Lesson 1.7 · Unit 1 · F-BF.4
The function that undoes the function
An inverse reverses what a function did: it takes the output back to the input. Not every function has one, and the condition that decides it is the same condition that will later force a restriction on the square root, the logarithm and the inverse trigonometric functions.
- \( f^{-1} \) is the function with \( f^{-1}(f(x)) = x \) and \( f(f^{-1}(x)) = x \).
- The notation is not a reciprocal. \( f^{-1}(x) \) is not \( \dfrac{1}{f(x)} \).
- To find it algebraically: write \( y = f(x) \), swap \( x \) and \( y \), solve for \( y \), and rename it \( f^{-1}(x) \).
- An inverse function exists only if \( f \) is one-to-one, meaning no output is repeated.
- The horizontal line test checks that: if any horizontal line meets the graph twice, no inverse function exists.
- A failing function can be fixed by restricting its domain, which is how \( \sqrt{x} \) becomes the inverse of \( x^2 \).
- The graphs are reflections in the line \( y = x \), because swapping coordinates is exactly that reflection.
- Domain and range swap: the domain of \( f^{-1} \) is the range of \( f \), and its range is the domain of \( f \).
Where students lose marks: reading \( f^{-1} \) as a reciprocal. For \( f(x) = x + 3 \), the inverse is \( x - 3 \), while the reciprocal is \( \dfrac{1}{x + 3} \). They are not remotely the same function, and the notation is genuinely ambiguous English that has to be learned rather than deduced.
The problem. (a) Find the inverse of \( f(x) = x^3 + 2 \) and verify it. (b) Find the inverse of \( g(x) = \dfrac{2x - 3}{x + 4} \). (c) Explain why \( h(x) = x^2 \) has no inverse function, and fix it. (d) State the domain and range of \( g^{-1} \) from part (b).
Step one: solve (a). Write \( y = x^3 + 2 \), then swap: \( x = y^3 + 2 \). Solve for \( y \): \( y^3 = x - 2 \), so \( y = \sqrt[3]{x - 2} \). Therefore \( f^{-1}(x) = \sqrt[3]{x - 2} \).
Step two: verify (a) by composing both ways. \( f^{-1}(f(x)) = \sqrt[3]{(x^3 + 2) - 2} = \sqrt[3]{x^3} = x \). Correct. \( f(f^{-1}(x)) = \left( \sqrt[3]{x - 2} \right)^3 + 2 = (x - 2) + 2 = x \). Correct. Both compositions give the identity, which is the definition, so the answer is confirmed. Numerically: \( f(2) = 10 \) and \( f^{-1}(10) = \sqrt[3]{8} = 2 \). Round trip closed.
Step three: set up (b). Write \( y = \dfrac{2x - 3}{x + 4} \) and swap: \( x = \dfrac{2y - 3}{y + 4} \).
Step four: clear the denominator. \( x(y + 4) = 2y - 3 \), so \( xy + 4x = 2y - 3 \).
Step five: collect the \( y \) terms and solve (b). Move every term with \( y \) to one side and everything else to the other: \( xy - 2y = -3 - 4x \). Factor: \( y(x - 2) = -3 - 4x \). Divide: \( y = \dfrac{-3 - 4x}{x - 2} \), which is tidier as \( y = \dfrac{4x + 3}{2 - x} \) after multiplying numerator and denominator by \( -1 \). So \( g^{-1}(x) = \dfrac{4x + 3}{2 - x} \).
Step six: verify (b) numerically. \( g(1) = \dfrac{2 - 3}{5} = -\dfrac{1}{5} \). \( g^{-1}\left( -\dfrac{1}{5} \right) = \dfrac{4\left(-\frac{1}{5}\right) + 3}{2 + \frac{1}{5}} = \dfrac{-\frac{4}{5} + \frac{15}{5}}{\frac{11}{5}} = \dfrac{\frac{11}{5}}{\frac{11}{5}} = 1 \). Correct. Factoring out the variable in step five is the move worth remembering: whenever the unknown appears in more than one term, collect and factor rather than trying to isolate directly.
