Functions — US Common Core Math Revision Notes

What you'll learn

Functions represent one of the most fundamental concepts in US Common Core Math, connecting algebraic thinking to real-world relationships. This revision guide covers function notation, domain and range, function operations, compositions, transformations, and inverse functions—all core components tested extensively on Common Core assessments. Mastery of functions provides the foundation for advanced algebra, trigonometry, and calculus.

Key terms and definitions

Function — a relation where each input (x-value) corresponds to exactly one output (y-value), often written as f(x) or y = f(x)

Domain — the complete set of possible input values (x-values) for which a function is defined

Range — the complete set of possible output values (y-values) that a function can produce

Function notation — the standard form f(x) read as "f of x," where f names the function and x represents the independent variable

Composition of functions — combining two functions where the output of one function becomes the input of another, written as (f ∘ g)(x) or f(g(x))

Inverse function — a function f⁻¹(x) that reverses the effect of the original function f(x), satisfying f(f⁻¹(x)) = x and f⁻¹(f(x)) = x

One-to-one function — a function where each output value corresponds to exactly one input value, required for a function to have an inverse

Transformation — a modification to a parent function that shifts, stretches, compresses, or reflects its graph

Core concepts

Identifying functions and function notation

A relation qualifies as a function only when each input maps to exactly one output. Multiple representations test this concept:

Vertical Line Test: On a graph, if any vertical line intersects the curve more than once, the relation is not a function. This test appears frequently on US Common Core Math assessments.

Mapping diagrams: Each element in the domain must point to exactly one element in the range. Multiple domain elements can map to the same range element, but not vice versa.

Ordered pairs: In a set like {(1,3), (2,5), (3,7)}, no x-value repeats. If (2,5) and (2,8) both appeared, this would not be a function.

Function notation f(x) serves multiple purposes:

Names the function (f, g, h commonly used)
Indicates the independent variable
Represents the output value
Enables evaluation: f(3) means "the output when x = 3"

For f(x) = 2x² - 5x + 1, the notation f(3) requires substituting 3 for every x: f(3) = 2(3)² - 5(3) + 1 = 18 - 15 + 1 = 4.

Domain and range determination

Domain restrictions arise from mathematical constraints:

Division by zero: For f(x) = 1/(x - 4), the domain excludes x = 4 because division by zero is undefined. Written as {x | x ≠ 4} or (-∞, 4) ∪ (4, ∞).

Square roots of negative numbers: For f(x) = √(x + 3), the expression under the radical must be non-negative: x + 3 ≥ 0, so x ≥ -3. Domain: [-3, ∞).

Combination restrictions: For f(x) = √(x - 2)/(x - 5), both restrictions apply: x ≥ 2 (from the square root) AND x ≠ 5 (from the denominator). Domain: [2, 5) ∪ (5, ∞).

Range determination requires analyzing possible outputs:

For f(x) = x² + 1, the minimum value occurs at the vertex. Since x² ≥ 0 for all real x, the range is [1, ∞).
For f(x) = -3x + 7, a linear function with negative slope, the range is all real numbers: (-∞, ∞).
For f(x) = 1/x, the function never equals zero, so the range is (-∞, 0) ∪ (0, ∞).

Operations with functions

Function arithmetic combines functions algebraically:

(f + g)(x) = f(x) + g(x) — add corresponding outputs

(f - g)(x) = f(x) - g(x) — subtract corresponding outputs

(f · g)(x) = f(x) · g(x) — multiply corresponding outputs

(f/g)(x) = f(x)/g(x) — divide corresponding outputs, where g(x) ≠ 0

For f(x) = x² - 1 and g(x) = x + 3:

(f + g)(x) = x² - 1 + x + 3 = x² + x + 2
(f · g)(x) = (x² - 1)(x + 3) = x³ + 3x² - x - 3
(f/g)(x) = (x² - 1)/(x + 3), domain excludes x = -3

The domain of combined functions is the intersection of individual domains, with additional restrictions for division.

Composition of functions

Function composition creates a new function by using one function's output as another's input. The notation (f ∘ g)(x) means f(g(x))—evaluate g first, then apply f to that result.

Order matters: (f ∘ g)(x) typically differs from (g ∘ f)(x).

