Relations and Functions Explained: Domain, Range, and Real Applications
PUC Mathematics | Relations & Functions | Daniel Classes
What Makes This Chapter Critical
Functions are the language of mathematics. Every model of the real world — population growth, stock prices, decay of radioactive materials — is described by a function.
Yet most students learn this chapter as if it’s pure abstraction: given a relation, determine if it’s a function; find the domain; find the range. Plug and chug.
The real question is: Why does it matter whether something is a function? Because functions have a special property: for every input, there’s exactly one output. That predictability is what makes them powerful for modeling reality.
I’ve taught this chapter to thousands of students. The ones who score well don’t memorize definitions. They understand *why* we distinguish functions from general relations.
Relations vs Functions
A Relation
A relation is any set of ordered pairs (x, y).
Example: {(1, 2), (1, 3), (2, 4), (3, 4)}
This is a valid relation. Notice that x = 1 maps to both y = 2 and y = 3. That’s fine for a relation.
A Function
A function is a special relation where each input (x) maps to exactly one output (y).
Same example: {(1, 2), (1, 3), (2, 4), (3, 4)}
This is NOT a function because x = 1 maps to two different outputs.
Modified example: {(1, 2), (2, 4), (3, 4)}
This IS a function. Every x maps to exactly one y.
The Vertical Line Test
Graphically, a relation is a function if every vertical line crosses its graph at most once. If any vertical line crosses the graph twice, it’s not a function.
Domain and Range
Domain
Definition: The set of all possible input values (x-values).
Example: For the function f(x) = √(x – 1), the domain is x ≥ 1 (because you can’t take the square root of a negative number).
Range
Definition: The set of all possible output values (y-values) that the function actually produces.
Example: For f(x) = √(x – 1), the range is y ≥ 0 (because square roots are non-negative).
Important distinction: Domain and range are not the same. Domain = inputs. Range = outputs the function actually produces.
Types of Functions
One-to-One (Injective)
Definition: Each output comes from exactly one input. If f(a) = f(b), then a = b.
Example: f(x) = 2x is one-to-one. If 2a = 2b, then a = b.
Non-example: f(x) = x² is not one-to-one. Both f(2) = 4 and f(-2) = 4.
Onto (Surjective)
Definition: Every element in the codomain is an output. The range equals the codomain.
Example: f: ℝ → ℝ where f(x) = 2x is onto (every real number is the output of some input).
Non-example: f: ℝ → ℝ where f(x) = x² is not onto (negative numbers are never outputs).
Bijective (One-to-One and Onto)
Definition: Every input maps to exactly one output, and every output comes from exactly one input.
Example: f(x) = 2x + 1 is bijective.
Importance: Bijective functions have inverses.
Function Composition
Definition: (f ∘ g)(x) = f(g(x))
You apply g first, then apply f to the result.
Example: f(x) = 2x, g(x) = x + 1
(f ∘ g)(x) = f(g(x)) = f(x + 1) = 2(x + 1) = 2x + 2
Important: f ∘ g ≠ g ∘ f (order matters).
Inverse Functions
Definition: If f is bijective, its inverse f⁻¹ undoes what f does: f⁻¹(f(x)) = x and f(f⁻¹(x)) = x.
Example: f(x) = 2x has inverse f⁻¹(x) = x/2.
How to find inverses:
- Replace f(x) with y: y = 2x
- Swap x and y: x = 2y
- Solve for y: y = x/2
- Write as f⁻¹(x) = x/2
How PUC Exams Test Functions
Type 1: Identify Functions (2-3 marks)
Example: “Is {(1, 2), (1, 3), (2, 4)} a function?”
Solution: No, because x = 1 maps to two values.
Type 2: Find Domain and Range (3-5 marks)
Example: “Find the domain and range of f(x) = √(x – 1).”
Solution: Domain: x ≥ 1. Range: y ≥ 0.
Type 3: Function Composition (3-5 marks)
Example: “If f(x) = 2x and g(x) = x + 1, find (f ∘ g)(3).”
Solution: (f ∘ g)(3) = f(g(3)) = f(4) = 8.
Type 4: Inverse Functions (3-5 marks)
Example: “Find the inverse of f(x) = 2x + 3.”
Solution: f⁻¹(x) = (x – 3)/2.
The Five Common Mistakes
Mistake 1: Confusing Domain with Range
Example: “Find the domain of f(x) = √(x – 1).”
Wrong: “The range is x ≥ 1.” [This is domain, not range]
Right: “The domain is x ≥ 1. The range is y ≥ 0.”
Fix: Domain = inputs. Range = outputs.
Mistake 2: Forgetting Restrictions in Domain
Example: “Find the domain of f(x) = 1/(x – 2).”
Wrong: “The domain is all real numbers.” [Can’t divide by zero]
Right: “The domain is all real numbers except x = 2.”
Fix: Check for: denominators (can’t be zero), square roots (can’t be negative), logarithms (can’t be of non-positive numbers).
Mistake 3: Using Wrong Order in Composition
Example: “Find (f ∘ g)(x) for f(x) = 2x, g(x) = x + 1.”
Wrong: “g first, so (f ∘ g)(x) = g(f(x)) = g(2x) = 2x + 1.” [Wrong order]
Right: “(f ∘ g)(x) = f(g(x)) = f(x + 1) = 2(x + 1) = 2x + 2.” [g first, then f]
Fix: In (f ∘ g), apply g first, then f.
Mistake 4: Incorrectly Finding Inverses
Example: “Find the inverse of f(x) = 2x + 3.”
Wrong: “f⁻¹(x) = 2x – 3.” [Not solving correctly]
Right: “y = 2x + 3 → x = 2y + 3 → y = (x – 3)/2 → f⁻¹(x) = (x – 3)/2.”
Fix: Follow the step-by-step process: replace f(x) with y, swap x and y, solve for y.
Mistake 5: Assuming All Relations Are Functions
Example: “Is {(1, 2), (2, 3), (1, 4)} a function?”
Wrong: “Yes, it’s a function.” [Not checking uniqueness]
Right: “No, x = 1 maps to two values (2 and 4).”
Fix: Always verify: does each input have exactly one output?
What to Do Next
To master domain and range: Find domain and range for 15 different functions (including square roots, fractions, logarithms).
To master composition: Compute (f ∘ g)(x) and (g ∘ f)(x) for 10 pairs of functions.
To master inverses: Find inverses for 10 bijective functions and verify f⁻¹(f(x)) = x.
The Problem Vault Alternative
The Problem Vault has 80+ Relations & Functions problems from actual PUC exams.