Find inverse functions: determine an input from a known output, read an inverse from a table or graph, find an inverse algebraically, verify it by composition, and restrict a domain to make a function invertible

Learning goal: Find inverse functions. Informally determine the input of a function when the output is known; read values of an inverse function from a graph or a table, given that the function has an inverse; find the inverse of a function algebraically, given that the function has an inverse; verify by composition that one function is the inverse of another; and produce an invertible function from a non-invertible function by restricting the domain.

Scope note. Parts b, c, d and e of F.BF.4 (reading an inverse from a graph or table, verifying by composition, finding an inverse algebraically, and restricting a domain) are marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate standards "for students who are planning on taking advanced courses". They are taught here in full, as extension content for students on that path.

Running a function backwards. A function f takes an input and produces an output. Many questions run the other way: the output is known and the input is wanted. If a taxi charges f(x) = 3x + 5 dollars for x miles and the fare was 26 dollars, the question "how far was the ride?" asks for the input that produced the output 26. Informally, undo the steps in reverse order: the function multiplied by 3 and then added 5, so subtract 5 to get 21 and divide by 3 to get 7 miles. The check f(7) = 21 + 5 = 26 confirms it. The Celsius temperature for F degrees Fahrenheit is C(F) = (5/9)(F − 32); if the output is 25 °C, undo by multiplying by 9/5 to get 45 and adding 32 to get 77 °F, and C(77) = (5/9)(45) = 25 checks. Determining an input from a known output in this informal way is the seed of the whole topic.

The inverse function. When a function f pairs each input with exactly one output and, in addition, each output comes from exactly one input, the pairing can be reversed to give a new function, the inverse of f, written f⁻¹. It takes f's outputs as its inputs and returns f's inputs as its outputs: f⁻¹(b) = a exactly when f(a) = b. The condition that each output comes from only one input is essential. The function f(x) = x² sends both 2 and −2 to 4, so "the input that gives 4" has two answers, and x² has no inverse on its full domain. On a graph, a function has an inverse exactly when no horizontal line crosses its graph more than once, because a horizontal line at height b marks every input with output b. Note that the superscript −1 in f⁻¹ names the inverse function; it is not the reciprocal 1/f(x), and confusing the two is the most damaging misconception in this lesson.

Reading an inverse from a table or a graph. A table of a function that has an inverse is also a table of the inverse, read from the output column back to the input column. Suppose a table lists f(1) = 3, f(2) = 7, f(3) = 12, f(4) = 18 and f(5) = 25. Then f⁻¹(12) = 3, f⁻¹(25) = 5 and f⁻¹(3) = 1; the value f⁻¹(10) is not in the table at all, and the table gives no way to find it. On a graph, the point (a, b) lies on the graph of f exactly when (b, a) lies on the graph of f⁻¹, so the graph of f⁻¹ is the reflection of the graph of f across the line y = x. To read f⁻¹(4) from the graph of f, find the point on f with vertical coordinate 4 and report its horizontal coordinate; if f passes through (2, 4), then f⁻¹(4) = 2. The phrase "given that the function has an inverse" is doing real work: if the horizontal line y = 4 crossed the graph twice, there would be two candidate answers and no inverse function to read.

Finding the inverse algebraically. Write y = f(x), swap the roles of x and y, and solve for y; the result is f⁻¹(x). For f(x) = 2x − 4: y = 2x − 4 becomes x = 2y − 4, so 2y = x + 4 and y = (x + 4)/2. Thus f⁻¹(x) = (x + 4)/2. Test one pair: f(3) = 2, and f⁻¹(2) = 6/2 = 3, so the input 3 is recovered. For g(x) = x³ + 1: x = y³ + 1 gives y³ = x − 1 and y = ∛(x − 1), so g⁻¹(x) = ∛(x − 1); g(2) = 9 and g⁻¹(9) = ∛8 = 2. For a rational function h(x) = (x + 3)/(x − 2), defined for x ≠ 2: x = (y + 3)/(y − 2), so x(y − 2) = y + 3, then xy − 2x = y + 3, gather the y terms as xy − y = 2x + 3, factor y(x − 1) = 2x + 3 and divide: h⁻¹(x) = (2x + 3)/(x − 1). Test: h(3) = 6/1 = 6 and h⁻¹(6) = 15/5 = 3. The step a student skips is the gathering of every y term on one side before factoring; without it the division cannot be done. Each of these functions has an inverse because each passes the horizontal line test: a line, a cubic that never turns, and a rational function whose graph never revisits a height.