For f(x) = 2x + 1 and g(x) = x² - 3:

(f ∘ g)(x) = f(g(x))

Start with g(x) = x² - 3
Substitute this entire expression into f: f(x² - 3) = 2(x² - 3) + 1
Simplify: 2x² - 6 + 1 = 2x² - 5

(g ∘ f)(x) = g(f(x))

Start with f(x) = 2x + 1
Substitute into g: g(2x + 1) = (2x + 1)² - 3
Expand: 4x² + 4x + 1 - 3 = 4x² + 4x - 2

The domain of a composition requires that x be in the domain of the inner function AND that the output of the inner function be in the domain of the outer function.

Inverse functions

An inverse function f⁻¹(x) undoes the operation of f(x). Only one-to-one functions have inverses.

Horizontal Line Test: A function is one-to-one if no horizontal line intersects its graph more than once. This test determines whether an inverse function exists.

Finding inverse functions algebraically:

Replace f(x) with y: y = f(x)
Swap x and y: x = f(y)
Solve for y
Replace y with f⁻¹(x)

For f(x) = (2x - 3)/5:

y = (2x - 3)/5
x = (2y - 3)/5
5x = 2y - 3 → 5x + 3 = 2y → y = (5x + 3)/2
f⁻¹(x) = (5x + 3)/2

Verification: Check that f(f⁻¹(x)) = x and f⁻¹(f(x)) = x.

Graphical relationship: The graphs of f and f⁻¹ are reflections across the line y = x. Domain and range swap: the domain of f becomes the range of f⁻¹, and vice versa.

Function transformations

Transformations modify parent functions systematically. Common Core assessments emphasize recognizing and describing these changes:

Vertical shifts: f(x) + k

k > 0 shifts up k units
k < 0 shifts down |k| units

Horizontal shifts: f(x - h)

h > 0 shifts right h units
h < 0 shifts left |h| units
Direction is opposite the sign inside parentheses

Vertical stretches/compressions: a·f(x)

|a| > 1 stretches vertically by factor |a|
0 < |a| < 1 compresses vertically by factor |a|
a < 0 also reflects across x-axis

Horizontal stretches/compressions: f(bx)

|b| > 1 compresses horizontally by factor 1/|b|
0 < |b| < 1 stretches horizontally by factor 1/|b|
b < 0 also reflects across y-axis

Combined transformations: f(x) = a·f(b(x - h)) + k applies all transformations. Process horizontal changes first (inside the function), then vertical changes (outside).

For g(x) = -2f(3(x + 1)) - 4 derived from parent function f(x):

Shift left 1 unit (x + 1)
Compress horizontally by factor 1/3 (multiply by 3)
Stretch vertically by factor 2 and reflect across x-axis (multiply by -2)
Shift down 4 units (subtract 4)

Worked examples

Example 1: Given f(x) = x² - 4x + 3 and g(x) = 2x - 1, find (f ∘ g)(3) and state the domain of (f/g)(x).

Solution: For (f ∘ g)(3):

Method 1 (evaluate g first): g(3) = 2(3) - 1 = 5, then f(5) = 5² - 4(5) + 3 = 25 - 20 + 3 = 8
Method 2 (find composition formula): (f ∘ g)(x) = f(2x - 1) = (2x - 1)² - 4(2x - 1) + 3 = 4x² - 4x + 1 - 8x + 4 + 3 = 4x² - 12x + 8, then evaluate at x = 3: 4(9) - 12(3) + 8 = 36 - 36 + 8 = 8

For domain of (f/g)(x):

f(x) has domain of all real numbers (polynomial)
g(x) has domain of all real numbers (linear)
Division requires g(x) ≠ 0: 2x - 1 ≠ 0 → x ≠ 1/2
Domain: {x | x ≠ 1/2} or (-∞, 1/2) ∪ (1/2, ∞)

Example 2: Find the inverse function of f(x) = √(x + 2) - 3, where x ≥ -2, and verify your answer.

Solution: Finding f⁻¹(x):

y = √(x + 2) - 3
x = √(y + 2) - 3
x + 3 = √(y + 2)
(x + 3)² = y + 2
y = (x + 3)² - 2
f⁻¹(x) = (x + 3)² - 2

Domain of f⁻¹(x): The range of f(x) becomes the domain of f⁻¹(x). Since √(x + 2) ≥ 0, we have f(x) ≥ -3, so domain of f⁻¹(x) is x ≥ -3.

