Lesson 1.6 · Unit 1 · F-BF.1, F-BF.4
Chaining functions together, and undoing them
Composition feeds the output of one function into another. Almost every formula in science is built this way. An inverse function runs the chain backward, and the fact that a function has an inverse only when it passes a simple test is the reason inverse trigonometric functions later need restricted ranges.
- \( (f \circ g)(x) = f\big(g(x)\big).\) Apply \( g \) first, then \( f \) to the result.
- To compose, substitute the whole of \( g(x) \) in place of each \( x \) in \( f \).
- Composition is not commutative: usually \( f(g(x)) \ne g(f(x)) \).
- The domain of \( f \circ g \) is the inputs where \( g \) is defined and where \( g(x) \) lies in the domain of \( f \).
- A function is one-to-one if no output is repeated, tested by the horizontal line test.
- Only a one-to-one function has an inverse. If it is not, restrict the domain to make it so.
- To find \( f^{-1} \), write \( y = f(x) \), swap \( x \) and \( y \), and solve for \( y \).
- Verify with \( f\big(f^{-1}(x)\big) = x \) and \( f^{-1}\big(f(x)\big) = x \). The domain and range of \( f \) become the range and domain of \( f^{-1} \).
Where students lose marks: reading \( f^{-1}(x) \) as \( \dfrac{1}{f(x)} \). The \( -1 \) is a label for the inverse function, not an exponent. For \( f(x) = 2x + 3 \), \( f^{-1}(x) = \dfrac{x - 3}{2} \), not \( \dfrac{1}{2x + 3} \).
The problem. Let \( f(x) = 2x + 3 \) and \( g(x) = x^2 - 1 \). (a) Find \( f(g(x)) \) and \( g(f(x)) \), and evaluate \( (f \circ g)(2) \). (b) Find the domain of \( f(x) = \dfrac{1}{x} \) composed with \( g(x) = x - 2 \). (c) Find the inverse of \( p(x) = \dfrac{x + 2}{x - 3} \) and verify it. (d) Restrict \( q(x) = x^2 - 4x \) to \( x \ge 2 \) and find its inverse.
Step one: compose in both orders for (a). \( f(g(x)) = 2(x^2 - 1) + 3 = 2x^2 + 1 \). \( g(f(x)) = (2x + 3)^2 - 1 = 4x^2 + 12x + 9 - 1 = 4x^2 + 12x + 8 \). They differ, as expected.
Step two: evaluate two ways. Directly: \( g(2) = 3 \), then \( f(3) = 9 \). From the formula: \( 2(2)^2 + 1 = 9 \). They agree.
Step three: the domain for (b). The composition is \( \dfrac{1}{x - 2} \). The inner function accepts all reals, but the result must avoid making the outer denominator zero: \( x - 2 \ne 0 \). Domain: \( x \ne 2 \).
Step four: swap and solve for (c). Write \( x = \dfrac{y + 2}{y - 3} \). Multiply out: \( x(y - 3) = y + 2 \), so \( xy - 3x = y + 2 \).
Step five: collect the \( y \) terms. \( xy - y = 3x + 2 \), so \( y(x - 1) = 3x + 2 \) and \( f^{-1}(x) = \dfrac{3x + 2}{x - 1} \).
Step six: verify. \( p\big(p^{-1}(x)\big) = \dfrac{\frac{3x+2}{x-1} + 2}{\frac{3x+2}{x-1} - 3} \). Multiply top and bottom by \( x - 1 \): \( \dfrac{3x + 2 + 2(x - 1)}{3x + 2 - 3(x - 1)} = \dfrac{5x}{5} = x \). Correct. Domains swap: \( p \) excludes \( x = 3 \) and never outputs 1, and \( p^{-1} \) excludes \( x = 1 \) and never outputs 3.
Step seven: restrict for (d). \( q(x) = x^2 - 4x = (x - 2)^2 - 4 \). The vertex is at \( x = 2 \), so \( x \ge 2 \) selects the right half, which is one-to-one. The range is \( [-4, \infty) \).
Step eight: invert the restricted piece. Swap: \( x = (y - 2)^2 - 4 \), so \( (y - 2)^2 = x + 4 \) and \( y - 2 = \pm\sqrt{x + 4} \). The original had \( y \ge 2 \) as its domain, so the inverse takes the positive root: \( q^{-1}(x) = 2 + \sqrt{x + 4} \), with domain \( [-4, \infty) \). Check: \( q(3) = 9 - 12 = -3 \), and \( q^{-1}(-3) = 2 + \sqrt{1} = 3 \). Correct. Choosing the left half instead would give \( 2 - \sqrt{x + 4} \); the restriction decides which inverse is meant.
Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.
- If \( f(x) = x + 5 \) and \( g(x) = 2x \), find \( (f \circ g)(3) \).
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\( g(3) = 6 \), then \( f(6) = 11 \). 11
- With the same functions, find \( (g \circ f)(3) \).
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\( f(3) = 8 \), then \( g(8) = 16 \). Note that it differs from the previous answer. 16
- If \( f(x) = x^2 \) and \( g(x) = x + 1 \), find \( f(g(x)) \).
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Substitute \( x + 1 \) for \( x \) in \( x^2 \). \( (x + 1)^2 = x^2 + 2x + 1 \)
- Find the inverse of \( f(x) = 4x - 8 \).
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\( x = 4y - 8 \), so \( y = \dfrac{x + 8}{4} \). Check: \( f(4) = 8 \), and \( f^{-1}(8) = \dfrac{16}{4} = 4 \). \( f^{-1}(x) = \dfrac{x + 8}{4} \)
- Is \( f(x) = x^2 \) one-to-one on all real numbers?
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\( f(2) = 4 = f(-2) \). Two inputs share an output, so it fails the horizontal line test. No
- Find the inverse of \( f(x) = x^3 - 5 \).
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\( x = y^3 - 5 \), so \( y^3 = x + 5 \) and \( y = \sqrt[3]{x + 5} \). Check: \( f(3) = 22 \), and \( \sqrt[3]{27} = 3 \) at input 22. \( f^{-1}(x) = \sqrt[3]{x + 5} \)
- If \( f(x) = \sqrt{x - 2} \) and \( g(x) = x^2 + 2 \), find \( f(g(x)) \) and its domain.
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\( f(g(x)) = \sqrt{x^2 + 2 - 2} = \sqrt{x^2} \). This is \( |x| \), not \( x \): the square root always returns the nonnegative root. The inside \( x^2 \ge 0 \) for every real \( x \), so the domain is all reals. Check at \( x = -3 \): \( g(-3) = 11 \), \( f(11) = \sqrt{9} = 3 = |-3| \). \( |x| \), domain all real numbers
- Find the inverse of \( f(x) = \dfrac{2x - 1}{x + 3} \).
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\( x = \dfrac{2y - 1}{y + 3} \), so \( xy + 3x = 2y - 1 \). Collect: \( xy - 2y = -1 - 3x \), so \( y(x - 2) = -(3x + 1) \). \( y = \dfrac{-(3x + 1)}{x - 2} = \dfrac{3x + 1}{2 - x} \). Check: \( f(1) = \dfrac{1}{4} \), and \( f^{-1}\left(\tfrac14\right) = \dfrac{\frac34 + 1}{2 - \frac14} = \dfrac{7/4}{7/4} = 1 \). \( f^{-1}(x) = \dfrac{3x + 1}{2 - x} \)
- Restrict \( f(x) = x^2 - 4x \) to \( x \ge 2 \) and find its inverse.
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Complete the square: \( f(x) = (x - 2)^2 - 4 \). With \( x \ge 2 \), swap: \( x = (y - 2)^2 - 4 \), so \( y - 2 = \sqrt{x + 4} \) (positive, because \( y \ge 2 \)). Check: \( f(5) = 25 - 20 = 5 \), and \( 2 + \sqrt{5 + 4} = 5 \). Correct. \( f^{-1}(x) = 2 + \sqrt{x + 4} \), domain \( [-4, \infty) \)
- Show that \( f(x) = x^2 \) and \( g(x) = \sqrt{x} \) satisfy \( f(g(x)) = x \) but are not
inverses on all reals.
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\( f(g(x)) = (\sqrt{x})^2 = x \), valid for \( x \ge 0 \), where \( g \) is defined. But \( g(f(x)) = \sqrt{x^2} = |x| \), which equals \( x \) only for \( x \ge 0 \). At \( x = -3 \): \( g(f(-3)) = \sqrt{9} = 3 \ne -3 \). Inverses need both compositions to return the input on the whole domain, and \( x^2 \) on all reals is not one-to-one. On \( [0, \infty) \) both hold. Both compositions must give \( x \); here only one does on all reals