Step seven: answer (c). The function \( h(x) = x^2 \) is not one-to-one: \( h(3) = 9 \) and \( h(-3) = 9 \). An inverse would have to send 9 back to its input, but there are two candidates and no rule can choose. The horizontal line \( y = 9 \) meets the graph twice, so the test fails. The fix is to restrict the domain to \( x \ge 0 \), which discards the left branch. On that restricted domain the function is one-to-one, and its inverse is \( \sqrt{x} \). Restricting to \( x \le 0 \) would work equally well and give \( -\sqrt{x} \); the choice of the nonnegative branch is a convention, made so that \( \sqrt{9} \) has one agreed meaning.
Step eight: answer (d) by swapping. The domain of \( g^{-1} \) is the range of \( g \), and its range is the domain of \( g \). \( g(x) = \dfrac{2x - 3}{x + 4} \) has domain all reals except \( -4 \). As \( x \) grows the value approaches \( \dfrac{2x}{x} = 2 \) without reaching it, so the range is all reals except 2. Therefore \( g^{-1} \) has domain all reals except 2 and range all reals except \( -4 \). Confirm directly from the formula: \( g^{-1}(x) = \dfrac{4x + 3}{2 - x} \) is undefined at \( x = 2 \), matching, and its own horizontal asymptote is at \( \dfrac{4x}{-x} = -4 \), matching too. The swap is not a shortcut to be trusted blindly, but when the direct reading agrees with it the answer is secure.
Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.
- Find the inverse of \( f(x) = x + 7 \).
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\( f^{-1}(x) = x - 7 \)
- Find the inverse of \( f(x) = 5x \).
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\( f^{-1}(x) = \frac{x}{5} \)
- Find the inverse of \( f(x) = 2x - 8 \).
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\( x = 2y - 8 \), so \( y = \dfrac{x + 8}{2} \). \( f^{-1}(x) = \frac{x+8}{2} \)
- Does \( f(x) = x^4 \) have an inverse function on all reals?
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An even power repeats outputs. No
- What line are \( f \) and \( f^{-1} \) reflections in?
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\( y = x \)
- Find the inverse of \( f(x) = \dfrac{x}{x - 1} \).
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Swap: \( x = \dfrac{y}{y - 1} \). Clear: \( x(y - 1) = y \), so \( xy - x = y \). Collect: \( xy - y = x \), so \( y(x - 1) = x \) and \( y = \dfrac{x}{x - 1} \). The function is its own inverse. Check: \( f(2) = \dfrac{2}{1} = 2 \), and \( f(3) = \dfrac{3}{2} \), then \( f\left( \dfrac{3}{2} \right) = \dfrac{3/2}{1/2} = 3 \). Round trip closed. \( f^{-1}(x) = \frac{x}{x-1} \), the same function
- Find the inverse of \( f(x) = \sqrt{x - 4} \) and state its domain.
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Swap: \( x = \sqrt{y - 4} \). Square both sides: \( x^2 = y - 4 \), so \( y = x^2 + 4 \). The domain restriction is essential. The original function outputs only nonnegative values, so its range is \( y \ge 0 \), which becomes the domain of the inverse. \( f^{-1}(x) = x^2 + 4 \) for \( x \ge 0 \). Without that restriction, \( x^2 + 4 \) would not be an inverse: at \( x = -3 \) it gives 13, but \( f(13) = \sqrt{9} = 3 \), not \( -3 \). Check with a valid value: \( f(20) = \sqrt{16} = 4 \), and \( f^{-1}(4) = 16 + 4 = 20 \). Correct. \( f^{-1}(x) = x^2 + 4 \), domain \( x \ge 0 \)
- Verify that \( f(x) = \dfrac{x - 5}{3} \) and \( g(x) = 3x + 5 \) are inverses.
