Functions in Precalculus
Chapter 2 — Precalculus
Functions are the language of mathematics. Everything in precalculus — and all of calculus — is built on this foundation. This guide covers domain/range, transformations, composition, and inverses.
Chapter 2 Practice Problems
80+ questions covering all function topics
Chapter 2 Topics
2.1 — What is a Function?
Definition, vertical line test, function notation f(x), evaluating functions
2.2 — Graphs of Functions
Reading graphs, intervals of increase/decrease, local max/min, even/odd functions
2.3 — Getting Information from Graphs
Domain and range from graphs, piecewise functions, interpreting function behavior
2.4 — Average Rate of Change
Slope of secant lines, difference quotient, preview of derivative concept
2.5 — Transformations
Vertical/horizontal shifts, reflections, stretches and compressions
2.6 — Combining Functions
Sum, difference, product, quotient of functions; domain of combined functions
2.7 — Composition of Functions
f ∘ g notation, evaluating compositions, domain of compositions
2.8 — Inverse Functions
Definition, horizontal line test, finding inverses algebraically, inverse graphs
Function Notation and Evaluation
If f(x) = x² − 3x + 1, then:
f(2) = (2)² − 3(2) + 1 = 4 − 6 + 1 = −1
f(a) = a² − 3a + 1
f(x + h) = (x+h)² − 3(x+h) + 1
f(x+h) − f(x) = 2xh + h² − 3h (difference quotient numerator)
The key rule
f(a + b) ≠ f(a) + f(b). You must substitute the entire expression (a + b) wherever x appears in the formula.
Finding Domain
Domain is all valid x-values. Most restrictions come from these situations:
√(expression)
Expression ≥ 0
f(x) = √(x − 4): x − 4 ≥ 0 → domain: [4, ∞)
1/(expression)
Expression ≠ 0
f(x) = 1/(x + 3): x ≠ −3 → domain: (−∞, −3) ∪ (−3, ∞)
ln(expression)
Expression > 0
f(x) = ln(2x): 2x > 0 → x > 0 → domain: (0, ∞)
√(expression) in denominator
Expression > 0
f(x) = 1/√x: x > 0 → domain: (0, ∞)
Polynomial
All real numbers
f(x) = x³ + 2x: domain (−∞, ∞)
Transformation Rules — Quick Reference
Note the counterintuitive direction for horizontal shifts: f(x + h) moves LEFT (not right), f(x − h) moves RIGHT.
| Form | Effect | Example |
|---|---|---|
| f(x) + k | Shift UP k units | f(x) + 3 → graph moves up 3 |
| f(x) − k | Shift DOWN k units | f(x) − 2 → graph moves down 2 |
| f(x + h) | Shift LEFT h units | f(x + 4) → graph moves left 4 |
| f(x − h) | Shift RIGHT h units | f(x − 1) → graph moves right 1 |
| a · f(x), a > 1 | Stretch VERTICALLY by a | 3f(x) → graph taller by factor 3 |
| a · f(x), 0 < a < 1 | Compress vertically | (1/2)f(x) → graph shorter |
| f(bx), b > 1 | Compress HORIZONTALLY | f(2x) → graph narrower |
| −f(x) | Reflect over x-axis | −f(x) → flip upside down |
| f(−x) | Reflect over y-axis | f(−x) → flip left-right |
Composition of Functions
(f ∘ g)(x) means f(g(x)) — apply g first, then f.
If f(x) = x² + 1 and g(x) = 2x − 3:
(f ∘ g)(x) = f(g(x)) = f(2x − 3) = (2x − 3)² + 1
= 4x² − 12x + 9 + 1 = 4x² − 12x + 10
Order matters
f ∘ g ≠ g ∘ f in general. The notation (f ∘ g)(x) reads right to left — g is applied first.
Inverse Functions
Steps to Find an Inverse
- 1.Replace f(x) with y
- 2.Swap x and y
- 3.Solve for y
- 4.Replace y with f⁻¹(x)
f(x) = 3x − 7
y = 3x − 7 → x = 3y − 7
x + 7 = 3y → y = (x + 7)/3
f⁻¹(x) = (x + 7)/3
Verifying an Inverse
f and f⁻¹ are inverses if and only if:
(f ∘ f⁻¹)(x) = x AND (f⁻¹ ∘ f)(x) = x
Horizontal Line Test
A function has an inverse (is one-to-one) if no horizontal line crosses its graph more than once. f(x) = x² fails this test — you must restrict the domain to [0, ∞) before inverting.
Practice Functions Problems
Chapter 2 in NailTheTest has 80+ practice problems covering all 8 sections of functions theory — domain/range, transformations, composition, and inverses. Free to start.