2. Functions, domains and inverse maps

Learning goals

Determine domains, compose functions in the correct order, interpret graph transformations, and justify the existence of an inverse. A formula alone does not specify a function: its domain and codomain are part of the data.

Functions and their operations

A function $f:A\to B$ assigns exactly one value in $B$ to each input in $A$. Its range is $f(A)=\{f(x):x\in A\}$, which can be smaller than its codomain $B$. The function is injective if $f(x)=f(y)$ implies $x=y$, and surjective if $f(A)=B$. A bijection is both.

For real formulas, square roots require nonnegative radicands, real logarithms require positive arguments, and denominators must not vanish. These constraints apply simultaneously. For functions $f:A\to B$ and $g:B\to C$, the composite $g\circ f:A\to C$ is $x\mapsto g(f(x))$. With partially specified real formulas, its domain consists of inputs in the domain of $f$ whose outputs lie in the domain of $g$.

For $a\ne0$, the graph of $f(a(x-h))+k$ is obtained by adjusting inputs and outputs. The input relation $u=a(x-h)$ means the old point $(u,f(u))$ becomes $(h+u/a,k+f(u))$. This equation is safer than memorizing a vague rule about stretching.

Theorem: inverse criterion

A function $f:A\to B$ has a two-sided inverse $g:B\to A$, satisfying $g\circ f=\operatorname{id}_A$ and $f\circ g=\operatorname{id}_B$, if and only if $f$ is bijective. The inverse is unique.

Proof. If an inverse exists and $f(x)=f(y)$, apply $g$ to get $x=y$. For $b\in B$, $b=f(g(b))$, so $f$ is onto. Conversely, surjectivity supplies at least one preimage of each $b$ and injectivity makes it unique; define $g(b)$ to be that preimage. The two identities follow from the definition. Any inverse must send $b$ to its unique preimage, proving uniqueness.

A strictly increasing real function is injective: $x<y$ forces $f(x)<f(y)$. Monotonicity alone does not identify its range. For instance $e^x$ is not onto $\mathbb R$ when viewed as a map into $\mathbb R$.

Worked examples

A restricted inverse. The map $f:[2,\infty)\to[0,\infty)$, $f(x)=(x-2)^2$, is strictly increasing and onto. Solving $y=(x-2)^2$ with $x\ge2$ gives $x=2+\sqrt y$. Therefore $f^{-1}(y)=2+\sqrt y$. On all of $\mathbb R$ the same quadratic is not injective; keeping both square-root signs does not produce an inverse function.

Composition domains. Let $f(x)=\sqrt{x-1}$ and $g(t)=1/(t-2)$ on their natural real domains. Then $(g\circ f)(x)=1/(\sqrt{x-1}-2)$ has domain $[1,\infty)\setminus\{5\}$. In the other order $(f\circ g)(x)=\sqrt{1/(x-2)-1}$ requires $(3-x)/(x-2)\ge0$, which holds exactly on $(2,3]$. The different domains reflect different functions.

Graded practice

1. Foundation. Find the natural real domain of $h(x)=\sqrt{4-x^2}/(x-1)$.
2. Core. For $f(x)=2x-3$ and $g(x)=x^2$, compute both composites and solve $(f\circ g)(x)=(g\circ f)(x)$.
3. Core. Find the range and inverse of $f:(-1,\infty)\to\mathbb R$, $f(x)=\log(x+1)$.
4. Proof. Suppose $g\circ f$ is injective. Prove that $f$ is injective. Must $g$ be injective on its entire domain?

Full solutions

1. The radicand condition is $-2\le x\le2$, and the denominator excludes $1$. Thus the domain is $[-2,1)\cup(1,2]$.
2. The composites are $2x^2-3$ and $(2x-3)^2=4x^2-12x+9$. Equating gives $x^2-6x+6=0$, so $x=3\pm\sqrt3$.
3. Here $\log$ denotes the natural logarithm, inverse to the real exponential. For each $y\in\mathbb R$, put $x=e^y-1>-1$. Then $\log(x+1)=\log(e^y)=y$, proving that the range is $\mathbb R$. Conversely, for every $x>-1$, $e^{\log(x+1)}-1=x$. The two identities verify the inverse $f^{-1}(y)=e^y-1$ on $\mathbb R$ directly, without inferring surjectivity from monotonicity and endpoint behavior alone.
4. If $f(x)=f(y)$, then $g(f(x))=g(f(y))$, so injectivity of the composite gives $x=y$. No conclusion holds about $g$ outside $f(A)$. For example, take $f:(0,\infty)\to\mathbb R$ to be inclusion and $g:\mathbb R\to[0,\infty)$ to be $g(t)=t^2$. The composite is injective but $g$ is not.

Summary and term definitions

A domain is the set of permitted inputs, a codomain is the declared target set, and the range is the actual output set. Restriction means using the same rule on a smaller domain. Composition feeds one output into another input. An inverse function exists for a bijection, while merely solving an equation may give several branches. Check the domain before algebraic simplification and state every branch restriction.