is $-1 < x < 0$ or $x > 1$.

## Common pitfalls

- **Wrong:** Multiplying both sides of a function inequality by a variable expression without flipping the inequality sign for negative values
  - Why it fails: Variable expressions can be positive or negative, and the inequality sign flips when multiplying by a negative
  - Correct: Rearrange all terms to one side to get an expression greater/less than zero, then test intervals around critical points
- **Wrong:** Forgetting to reject extraneous solutions after solving a radical or rational function equation
  - Why it fails: Algebraic steps like squaring both sides introduce solutions that do not satisfy the original equation
  - Correct: Always check all solutions against the original equation and its domain before writing your final answer
- **Wrong:** Including points where the function is undefined in the solution set of an inequality
  - Why it fails: Points of discontinuity or vertical asymptotes are not part of the domain, so they cannot satisfy the inequality
  - Correct: Exclude all undefined points from your solution, use open intervals around asymptotes
- **Wrong:** Incorrect substitution of composite inputs in function equations
  - Why it fails: Mixing up substitution order leads to incorrect starting equations and wrong solutions
  - Correct: Substitute the inner input step by step, expand and simplify carefully before solving

## Cheatsheet

| Strategy | Use Case | Key Exam Tip |
| --- | --- | --- |
| Algebraic substitution | Solve $f(g(x)) = k$ | Check domain after solving |
| Interval testing | All function inequalities | Rearrange to 0 on one side first |
| Graphical intersection | Find $f(x) > g(x)$ | Mark roots/asymptotes before reading intervals |
| Equating coefficients | Find unknown function $f$ | Match all powers, including constants |

## What's next

This sub-topic builds on core function properties you will use throughout the entire IB AA HL course, from polynomial algebra to differential calculus. Mastery of solving function equations and identifying extraneous solutions prepares you for more advanced topics like composite functions, inverse functions, and finding roots of higher-degree polynomials in upcoming units. You will also regularly apply these skills when solving optimization problems in calculus, where you need to determine where functions change sign to identify critical points, maxima, and minima. The habit of checking domain restrictions you develop here will help you avoid common mistakes in nearly every other topic you study for the exam.

- [Composite and Inverse Functions](https://www.owlsprep.com/study/ib-math-aa-hl-u2-composite-and-inverse-functions/)
- [Geometry & Trigonometry](https://www.owlsprep.com/study/ib-math-aa-hl-u3-overview/)
- [Radians and the unit circle](https://www.owlsprep.com/study/ib-math-aa-hl-u3-radians-and-the-unit-circle/)

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