Overview
Functions are one of the most heavily tested topics in the SAT Advanced Math domain. Questions assess function notation — reading and evaluating f(x), understanding f(x+h) as a substitution, and working with composite and inverse functions. The SAT also tests domain and range and piecewise-defined functions. Because functions underlie nearly all Advanced Math topics, strong fluency here pays dividends across multiple question types.
Key Points
1. What Is a Function?
A function assigns exactly one output to every valid input. The vertical line test confirms a relation is a function: if any vertical line crosses the graph more than once, the relation is not a function.
f: input (x) → exactly one output (f(x))
2. Evaluating f(a) and f(x+h)
Rule: Replace every instance of x in the function’s formula with the given input.
If f(x) = 3x² - 2x + 1, then:
f(4) = 3(4)² - 2(4) + 1 = 48 - 8 + 1 = 41
f(x+h) = 3(x+h)² - 2(x+h) + 1
= 3(x² + 2xh + h²) - 2x - 2h + 1
Common error: Substituting only in the first occurrence of x instead of every occurrence.
3. Domain and Range
| Concept | Definition | SAT Focus |
|---|---|---|
| Domain | All valid input (x) values | Exclude values causing division by zero or negative radicands |
| Range | All resulting output (y) values | Read from graph; note minimum/maximum of quadratics |
Examples of domain restrictions:
f(x) = 1/(x-3) → domain: x ≠ 3
g(x) = √(x-5) → domain: x ≥ 5
h(x) = x² → domain: all real numbers
4. Composite Functions
(f ∘ g)(x) = f(g(x))
Order rule: Apply g first (it is nearest to x), then feed the result into f.
Step-by-step:
Step 1: Evaluate g(x) (the inner function).
Step 2: Use the result as the input to f (the outer function).
Example:
f(x) = 2x + 1, g(x) = x²
(f ∘ g)(3) = f(g(3)) = f(9) = 2(9) + 1 = 19
(g ∘ f)(3) = g(f(3)) = g(7) = 49 ← order matters!
5. Inverse Functions
f⁻¹(x): the function that "undoes" f
- The domain of
f⁻¹equals the range off. - The range of
f⁻¹equals the domain off. - A function has an inverse only if it is one-to-one (passes the horizontal line test).
Method to find the inverse algebraically:
Step 1: Replace f(x) with y.
Step 2: Swap x and y.
Step 3: Solve for y.
Step 4: Replace y with f⁻¹(x).
Example:
f(x) = 3x - 6
y = 3x - 6 → x = 3y - 6 → y = (x+6)/3
f⁻¹(x) = (x + 6) / 3
6. Piecewise Functions
A piecewise function applies different rules on different parts of the domain:
f(x) = { x² if x < 0
{ 2x + 1 if x ≥ 0
Key step: Identify which piece applies to the given input before evaluating.
f(-3) → uses x²: f(-3) = 9
f(4) → uses 2x+1: f(4) = 9
Pitfalls and Common Mistakes
Pitfall 1: Reversing the Order of Composite Functions
Description: Students evaluate f first and g second when the problem asks for (f ∘ g)(x) = f(g(x)).
Example: For (f ∘ g)(3), computing f(3) and using it as input to g instead of computing g(3) first.
Fix: The inner function is always the one closest to x in the notation. In f(g(x)), g is inner — apply g first, always.
Pitfall 2: Assigning the Wrong Domain to an Inverse Function
Description: Students use the domain of f as the domain of f⁻¹.
Example: If f has domain x ≥ 0 and range y ≥ 2, then f⁻¹ has domain x ≥ 2 (not x ≥ 0).
Fix: Remember — domain and range swap when you invert a function. The domain of f⁻¹ is the range of f, and vice versa.
Pitfall 3: Partial Substitution in f(x+h)
Description: When computing f(x+h), students replace only the first x with (x+h) rather than every occurrence.
Example: For f(x) = x² + x, computing f(x+h) = (x+h)² + x instead of (x+h)² + (x+h).
Fix: Before substituting, underline or circle every x in the formula. Replace each one with the given expression (in parentheses to preserve signs).
Pitfall 4: Using a Non-One-to-One Function to Find an Inverse
Description: Attempting to find the inverse of a parabola f(x) = x² over all real numbers, which is not one-to-one.
Example: f(x) = x² has both f(2) = 4 and f(-2) = 4, so it fails the horizontal line test and has no inverse over all reals.
Fix: Apply the horizontal line test. If any horizontal line crosses the graph more than once, restrict the domain before finding the inverse, or note that the SAT question will specify a restricted domain.
Related Entries
- Function_Transformations — Translations and reflections change how
f(x)is written and graphed; understandingf(x)notation is prerequisite. - Rational_Expressions_Equations — Functions defined by rational expressions require domain analysis to exclude denominator zeros.
- Radicals_Rational_Exponents — Radical functions
f(x) = √(x-k)require non-negative radicands, directly setting domain restrictions. - Quadratic_Functions_Parabolas — Quadratic functions demonstrate range and the importance of domain restriction for inverses.
- Linear_Functions_Graphs — Linear functions are the simplest case of function notation; all rules (evaluation, composition, inverse) apply directly.
Quick Reference Card
| Concept | Rule |
|---|---|
| Evaluate f(a) | Replace every x with a |
| Evaluate f(x+h) | Replace every x with (x+h), use parentheses |
| Composite (f∘g)(x) | Apply g first, then f: f(g(x)) |
| Inverse f⁻¹(x) | Swap x and y, solve for y |
| Domain of f⁻¹ | Equals the range of f |
| Range of f⁻¹ | Equals the domain of f |
| Piecewise | Match input to correct interval, then evaluate |
| Horizontal line test | Passes → one-to-one → inverse exists |
| Vertical line test | Passes → relation is a function |
| Key trap | Composite order reversed: apply inner function first |