Function notation, domains and ranges

1. A function is a rule with an input set

A function assigns exactly one output to each allowed input. Its definition therefore needs both a rule and a domain. The expression f(x)=x² is not enough to settle every question about f: a function on all real numbers behaves differently from the same rule restricted to x≥0. Both satisfy the vertical line test, but only the restricted version has a single-valued inverse. Distinguish the independent variable x from the output f(x); f is the name of the mapping, not a number to multiply by x.

For f(x)=3x²−2x+1, evaluate a function by replacing every occurrence of x with the entire input in brackets. Thus f(−2)=3(−2)²−2(−2)+1=17. For an algebraic input, f(a+1)=3(a+1)²−2(a+1)+1=3a²+4a+2. This is not f(a)+1, which equals 3a²−2a+2. The two expressions describe different operations: one changes the input before applying f, while the other changes its output afterwards. This distinction later controls horizontal and vertical graph translations.

Function equations ask a different question from evaluation. To solve f(x)=1 here, write 3x²−2x+1=1, then x(3x−2)=0. The inputs are x=0 and x=2/3, subject to the stated domain. If the domain were [1,4], neither would be allowed, so the restricted function would have no preimage of 1. Do not silently replace the given domain with the largest algebraically possible one. As a check, substitute each candidate into the original rule rather than only into a transformed equation. In written solutions, distinguish an output statement such as f(−2)=17 from an input solution such as x=−2.

2. Finding a maximal real domain

The maximal real domain consists of all real inputs for which the stated expression is defined. A denominator cannot be zero, an even root needs a non-negative radicand, and a logarithm needs a strictly positive argument. Apply every condition simultaneously: finding one restriction does not remove the need to inspect the rest of the expression. A prescribed domain can be smaller than this maximal domain, especially when a function models a physical situation.

Consider g(x)=√(x+2)/(x−1). The square root requires x+2≥0, so x≥−2. The denominator excludes x=1. Therefore dom g=[−2,1)∪(1,∞). The input −2 is included because √0 is defined and its denominator is nonzero; the input 1 is excluded because division by zero is undefined. For h(x)=ln(5−2x), require 5−2x>0, giving x<5/2. The inequality reverses if you divide −2x>−5 by −2. The boundary x=5/2 is excluded because ln0 is not a real number.

A more demanding example is p(x)=√((x−1)/(x+2)). The quotient under the root must be non-negative and x≠−2. Critical values −2 and 1 divide the number line into three intervals. At x=−3 the quotient is positive; at x=0 it is negative; at x=2 it is positive. Hence dom p=(−∞,−2)∪[1,∞). It is incorrect to demand separately that numerator and denominator both be positive: they may both be negative and still give a valid radicand. Mark the zero at x=1 as included and the pole at x=−2 as excluded. A sign chart makes the reasoning auditable and avoids guessing from the appearance of a graph.

A useful final check is to test one allowed input and each excluded boundary directly in the original expression; this catches restrictions lost during rearrangement.

3. Range is an output set, not another domain

The range is the collection of outputs actually achieved from the domain. It must be found after the domain is established. For simple quadratics, completing the square identifies the turning value, but endpoint restrictions determine whether that turning value and other extrema are reached. An interval extending forever in x does not automatically imply that outputs extend forever in both directions.

Let f(x)=x²−4x+7=(x−2)²+3 with domain [−1,4]. The vertex x=2 belongs to the domain and gives the minimum f(2)=3. Evaluate both endpoints: f(−1)=12 and f(4)=7. The largest value is 12, so the range is [3,12]. It would be wrong to use [3,∞), the range for an unrestricted quadratic. If the domain changes to (−1,4], the value 12 is no longer attained: solving f(x)=12 gives x=−1 or 5, and neither is in the new domain. The range becomes [3,12). This extra equation check is valuable when an excluded endpoint's output might also occur elsewhere.

For q(x)=2/(x−3)+1 over all allowed real inputs, q(x) cannot equal 1 because 2/(x−3) cannot equal zero. Conversely, every y≠1 yields x=3+2/(y−1), so every other output is possible. Therefore ran q=R\{1}. This algebraic argument does more than assert that a horizontal asymptote is never crossed: some other functions do cross their horizontal asymptotes. When finding a range, use the actual rule, its monotonic branches, endpoint values and any stationary points. For a graph, project its included points onto the vertical axis, keeping track of open circles, isolated points and unbounded branches.

