Functions: Domain and Range
Domain and range describe the inputs a function is allowed to take and the outputs it produces, and Edexcel IGCSE Maths Higher papers test both directly. This page covers function notation, evaluating and solving with functions, deciding whether a mapping is a function, finding the range from a given domain, and spotting the values that must be excluded from a domain. Work through the worked examples and diagrams, then test every skill on the auto-marked practice questions below.
What Is a Function?
A function is a rule that turns each input number into exactly one output number. Picture a machine: a number goes in, the rule is applied, and a single number comes out.
Function notation
The rule "double, then add 7" is written \(f(x) = 2x + 7\), or equivalently \(f : x \mapsto 2x + 7\). Both say: take an input \(x\), double it, add 7. Evaluating a function is substitution: \(f(3)\) means replace \(x\) with 3, giving \(f(3) = 2 \times 3 + 7\), so \(f(3) = 13\).
Any letter can name a function: \(g\), \(h\) and \(\mathrm{p}\) work exactly like \(f\).
Which mappings count as functions?
- One-to-one: every input has its own output. This is a function.
- Many-to-one: several inputs share an output (squaring does this: \(3\) and \(-3\) both give \(9\)). Still a function, because each input has exactly one output.
- One-to-many: one input produces more than one output. Not a function.
On a graph, use the vertical line test: if every vertical line crosses the graph at most once, the graph shows a function. If any vertical line crosses it twice or more, it does not.
How to Find the Domain and Range
Range from a listed domain
Substitute every value of the domain into the function and list the outputs. Duplicates are written once: if \(f(x) = x^2\) has domain \(\{-3, 3\}\), the range is just \(\{9\}\).
Range over an interval
- Substitute both ends of the domain into the function.
- The two outputs are the ends of the range; write it as an inequality in \(f(x)\).
- Check which output is the larger one. A negative gradient sends the smallest input to the largest output, so never assume the order.
Range of a squared function
A square is never negative, so \(x^2 \geq 0\) for every real \(x\). Adding a constant shifts every output: \(x^2 + 5\) has range \(f(x) \geq 5\). In completed-square style, \((x-a)^2 + b\) has least value \(b\) (at \(x = a\)), so its range is \(f(x) \geq b\). A subtracted square flips it: \(b - x^2\) has greatest value \(b\), so its range is \(f(x) \leq b\).
Values excluded from the domain
Two operations are impossible with real numbers: dividing by zero, and square rooting a negative number. To find the excluded inputs:
- Fractions: set the denominator equal to zero and solve. Those \(x\) values are excluded. A squared bracket in the denominator still gives one excluded value; a difference of two squares such as \(x^2 - 25\) gives two.
- Square roots: the expression under the root must be at least zero. Solve for the allowed inputs; everything else is excluded. Note that \(\sqrt{0} = 0\) is allowed, so the boundary value itself is fine.
- Roots in a denominator: for \(\dfrac{1}{\sqrt{x}}\)-style functions the boundary is excluded too, because the root would be zero and you would divide by it.
A polynomial such as \(f(x) = x^2 + 3x - 5\) uses neither operation, so it has no excluded values.
Worked Examples: Domain, Range and Function Notation
💡 Example 1: Is the mapping a function?
Each mapping diagram below shows a rule applied to a small set of inputs. Decide whether each one is a function.
\(x \mapsto x + 4\)
One arrow leaves each input: one-to-one, a function ✓
\(x \mapsto \pm\sqrt{x}\)
Input 4 fires two arrows: one-to-many, not a function ✗
What's happening?
Count the arrows leaving each input. Exactly one arrow from every input means a function. It does not matter how many arrows arrive at an output; only the arrows leaving each input decide it.
💡 Example 2: The vertical line test
Use the vertical line test to decide whether each graph shows a function.
A parabola
Every vertical line crosses once: a function (many-to-one) ✓
A sideways curve
The red line crosses twice: not a function ✗
What's happening?
A vertical line marks one input value. If the graph meets it twice, that input has two outputs, which breaks the function rule. The parabola is safe everywhere; the sideways curve fails wherever the red line catches both branches.
💡 Example 3: Evaluate and solve
Given \(f(x) = 5x + 2\), find \(f(4)\) and \(f(-3)\), then solve \(f(x) = 32\).