Verification (checking f(f⁻¹(x)) = x): f(f⁻¹(x)) = f((x + 3)² - 2) = √((x + 3)² - 2 + 2) - 3 = √(x + 3)² - 3 = |x + 3| - 3 Since x ≥ -3, we have x + 3 ≥ 0, so |x + 3| = x + 3 Therefore: (x + 3) - 3 = x ✓

Example 3: The function h(x) is obtained by transforming the parent function f(x) = x² according to these steps: reflect across the x-axis, shift right 3 units, and shift up 5 units. Write the equation for h(x) and identify its vertex.

Solution: Applying transformations in order:

Reflect across x-axis: -f(x) = -x²
Shift right 3 units: -f(x - 3) = -(x - 3)²
Shift up 5 units: -f(x - 3) + 5 = -(x - 3)² + 5

Therefore: h(x) = -(x - 3)² + 5

Vertex: For the form a(x - h)² + k, the vertex is (h, k) = (3, 5). This parabola opens downward with maximum value 5 at x = 3.

Common mistakes and how to avoid them

Mistake 1: Confusing f(x + 2) with f(x) + 2. Correction: f(x + 2) means substitute (x + 2) for every x in the function definition; f(x) + 2 means evaluate the function at x, then add 2. For f(x) = x², f(x + 2) = (x + 2)² = x² + 4x + 4, while f(x) + 2 = x² + 2.

Mistake 2: Reversing the order of composition. Correction: (f ∘ g)(x) means f(g(x))—always work from right to left, evaluating the inner function first. Write out g(x) explicitly, then substitute that entire expression into f.

Mistake 3: Incorrectly identifying transformation directions, especially thinking f(x - 3) shifts left. Correction: The transformation f(x - h) shifts RIGHT when h is positive. Think "opposite of the sign inside." Similarly, f(x + 2) = f(x - (-2)) shifts left 2 units.

Mistake 4: Forgetting domain restrictions when finding inverse functions. Correction: The domain of f⁻¹(x) equals the range of f(x), not necessarily all real numbers. Restrict the domain based on the original function's behavior, especially for squared terms in the inverse.

Mistake 5: Assuming all functions have inverses. Correction: Apply the horizontal line test first. Functions like f(x) = x² (without restricted domain) fail this test and do not have inverse functions over all real numbers.

Mistake 6: Incorrectly combining domain restrictions in function operations. Correction: For combined functions like (f + g)(x), take the intersection of domains. For (f/g)(x), also exclude any x-values where g(x) = 0. List all restrictions explicitly.

Exam technique for Functions

Command word recognition: "Evaluate" requires calculating a specific numerical value; "find" or "determine" often means expressing an algebraic result; "state the domain" requires interval or set notation; "describe the transformation" needs precise language (direction, magnitude, and type of transformation).

Multi-step problems: Function questions often combine multiple skills—composition followed by evaluation, or transformations followed by identifying key features. Read the entire question before starting to identify what the final answer should look like (a number, an equation, a set, etc.). Show intermediate steps clearly, as partial credit depends on visible work.

Notation precision: Write f⁻¹(x) for inverse functions, not 1/f(x) (which means the reciprocal). Use proper interval notation with square brackets for inclusive endpoints and parentheses for exclusive endpoints. For domain restrictions, both set-builder notation {x | x ≠ 3} and interval notation (-∞, 3) ∪ (3, ∞) are acceptable.

Graph interpretation: When given a graph, extract information systematically—identify domain and range from the visible extent, check for discontinuities or asymptotes, verify if the relation is a function using the vertical line test, and check if it's one-to-one using the horizontal line test before discussing inverses.

Quick revision summary

Functions map each input to exactly one output, verified by the vertical line test. Master function notation f(x), domain (input restrictions from division by zero, square roots), and range (possible outputs). Operations combine functions algebraically; composition (f ∘ g)(x) = f(g(x)) nests functions. Inverse functions f⁻¹(x) reverse operations; only one-to-one functions have inverses. Transformations modify parent functions: f(x - h) + k shifts horizontally and vertically; a·f(bx) stretches or compresses. Always verify domains, show intermediate steps, and use precise notation.