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Both compositions must give the identity. \( f(g(x)) = \dfrac{(3x + 5) - 5}{3} = \dfrac{3x}{3} = x \). Correct. \( g(f(x)) = 3 \cdot \dfrac{x - 5}{3} + 5 = (x - 5) + 5 = x \). Correct. Why both are needed. One composition alone is not enough in general. A function can undo another in one direction only, when domains are restricted, so the definition requires both and a complete verification shows both. Numerical confirmation: \( g(2) = 11 \) and \( f(11) = \dfrac{6}{3} = 2 \). Verified: both compositions equal \( x \)
- Explain why a function must be one-to-one to have an inverse function.
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An inverse must send each output back to the input it came from. If two different inputs produced the same output, the inverse would have to send that one output back to two different values, and a function is not permitted to do that. The definition it violates. A function assigns exactly one output to each input. The inverse relation would assign two, so it is a relation but not a function. A concrete case. For \( f(x) = x^2 \), both 4 and \( -4 \) map to 16. Asking for \( f^{-1}(16) \) has no single right answer, and choosing 4 by convention is what defines \( \sqrt{\cdot} \) rather than what the inverse would naturally be. Where this recurs this year. The same problem forces every restriction of the year: \( \sqrt{x} \) exists as a function only because \( x^2 \) was restricted to \( x \ge 0 \); the logarithm works without restriction because the exponential is already one-to-one; and the inverse sine, cosine and tangent of unit 10 need narrow restricted ranges because the trigonometric functions repeat their outputs infinitely often. The pattern worth extracting. Whenever a course introduces an inverse with an odd-looking restriction attached, the restriction is not arbitrary. It is the minimum needed to make the original function one-to-one, and it was chosen once, by convention, so that the inverse has a single agreed value. Two inputs sharing an output would force the inverse to assign two outputs to one input, which no function may do
- A temperature conversion is \( F(c) = \dfrac{9}{5}c + 32 \). Find its inverse, interpret
both functions, and find the temperature at which the two scales agree.
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Find the inverse. Write \( f = \dfrac{9}{5}c + 32 \) and solve for \( c \): \( f - 32 = \dfrac{9}{5}c \), so \( c = \dfrac{5}{9}(f - 32) \). As a function of its input, \( F^{-1}(f) = \dfrac{5}{9}(f - 32) \). Interpret. \( F \) converts Celsius to Fahrenheit; \( F^{-1} \) converts Fahrenheit back to Celsius. The inverse undoes the conversion, which is exactly what the word means here. Verify both ways. \( F(100) = \dfrac{9}{5}(100) + 32 = 180 + 32 = 212 \), the boiling point. Correct. \( F^{-1}(212) = \dfrac{5}{9}(180) = 100 \). Round trip closed. \( F(0) = 32 \) and \( F^{-1}(32) = 0 \), the freezing point. Correct. Find where the scales agree. The two scales give the same number when \( F(c) = c \): \( \dfrac{9}{5}c + 32 = c \). Multiply through by 5: \( 9c + 160 = 5c \). So \( 4c = -160 \) and \( c = -40 \). Check. \( F(-40) = \dfrac{9}{5}(-40) + 32 = -72 + 32 = -40 \). Correct. Minus forty degrees is the same temperature on both scales, which is why weather reports at that temperature omit the unit without ambiguity. The graphical meaning. Solving \( F(c) = c \) finds where the graph of \( F \) meets the line \( y = x \). Since \( F \) and \( F^{-1} \) are reflections in that line, any point where a function meets \( y = x \) is a point the function and its inverse share. So \( (-40, -40) \) lies on both graphs, and it is their only intersection here because both are lines with different slopes. A note on the model. The conversion is exact by definition rather than approximate, since both scales are defined by fixed reference points. That is unusual: most formulas in an applied problem are models with error, and this one is not. \( F^{-1}(f) = \frac{5}{9}(f - 32) \); the scales agree at \( -40^\circ \)