Week 7, Session 1: Function Notation, Domain & Range

Functions are the language of mathematical relationships. Every SAT® Math section includes 4–6 function questions. Mastering substitution, domain, range, and composition gives you reliable, high-value points.

Week 7, Session 1 Learning Goals
By the end of this session you will:

1. Evaluate any function at a numerical or algebraic input with full confidence.
2. Find the domain of functions restricted by denominators, even roots, or both.
3. Determine the range from a formula or graph.
4. Evaluate composite functions f(g(x)) step by step without confusion.
5. Find and verify the inverse of a linear or simple rational function.

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Part 1 — Evaluating Functions

f(x) is a rule: replace every occurrence of x with the given input.

Example 1 — Numerical input:
f(x) = 3x²−5x+2. Find f(4).
Step 1: Replace x with 4: 3(4)²−5(4)+2
Step 2: Compute: 3(16)−20+2 = 48−20+2 = 30 ✓

Example 2 — Algebraic input (SAT® favorite!):
f(x) = x²−2x+1. Find f(a+1).
Step 1: Replace every x with (a+1): (a+1)²−2(a+1)+1
Step 2: Expand: a²+2a+1−2a−2+1
Step 3: Combine: = a² (the algebra simplifies beautifully!)

SAT® Alert: When the input is an expression like (a+1), always wrap it in parentheses on substitution. Dropping the parentheses is the #1 algebraic error on these questions.

Example 3 — Solving for an input (reverse direction):
g(x) = 2x+5. For what value of x does g(x) = 13?
Set 2x+5 = 13 → 2x = 8 → x = 4.
Check: g(4) = 2(4)+5 = 13 ✓

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Part 2 — Domain Restrictions

The domain is all x-values that produce a real, defined output.

Rule 1 — Denominator cannot be zero:
f(x) = 5/(x−3): exclude x = 3.
h(x) = 3/(x²−9) = 3/[(x−3)(x+3)]: exclude x = 3 AND x = −3.

Rule 2 — Even-root radicand must be non-negative:
f(x) = √(2x−4): require 2x−4 ≥ 0 → x ≥ 2. Domain: [2, ∞).
g(x) = √(9−x²): require 9−x² ≥ 0 → −3 ≤ x ≤ 3. Domain: [−3, 3].

Rule 3 — No restrictions for polynomials, absolute value, or odd roots:
f(x) = x³+2x−7: domain = all real numbers.
g(x) = ∛(x−5): domain = all real numbers (cube roots accept negatives).

Example 4 — Combined restriction:
Find the domain of f(x) = √(x+3)/(x−1).
Radical: x+3 ≥ 0 → x ≥ −3.
Denominator: x ≠ 1.
Domain: x ≥ −3 and x ≠ 1.

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Part 3 — Range

The range is the complete set of possible output values.

• f(x) = x²: outputs are always ≥ 0. Range: [0, ∞).
• f(x) = x²+3: shifted up 3. Range: [3, ∞).
• f(x) = −(x+1)²+5: opens downward, maximum = 5. Range: (−∞, 5].
• f(x) = 3x+2 (linear): range = all real numbers.

Key SAT® shortcut: For vertex form f(x) = a(x−h)²+k, the value k is the minimum (a>0) or maximum (a<0) of the range. Read it directly — no calculation needed.

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Part 4 — Composite Functions

f(g(x)) means: apply g first, then apply f to that result. Always work from the INSIDE OUT.

Example 5 — Step-by-step composite:
f(x) = 2x+1, g(x) = x²−3. Find f(g(4)).
Step 1: g(4) = 16−3 = 13.
Step 2: f(13) = 2(13)+1 = 27.

Now find f(g(x)) as a general expression:
f(g(x)) = f(x²−3) = 2(x²−3)+1 = 2x²−5.

SAT® Tip: "f(g(2))" and "g(f(2))" are different calculations. The inner function always executes first.

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Part 5 — Inverse Functions

f⁻¹(x) reverses input and output. If f(2) = 7, then f⁻¹(7) = 2.

Four-step method to find f⁻¹(x):
Step 1: Write y = f(x), e.g., y = 3x−4.
Step 2: Swap x and y: x = 3y−4.
Step 3: Solve for y: 3y = x+4 → y = (x+4)/3.
Step 4: Write result: f⁻¹(x) = (x+4)/3.

Verify: f(f⁻¹(x)) = 3·(x+4)/3 − 4 = x+4−4 = x ✓

Example 6:
f(x) = 5x+2 → swap → x = 5y+2 → y = (x−2)/5.
f⁻¹(x) = (x−2)/5.
Check: f(3) = 17. f⁻¹(17) = (17−2)/5 = 3 ✓

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Common Errors Table

Before Week 7, Session 2, answer at least 10 of the 12 lesson-practice questions correctly without notes. For each miss, record one precise cause: substitution parentheses, domain denominator rule, domain radical rule, range direction, composite order, composite algebra, inverse swap, solving after swap, piecewise selection, input-output confusion, notation misread, or algebraic arithmetic error.