Questions: Domain and Range

Question 1

What is the domain of f(x) = √(x + 3) / (x − 1)?

• A. All real numbers
• B. x ≥ −3, written [−3, ∞)
• C. x ≥ −3 and x ≠ 1, written [−3, 1) ∪ (1, ∞)
• D. x > 0

Answer: (C)

Two restrictions apply simultaneously. The square root requires x + 3 ≥ 0, giving x ≥ −3. The denominator requires x − 1 ≠ 0, excluding x = 1. Both must hold: the domain is all x ≥ −3 except x = 1. Option B misses the denominator restriction; Option A assumes all inputs are valid; Option D ignores both restrictions. When multiple constraints exist, every one of them must be satisfied.

Question 2

A student claims the range of f(x) = x² is all real numbers because 'x can be any real number.' What error has the student made?

• A. None — x² does produce all real numbers as outputs
• B. The student has confused the domain with the range — x can be any real number (the domain), but x² is never negative (the range is [0, ∞))
• C. The student correctly identified the domain but should express it in interval notation
• D. The student is wrong because x cannot equal 0

Answer: (B)

The student described the domain (inputs: any real number) but called it the range. The range is the set of actual output values. Since squaring any real number — positive, negative, or zero — always produces a non-negative result, the range is [0, ∞). Negative numbers are simply unreachable outputs. This is the core distinction: domain is where the function accepts inputs; range is where its outputs actually land.

Question 3

The range of f(x) = x² is most real numbers.

• T. True
• F. False

Answer: False

The domain of f(x) = x² is all real numbers (any real input can be squared), but the range is only [0, ∞). Squaring any real number — whether positive, negative, or zero — always yields a non-negative result. There is no input x such that x² < 0, so negative numbers are not in the range. This is a common confusion: a function with an unrestricted domain can still have a restricted range.

Question 4

To find the range of a function algebraically, one strategy is to write y = f(x), solve for x in terms of y, and determine which values of y allow a real solution for x.

• T. True
• F. False

Answer: True

This is the standard algebraic method for finding range. For y = x², solving gives x = ±√y, which has a real solution only when y ≥ 0, confirming range = [0, ∞). The strategy works because you are asking: 'which output values y are actually achievable?' — i.e., which y allow a valid input x to exist. When this algebraic approach is cumbersome, graphical inspection (reading all y-values from the graph) is often easier.

Question 5

Explain why finding the range of a function is generally more difficult than finding its domain.

Model answer: Finding the domain requires identifying which inputs cause the function to fail — division by zero, square roots of negatives, or logarithms of non-positives — which reduces to setting up simple inequalities or exclusions. Finding the range requires determining which outputs the function actually produces across all valid inputs. There is no simple algebraic check analogous to 'set denominator ≠ 0': you must reason about the function's behavior as a whole, either by analyzing the formula algebraically (solving y = f(x) for x and asking which y allow a real solution) or by reading all y-values from a graph. The range depends on the function's specific structure, not just which inputs are forbidden.

Domain restrictions come from a short list of operations that fail on specific inputs — easily checked by inspection. Range requires understanding what the function actually does to its inputs across the entire domain. For example, f(x) = x² has domain all reals (no failures) but range [0, ∞) — you can only discover this by observing that squaring always yields non-negative results, which requires understanding the function's behavior, not just its failure modes.