The two words
A function takes inputs and produces outputs. The words domain and range simply name those two collections.
- Domain
- The complete set of inputs a function is allowed to take, that is, all the values you may put in for \(x\).
- Range
- The complete set of outputs the function actually produces, that is, all the values \(f(x)\) can come out as.
A handy way to keep them straight: domain is the door you go in, range is what you get back. The domain is always about \(x\), the horizontal direction, and the range is always about \(f(x)\), the vertical direction.
Where this is assessed
Stating a domain and range is Criterion A (knowing and understanding). Using correct set or inequality notation is Criterion C (communicating), and deciding a sensible domain for a real situation, such as a length that cannot be negative, reaches into Criterion D (applying maths in context).
Reading them from a rule
Sometimes the domain is handed to you as a short list, and you find the range by putting each input through the rule. For example, if \(f(x) = x^2\) has the domain \(\{-2,\ -1,\ 0,\ 1,\ 2\}\), you square each one to get the outputs \(4,\ 1,\ 0,\ 1,\ 4\). Collecting these into a set, and listing each value once, the range is \(\{0,\ 1,\ 4\}\).
When no list is given, the domain is usually "all real numbers" apart from any input that would break the maths. Two things break it at this level: dividing by zero, and taking the square root of a negative number. For instance, \(f(x) = \dfrac{1}{x - 2}\) works for every real number except \(x = 2\), because \(x = 2\) makes the bottom zero. So its domain is "all real \(x\) with \(x \neq 2\)".
Squares are never negative
For \(f(x) = x^2\) over all real numbers, every output is zero or positive, so the range is \(f(x) \geq 0\). Spotting patterns like this saves you from testing endless values.
The function \(f(x) = x^2 + 1\) has the domain \(\{-2,\ -1,\ 0,\ 1,\ 2\}\). List the range.
Reading them from a graph
On a graph, the domain is how far the curve spreads left to right, and the range is how far it spreads bottom to top. You read the domain off the \(x\)-axis and the range off the \(y\)-axis.
Suppose the straight line \(y = 2x + 1\) is drawn only for \(-1 \leq x \leq 3\). The domain is exactly that stretch of inputs: \(-1 \leq x \leq 3\). For the range, find the outputs at the two ends: at \(x = -1\), \(y = 2(-1) + 1 = -1\); at \(x = 3\), \(y = 2(3) + 1 = 7\). Since a straight line climbs steadily between the ends, every value in between is reached, so the range is \(-1 \leq y \leq 7\).
Ends are not always the largest and smallest
Reading the range off the endpoints works for a straight line, but not for a curve that turns. A parabola like \(y = x^2\) dips to a lowest point in the middle, so check any turning point as well as the ends before you state the range.
Check yourself
Try these before opening the answers. Show the working, not just the final value.
1. State the domain of \(f(x) = \dfrac{1}{x - 2}\). +
The only forbidden input is the one that makes the bottom zero. Set \(x - 2 = 0\), giving \(x = 2\). Every other real number is allowed, so the domain is all real \(x\) with \(x \neq 2\).
2. If \(f(x) = x^2\) has domain \(\{-3,\ 0,\ 3\}\), list the range. +
Square each input: \((-3)^2 = 9\), \(0^2 = 0\), \(3^2 = 9\). Collecting these and listing each once gives range \(= \{0,\ 9\}\).
3. What is the range of \(f(x) = x^2\) when the domain is all real numbers? +
Squaring any real number gives zero or a positive result, and it can be as large as you like, but never negative. So the range is \(f(x) \geq 0\).
Part of the Standard Maths library. Spotted an error or want a topic added? That feedback makes the notes better.