Evaluating is substitution:
\[ \begin{array}{rcl} f(4) &=& 5 \times 4 + 2 \\ &=& 22 \\[4pt] f(-3) &=& 5 \times (-3) + 2 \\ &=& -13 \end{array} \]Solving \(f(x) = 32\) is an equation:
\[ \begin{array}{rcl} 5x + 2 &=& 32 \\ 5x &=& 30 \\ x &=& 6 \end{array} \]What's happening?
\(f(4)\) asks for an output: substitute in. \(f(x) = 32\) gives you the output and asks which input produced it: set the rule equal to 32 and solve. Keep negatives in brackets when substituting.
💡 Example 4: Range from a listed domain
Find the range of \(g(x) = x^2 + 1\) for the domain \(\{-3, -1, 1, 2\}\).
Range: \(\{2, 5, 10\}\)
What's happening?
Push every domain value through the function and collect the outputs. \(-1\) and \(1\) both give 2, and a set lists each value once, so the range has three members, not four.
💡 Example 5: Range over an interval
Find the range of \(f(x) = 3x - 1\) for the domain \(0 \leq x \leq 3\).
Range: \(-1 \leq f(x) \leq 8\)
What's happening?
The domain runs along the x-axis; the graph carries it up to the line; the range is the strip of the y-axis the segment covers. If the gradient were negative, the smallest input would give the largest output, so always check both ends.
💡 Example 6: Range of a squared function
Find the range of \(h(x) = (x - 2)^2 + 3\), where the domain is all real numbers.
\((x-2)^2 \geq 0\) for every real \(x\), and it equals 0 only at \(x = 2\).
So the least value of \(h(x)\) is \(0 + 3 = 3\).
Range: \(h(x) \geq 3\)
What's happening?
The squared bracket bottoms out at zero, so the whole function bottoms out at the constant on the end. The graph confirms it: the lowest point is \((2, 3)\) and every output from 3 upwards is produced. For \(3 - x^2\)-style functions the parabola opens downwards and 3 becomes the greatest value instead.
💡 Example 7: Values excluded from the domain
State which values must be excluded from the domain of \(g(x) = \dfrac{1}{x + 3}\) and of \(h(x) = \sqrt{x - 1}\).
\(y = \dfrac{1}{x+3}\)
\(y = \sqrt{x-1}\)
For \(g\), set the denominator equal to zero:
\[ \begin{array}{rcl} x + 3 &=& 0 \\ x &=& -3 \end{array} \]Exclude \(x = -3\).
For \(h\), the radicand must not be negative: \(x - 1 \geq 0\) gives \(x \geq 1\) allowed, so exclude \(x < 1\).
What's happening?
The graphs tell the same story. The reciprocal curve splits into two branches either side of the dashed line at \(x = -3\): the function simply has no value there. The root curve starts at \((1, 0)\) and exists only to the right. Note \(\sqrt{0} = 0\) is allowed, so \(x = 1\) itself stays in the domain: the excluded region is strictly \(x < 1\). Only when the root sits in a denominator, as in \(\dfrac{1}{\sqrt{x-1}}\), does the boundary get excluded too, giving \(x \leq 1\).
🔑 Key Points
- A function turns each input into exactly one output.
- One-to-one and many-to-one mappings are functions; one-to-many is not.
- Vertical line test: a graph shows a function if no vertical line crosses it more than once.
- Domain = allowed inputs (x-axis). Range = outputs produced (y-axis).
- Evaluating \(f(3)\) is substitution; solving \(f(x) = k\) is an equation.
- \((x-a)^2 + b\) has least value \(b\); its range is \(f(x) \geq b\).
- Excluded values come from a denominator equal to zero or a negative under a square root.
⚠️ Common Pitfalls
- Swapping domain and range: the domain is the inputs, the range is the outputs.
- Calling many-to-one "not a function". Sharing an output is fine; only one input with two outputs breaks the rule.
- Substituting a negative without brackets: \(f(-3)\) into \(x^2\) is \((-3)^2 = 9\), not \(-9\).
- Writing the range of \(x^2 + 6\) as \(f(x) \geq 0\). The square is at least 0, so the function is at least 6.
- Missing the second excluded value of \(\dfrac{1}{x^2 - 25}\): both \(x = 5\) and \(x = -5\) make the denominator zero.
- Excluding the boundary of a square root: \(\sqrt{0}\) is allowed, so \(\sqrt{x - 4}\) excludes only \(x < 4\), not \(x \leq 4\).