Learning goals: Compute compositions with domain restrictions; find and verify inverse functions; distinguish inverses from reciprocals.
Combining functions
Functions can be added, subtracted, multiplied, and divided where the required outputs exist. Division also excludes inputs where the denominator function is zero. Composition is different: it feeds the output of one function into another. The notation \((f\circ g)(x)=f(g(x))\) means apply \(g\) first, then \(f\). Order matters, as it does when putting on socks and shoes.
Let \(f(x)=2x+3\) and \(g(x)=x^2\). Then \(f(g(x))=2x^2+3\), while \(g(f(x))=(2x+3)^2\). At one, these give five and twenty-five. The symbols look similar, but the input-output processes differ.
Domain of a composition
For \(f(g(x))\), the input must belong to the domain of \(g\), and the resulting \(g(x)\) must belong to the domain of \(f\). If \(f(u)=\sqrt{u}\) and \(g(x)=5-2x\), then \(f(g(x))=\sqrt{5-2x}\) requires \(x\le5/2\). If \(g(x)=1/(x-1)\), then \(\sqrt{g(x)}\) requires the reciprocal to be nonnegative as well as defined; the domain is \((1,\infty)\), not merely all real numbers except one.
What an inverse does
An inverse function reverses an input-output assignment. If \(f(2)=9\), then \(f^{-1}(9)=2\), when an inverse exists. The notation \(f^{-1}\) is not the reciprocal \(1/f\). An inverse can be a function only when the original function is one-to-one: distinct inputs must have distinct outputs. On a graph, use the horizontal line test. Domain and range exchange roles, and inverse graph points swap coordinates, producing reflection across \(y=x\).
For \(f(x)=3x-4\), write \(y=3x-4\), solve for the original input, and interchange labels: \[x=\frac{y+4}{3}\quad\Longrightarrow\quad f^{-1}(x)=\frac{x+4}{3}.\] Verify both compositions: \(f(f^{-1}(x))=x\) and \(f^{-1}(f(x))=x\) on their appropriate domains. This check is stronger than simply recognizing a familiar formula.
Restricting a domain
The function \(x^2\) is not one-to-one on all real numbers because opposite inputs share an output. Restrict it to \(x\ge0\); it then has inverse \(\sqrt{x}\) with domain and range both \([0,\infty)\). Restricting instead to \(x\le0\) gives inverse \(-\sqrt{x}\). The domain choice determines the inverse, so do not omit it.
Application: reverse a conversion
The temperature rule \(F(C)=\frac95C+32\) takes Celsius to Fahrenheit. Subtract 32 and multiply by \(5/9\) to reverse the process: \(C(F)=\frac59(F-32)\). The order of inverse operations reverses the order of the original operations. As an investigation, build a three-step “function machine,” exchange outputs with a partner, and recover inputs. Explain when an operation such as squaring makes recovery ambiguous and what domain restriction fixes it.
Practice: attempt before reading the solutions
Show your reasoning, state restrictions, and check your result. For explanation questions, use complete mathematical sentences.
- For \(f(x)=x+4\), \(g(x)=3x\), find \(f(g(2))\).
- For \(f(x)=x^2\), \(g(x)=x-1\), compute both compositions.
- Find the domain of \(\sqrt{3x-6}\).
- Find the inverse of \(f(x)=5x+2\).
- Explain why \(f(x)=|x|\) has no inverse function on all real inputs.
- Find the inverse of \(f(x)=(x-2)^2\) restricted to \(x\ge2\).
- If \(f(7)=11\) and \(f\) is invertible, what is \(f^{-1}(11)\)?
- Find the domain of \(h(x)=\sqrt{1/(x+2)}\).
Worked solutions and reasoning
- First compute \(g(2)=6\), then \(f(6)=10\). Composition follows the inner function first.
- \((f\circ g)(x)=(x-1)^2\), whereas \((g\circ f)(x)=x^2-1\). They differ because the operations occur in different orders.
- Require \(3x-6\ge0\), so \(x\ge2\). The square-root composition has domain \([2,\infty)\).
- Solve \(y=5x+2\) for \(x\): \(x=(y-2)/5\). Thus \(f^{-1}(x)=(x-2)/5\), with all real numbers as domain and range.
- For example, both \(-3\) and \(3\) map to three. The reverse assignment would give two outputs for input three, so it is not a function.
- From \(y=(x-2)^2\) and \(x-2\ge0\), obtain \(x=2+\sqrt y\). Thus \(f^{-1}(x)=2+\sqrt x\), with domain \([0,\infty)\) and range \([2,\infty)\).
- It is seven. The inverse reverses the original ordered pair: \((7,11)\) becomes \((11,7)\).
- The fraction must be defined and nonnegative. Its numerator is positive, so the denominator must be positive: \(x+2\gt0\). Domain: \((-2,\infty)\).
Reflection
Choose one problem you initially found difficult. Explain the key idea, identify a possible mistake, and write a new problem that uses the same idea. Solve your new problem and check it independently.
Unit synthesis investigation: A restricted inverse with a real purpose
A square display has side length \(s\ge2\) meters and area \(A(s)=s^2\). A two-meter-wide border is not involved; the condition is simply a minimum side length. Find the domain and range of A, construct its inverse with domain and range, and verify both compositions on their proper domains. If a vendor reports area three square meters, explain why a square exists mathematically but does not satisfy this model.
Worked discussion — open after attempting
The domain is \([2,\infty)\) and range \([4,\infty)\). On this domain squaring is one-to-one, so \(A^{-1}(u)=\sqrt u\), with domain \([4,\infty)\) and range \([2,\infty)\). For \(s\ge2\), \(A^{-1}(A(s))=\sqrt{s^2}=s\) because s is nonnegative. For \(u\ge4\), \(A(A^{-1}(u))=(\sqrt u)^2=u\). An area of three corresponds to side length \(\sqrt3\), below two, so it violates the display constraint. The inverse is not \(1/u^2\); it reverses the side-to-area assignment. This example shows how context selects a domain before inverse algebra begins.