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
Compare these steps with your own work. The solution numbers match the practice questions above.
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First compute \(g(2)=6\), then \(f(6)=10\). Composition follows the inner function first.
The inner function triples the input before the outer function adds four. In symbols, \(f(g(x))=3x+4\), so the value at two is ten. Reversing the order would instead give \(g(f(2))=18\).
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\((f\circ g)(x)=(x-1)^2\), whereas \((g\circ f)(x)=x^2-1\). They differ because the operations occur in different orders.
Expand the first composition to obtain \(x^2-2x+1\); the second is \(x^2-1\). For example, at zero they give one and negative one. A single differing input is enough to show that the two functions are not equal.
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Require \(3x-6\ge0\), so \(x\ge2\). The square-root composition has domain \([2,\infty)\).
The inner linear expression has no restriction of its own, but its output must be an allowed input of the square root. Solve \(3x-6\ge0\). Equality is allowed because a square root of zero is defined.
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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.
Check both compositions: \(f((x-2)/5)=x\) and \((f(x)-2)/5=x\). The inverse undoes “multiply by five, then add two” in reverse order: subtract two, then divide by five.
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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.
An inverse function requires the original function to be one-to-one. The horizontal line at output three meets the absolute-value graph twice. Restricting the original domain, for example to \(x\ge0\), can remove this ambiguity.
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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 \([2,\infty)\).
The restriction determines the positive square-root branch: \(\sqrt{(x-2)^2}=x-2\) only because \(x\ge2\). The original range becomes the inverse domain. Without the restriction, reversing the parabola would produce two possible outputs for most inputs.
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It is seven. The inverse reverses the original ordered pair: \((7,11)\) becomes \((11,7)\).
The original function sends seven to eleven, so the inverse sends eleven back to seven. This use of \(f^{-1}\) means an inverse function, not the reciprocal \(1/f\); those operations have different meanings.
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The fraction must be defined and nonnegative. Its numerator is positive, so the denominator must be positive: \(x+2\gt0\). Domain: \((-2,\infty)\).
If \(x<-2\), the fraction is negative and has no real square root. At \(x=-2\), the denominator is zero. If \(x>-2\), the fraction is positive, so every such input is valid; the endpoint is excluded.
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.