4. Reading domain and range from a graph

A graph communicates which ordered pairs belong to a relation. Its domain is the horizontal projection of those points and its range is the vertical projection. A solid endpoint is included; an open endpoint is excluded unless another point elsewhere supplies the same input or output. Arrows show continuation, not a finite endpoint. A vertical line through any allowed input must meet a function graph once, although a horizontal line may meet it several times.

Imagine a line segment y=2x+1 for −2≤x<1, together with an isolated point (1,−3). The segment contributes domain [−2,1) and range [−3,3). The isolated point adds input 1 but adds no new output because −3 already occurs at x=−2. Thus the complete domain is [−2,1] and range [−3,3). The graph is still a function: at x=1 the open segment endpoint is not present, leaving only the solid point (1,−3). Its value f(1)=−3 differs from the value approached along the segment, demonstrating why an open circle cannot be read as the function value.

Now replace the isolated point with (0,5). Input 0 already has the segment output 1, so the relation gives two outputs for one input and is not a function. Its domain and range can still be described, but function notation for a unique f(0) is inappropriate. In an examination sketch, label axes, relevant intercepts and endpoint coordinates; do not rely on a drawing's scale to communicate an exact value. For unbounded intervals use parentheses next to infinity, because infinity is not an included real endpoint. When a question asks for domain and range, name the variables or write dom f and ran f so the examiner can tell which set is which.

5. Domains in practical models

A mathematical rule may exist at inputs that make no sense in its context. A practical domain combines algebraic restrictions with information about time, length, counts or the interval over which a model is intended to apply. State units and distinguish continuous quantities from discrete ones. A formula for the cost of tickets does not permit 2.4 tickets merely because the algebra accepts x=2.4.

A rectangular garden has perimeter 28 m. If its width is x metres, its length is 14−x and its area is A(x)=x(14−x). Both dimensions must be positive, giving 0<x<14. Completing the square gives A(x)=49−(x−7)². Therefore the area range is (0,49] square metres: zero area is approached but excluded, while a 7 m by 7 m square achieves 49. If a path requires the width to lie between 3 m and 5 m inclusive, A increases across that interval and the range becomes [33,45]. The vertex still exists algebraically, but is outside the permitted design interval.

For a machine that packages whole boxes, suppose revenue is R(n)=18n for n∈{0,1,…,40}. The range is the discrete set {0,18,36,…,720}, not the continuous interval [0,720]. Graph it with separate points when discreteness matters. For a temperature model T(t)=20+60e^(−0.4t) used for t≥0 minutes, the initial output is 80°C and the temperature approaches 20°C without reaching it in finite time, so the model range is (20,80]. A model is a statement with assumptions: a long-run limiting temperature predicted by the formula does not by itself establish that the same physical conditions persist indefinitely.

6. Communicating sets and checking conclusions

Interval notation compresses a set without losing endpoint information. Square brackets include finite endpoints and round brackets exclude them. The union symbol joins separate pieces. The set R\{a} means all real numbers except a. Translate your final set back into an inequality or test values to catch notation slips before moving on. A correct method followed by a reversed bracket changes the mathematical answer.

For f(x)=√(9−x²), require 9−x²≥0, hence −3≤x≤3. The output is non-negative, reaches 3 at x=0 and reaches 0 at x=±3, so dom f=[−3,3] and ran f=[0,3]. Squaring y=√(9−x²) gives x²+y²=9, but the original rule describes only the upper semicircle. If you infer the range [−3,3] from the squared equation, you have introduced outputs that do not satisfy the original square-root definition. Reversing a transformation in algebra always requires checking whether it preserved all restrictions.

For f(x)=1/(x²+1), the denominator is at least 1 for real x and never zero. Thus the domain is R. Outputs satisfy 0<f(x)≤1, with 1 reached at x=0 and 0 never reached. To confirm every value in (0,1] occurs, solve x²=1/y−1; its right-hand side is non-negative precisely when 0<y≤1. This verifies the range rather than just bounding it. In a final answer, separate the algebraic domain from any contextual restriction, and justify an excluded range boundary with a limit or impossibility argument. These habits support criterion 4 while also demonstrating mathematical communication and reasoning across the course.