Functions — Notation, Domain, Transformations, and Inverses

Functions are a compact way to describe an input-output rule. They show up in algebra, modelling, graphs, and later calculus. The main job is to keep track of which inputs are allowed, what each input produces, and how a change to a formula changes its graph.

Relations and Functions

A relation is any set of ordered pairs. A function is a relation in which every input is paired with exactly one output. Different inputs may have the same output; the rule only forbids one input from producing two different outputs.

RelationFunction?Reason
{(1, 4), (2, 4), (3, 9)}YesEach first coordinate appears once.
{(1, 4), (1, 5), (2, 7)}NoInput 1 is assigned both 4 and 5.

On a graph, use the vertical line test: if any vertical line meets the graph more than once, the relation is not a function of x. A circle fails this test because some x-values have an upper and a lower y-value.

Function Notation and Evaluation

Write f(x) for “the output of function f at input x.” It is not multiplication. For example, if f(x) = 3x − 2, then f(5) means substitute 5 for every x.

Worked example: evaluate a function

Given f(x) = 3x − 2, find f(−4).

f(−4) = 3(−4) − 2
      = −12 − 2
      = −14

Therefore, f(−4) = −14. For an expression such as f(a + 1), substitute the whole expression: f(a + 1) = 3(a + 1) − 2 = 3a + 1.

Domain and Range

The domain is the set of allowed inputs. The range is the set of outputs that actually occur. Unless a context gives a restriction, begin with all real numbers and remove inputs that make the expression undefined.

RuleDomain restrictionDomain
f(x) = 1/(x − 3)Denominator cannot be 0, so x ≠ 3.All real x except 3
g(x) = √(x + 2)For real outputs, x + 2 ≥ 0.x ≥ −2

Worked example: state domain and range

Find the domain and range of h(x) = √(x − 1) + 4.

For the square root to be real:
x − 1 ≥ 0
x ≥ 1

The smallest square-root value is 0.
h(x) = √(x − 1) + 4 ≥ 0 + 4
h(x) ≥ 4

The domain is x ≥ 1, and the range is h(x) ≥ 4. In interval notation these are [1, ∞) and [4, ∞).

Graph Transformations

Start with a parent function f(x). In g(x) = af(x − h) + k, the changes are predictable:

  • f(x − h): shift right h units; f(x + h) shifts left.
  • + k outside the function: shift up k units; a negative k shifts down.
  • |a| > 1: vertical stretch; 0 < |a| < 1: vertical compression.
  • a < 0: reflect in the x-axis.

Notice that horizontal changes happen inside the brackets, so their direction can feel backwards. Read the expression x − h as “move right h.”

Translation of a parabola two units right and one unit up Coordinate axes show the parent parabola y equals x squared with vertex at zero, zero in muted gray. A highlighted parabola y equals open parenthesis x minus two close parenthesis squared plus one has vertex at two, one. Dashed arrows show the vertex moving two units right and one unit up. x y −2−101234 123 y = x² y = (x − 2)² + 1 (0, 0) (2, 1)
The graph of y = (x − 2)2 + 1 comes from y = x2: move every point 2 units right, then 1 unit up. The vertex moves from (0, 0) to (2, 1).

Worked example: describe a transformation

Describe the changes from f(x) = x2 to g(x) = −2(x − 3)2 + 5.

g(x) = −2(x − 3)² + 5

x − 3   → shift right 3
−2 outside the square → reflect in the x-axis and stretch vertically by factor 2
+5      → shift up 5

Parent vertex: (0, 0)
New vertex:    (3, 5)

Composition of Functions

Composition feeds the output of one function into another. The notation (f ∘ g)(x) means f(g(x)), so do g first. Order matters in general: (f ∘ g)(x) is usually not the same as (g ∘ f)(x).

Worked example: compose two rules

Let f(x) = 2x + 1 and g(x) = x2 − 4. Find (f ∘ g)(x).

(f ∘ g)(x) = f(g(x))
           = f(x² − 4)
           = 2(x² − 4) + 1
           = 2x² − 8 + 1
           = 2x² − 7

Inverse Functions and Restrictions

An inverse function reverses the original rule. If f sends a to b, then f−1 sends b to a. The exponent −1 means inverse here, not reciprocal: f−1(x) is not 1/f(x).

A function has an inverse function only when it is one-to-one: every output comes from exactly one input. On a graph, it must pass the horizontal line test. The inverse graph is the reflection of the original graph in the line y = x.

Worked example: find and check an inverse

Find the inverse of f(x) = 3x − 6.

1. Write y = 3x − 6.
2. Swap x and y:  x = 3y − 6.
3. Solve for y:
   x + 6 = 3y
   (x + 6)/3 = y
4. Rename y:
   f⁻¹(x) = (x + 6)/3

Check:
f(f⁻¹(x)) = 3((x + 6)/3) − 6
            = x + 6 − 6
            = x

The quadratic q(x) = x2 is not one-to-one over all real numbers because q(−2) = q(2) = 4. Restrict its domain to x ≥ 0 before finding an inverse.

y = x², with x ≥ 0
x = y²
y = √x       (choose the non-negative root because y is in the restricted domain)

q⁻¹(x) = √x, with domain x ≥ 0

Common Mistakes

  • Calling every relation a function: check whether an input repeats with different outputs.
  • Ignoring a domain restriction: never allow division by zero or an even root of a negative number when working over the reals.
  • Reversing a horizontal shift: (x − 3) moves right 3, not left 3.
  • Doing composition in the wrong order: in f(g(x)), calculate g first.
  • Forgetting to restrict a quadratic: without a one-to-one domain restriction, its inverse is not a function.

Practice Set

  1. For p(x) = 4x2 − 1, find p(−2).
  2. State the domain of r(x) = 5/(x + 4).
  3. Describe the transformation from y = |x| to y = |x + 2| − 3.
  4. Let f(x) = x + 5 and g(x) = 2x. Find (g ∘ f)(x).
  5. Find the inverse of m(x) = (x − 4)/2.

Answers

  1. p(−2) = 4(−2)2 − 1 = 4(4) − 1 = 15.
  2. x + 4 ≠ 0, so x ≠ −4.
  3. Shift 2 units left and 3 units down.
  4. (g ∘ f)(x) = g(x + 5) = 2(x + 5) = 2x + 10.
  5. y = (x − 4)/2; swap to get x = (y − 4)/2; then 2x = y − 4 and m−1(x) = 2x + 4.