Unit 4 · 11th Grade Mathematics

Rational & Radical
Functions.

Inverse functions, rational expressions, radical equations, and extraneous solutions.

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Unit 4 — Lesson Video

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Topic A: Inverse Functions & Radical Functions

Lessons 1–5 · Inverses of higher-degree functions, radical expressions and graphs.

01–02 — Lessons 1–2

Inverse Functions (Higher Degree)

Extending inverse function concepts to non-linear functions.

  • A function must be one-to-one (passes horizontal line test) to have an inverse.
  • Domain restriction: limit the domain so a function becomes invertible.
  • f⁻¹(f(x)) = x and f(f⁻¹(x)) = x — they undo each other.
  • Graphs of f and f⁻¹ are reflections over the line y = x.
f(x) = x² on x ≥ 0 has inverse f⁻¹(x) = √x. (Without restriction, x² fails the HLT.)
Practice — Lessons 1–2
Q1 · Multiple Choice

Why does f(x) = x² (over all reals) NOT have an inverse function?

Q2 · Multiple Choice

The graphs of f and f⁻¹ are reflections over which line?

03–05 — Lessons 3–5

Radical Functions

Radical functions involve roots (√, ∛, etc.). They are the inverses of power functions.

  • f(x) = √x: domain x ≥ 0, range y ≥ 0. Starts at origin, curves right.
  • f(x) = ∛x: domain all reals (cube root handles negatives).
  • Transformations: f(x) = a√(x − h) + k shifts/stretches like other functions.
  • Domain restricted by the radicand: set expression under √ ≥ 0.
f(x) = √(x − 3): domain x ≥ 3 (need x−3 ≥ 0). Starts at (3, 0).
Practice — Lessons 3–5
Q3 · Short Answer

What is the domain of f(x) = √(2x − 8)? Write as x ≥ ?

Q4 · Multiple Choice

Which radical function has domain = all real numbers?

Topic B: Rational Functions

Lessons 6–11 · Domain, asymptotes, simplification, and graphing.

06–08 — Lessons 6–8

Domain, Asymptotes & Features

Rational functions are ratios of polynomials: f(x) = p(x)/q(x). They have restrictions where q(x) = 0.

  • Domain: all x where denominator ≠ 0.
  • Vertical asymptote: x = a where denominator = 0 (and factor doesn't cancel).
  • Hole: factor cancels from numerator and denominator at x = a.
  • Horizontal asymptote: determined by comparing degrees of numerator and denominator.
f(x) = (x+2)/[(x−3)(x+1)]: domain excludes x=3 and x=−1; VAs at x=3 and x=−1.
Practice — Lessons 6–8
Q5 · Short Answer

Where is the vertical asymptote of f(x) = 1/(x − 5)?

Q6 · Multiple Choice

For f(x) = (x−2)/(x²−4) = (x−2)/[(x−2)(x+2)], what exists at x = 2?

09–11 — Lessons 9–11

Operations with Rational Expressions

Add, subtract, multiply, and divide rational expressions like fractions.

  • Multiply: factor, cancel common factors, then multiply across.
  • Divide: multiply by the reciprocal of the divisor.
  • Add/Subtract: find a common denominator, combine numerators.
  • Always state domain restrictions (values that make any denominator zero).
(x+1)/(x−2) · (x−2)/(x+3) = (x+1)/(x+3), x ≠ 2, x ≠ −3.
Practice — Lessons 9–11
Q7 · Multiple Choice

To add 1/(x+2) + 1/(x−3), the LCD is:

Q8 · Multiple Choice

To divide (x+1)/x ÷ (x+1)/x², you should:

Topic C: Solving Rational & Radical Equations

Lessons 12–18 · Solving equations, checking for extraneous solutions.

12–15 — Lessons 12–15

Solving Rational Equations

Multiply both sides by the LCD to clear denominators, then solve.

  • Step 1: Find the LCD of all denominators.
  • Step 2: Multiply every term by the LCD.
  • Step 3: Solve the resulting equation.
  • Step 4: Check for extraneous solutions — values that make the original denominator zero.
1/(x−1) = 2/(x+1) → cross-multiply: x+1 = 2(x−1) → x+1 = 2x−2 → x = 3. Check: x≠1, x≠−1 ✓
Practice — Lessons 12–15
Q9 · Multiple Choice

After solving a rational equation, you get x = 4. The original equation had (x − 4) in the denominator. Is x = 4 valid?

Q10 · Multiple Choice

Why do you always check solutions when solving rational equations?

16–18 — Lessons 16–18

Solving Radical Equations

Isolate the radical, then raise both sides to a power to eliminate it.

  • Step 1: Isolate the radical on one side.
  • Step 2: Square (or cube, etc.) both sides.
  • Step 3: Solve the resulting equation.
  • Step 4: Check all solutions — squaring can introduce extraneous solutions.
√(x+5) = 3 → square both sides: x+5 = 9 → x = 4. Check: √9 = 3 ✓
Practice — Lessons 16–18
Q11 · Short Answer

Solve: √(2x − 3) = 5. What is x?

Q12 · Multiple Choice

When solving √(x+3) = x−3, you get x = 6 and x = 1. Check: √(1+3) = 2 ≠ 1−3 = −2. So...