Functions: domain, range and compositionEdexcel International A Level Maths: Revision notes

A function is a rule that gives exactly one output for each input in its domain. It is written $f:x\mapsto x^2-4x+7$ or $f(x)=x^2-4x+7$. The inputs form the domain; the set of outputs is the range. A function is one-one if each output comes from exactly one input, and many-one if two or more inputs give the same output. For $f(x)=x^2-4x+7$, $f(1)=f(3)=4$, so $f$ is many-one. A one-to-many rule is not a function.

The domain may be stated, or you find the largest set where the rule works: avoid division by zero, and square roots of negatives. For $\sqrt{x-2}+1$ the domain is $x\geqslant2$. To find the range, complete the square, sketch the graph, or use the extremes of the domain. $x^2-4x+7=(x-2)^2+3$ has range $f(x)\geqslant3$; restricting the domain to $x\geqslant4$ makes $f$ increasing, so the range becomes $f(x)\geqslant f(4)=7$. Restricting a domain can turn a many-one function into a one-one function.

The composite function $\mathrm{fg}(x)=f(g(x))$ means 'do $g$ first, then $f$'. With $f(x)=2x+3$ and $g(x)=x^2-1$: $\mathrm{fg}(x)=2(x^2-1)+3=2x^2+1,\qquad \mathrm{gf}(x)=(2x+3)^2-1.$ In general $\mathrm{fg}\neq\mathrm{gf}$. To evaluate $\mathrm{gf}(1)$: $f(1)=5$, then $g(5)=24$.

For $\mathrm{fg}$ to be defined, the range of $g$ must lie within the domain of $f$. With $f(x)=\sqrt{x-2}+1$ ($x\geqslant2$) and $g(x)=\frac{4}{x-1}$ we need $g(x)\geqslant2$, so $\frac{4}{x-1}\geqslant2$, giving $1<x\leqslant3$. Also $\mathrm{gf}(x)=\frac{4}{\sqrt{x-2}}$ must exclude $x=2$, so its domain is $x>2$. The range of the composite comes from the range of the inner function passed through the outer.