Verifying by composition. Two functions f and g are inverses of each other when composing them in either order returns the input unchanged: f(g(x)) = x for every x in the domain of g, and g(f(x)) = x for every x in the domain of f. For f(x) = 2x − 4 and g(x) = (x + 4)/2: f(g(x)) = 2 · (x + 4)/2 − 4 = x + 4 − 4 = x, and g(f(x)) = ((2x − 4) + 4)/2 = 2x/2 = x. Both compositions collapse to x, so g is the inverse of f. The test also exposes wrong candidates. A student who guesses that the inverse of 2x − 4 is x/2 + 4, undoing the operations in the wrong order, computes f(x/2 + 4) = 2(x/2 + 4) − 4 = x + 8 − 4 = x + 4, which is not x, and the candidate is rejected; at x = 6 the composition gives 10 rather than 6. Verification by composition is the proof; the swap-and-solve procedure is only the method of discovery.

Restricting the domain to produce an invertible function. A function that fails the horizontal line test can be made invertible by keeping only part of its domain, chosen so that no output is repeated. The function f(x) = x² is not invertible on all real numbers, but on the domain x ≥ 0 it is: each nonnegative output 4 comes from exactly one nonnegative input, 2. The inverse of this restricted function is f⁻¹(x) = √x, the nonnegative square root, defined for x ≥ 0. On the graph, restricting to x ≥ 0 erases the left arm of the parabola, and the surviving right arm reflects across y = x onto the graph of √x, which passes through (4, 2) just as the arm passes through (2, 4). Restricting instead to x ≤ 0 is an equally valid choice, and its inverse is −√x; what matters is that a choice is made and stated. The same idea handles any parabola. For p(x) = x² − 6x + 10 = (x − 3)² + 1, with vertex (3, 1), the natural restriction is x ≥ 3, the right arm. Then x = (y − 3)² + 1 gives (y − 3)² = x − 1, and since y ≥ 3 the root is the positive one, y = 3 + √(x − 1). So p⁻¹(x) = 3 + √(x − 1) for x ≥ 1, and p(5) = 25 − 30 + 10 = 5 with p⁻¹(5) = 3 + √4 = 5 confirms the pair. Had the restriction been x ≤ 3, the inverse would be 3 − √(x − 1); the restriction decides which root, which is why it must be stated before the algebra begins.

Worked reading of an inverse from a table and a graph: A table gives the volume of water in a tank as a function of time: V(0) = 100, V(2) = 160, V(4) = 220, V(6) = 280, V(8) = 340 liters. The tabulated volumes rise by the same 60 liters every 2 minutes; modeling the tank as filling at that steady rate between readings, V is increasing, so it has an inverse, and V⁻¹ gives the time at which a volume was reached: V⁻¹(220) = 4 minutes and V⁻¹(340) = 8 minutes. On the graph of V, the point (6, 280) becomes (280, 6) on the graph of V⁻¹. Informally, the input for output 250 is found by noticing 250 is halfway between 220 and 280 and the function rises 30 liters per minute, so the time is 5 minutes; V(5) = 100 + 30 · 5 = 250 confirms it.

Worked algebraic inverse with verification: Find the inverse of f(x) = 5 − 3x and verify it. Swap: x = 5 − 3y, so 3y = 5 − x and y = (5 − x)/3, giving f⁻¹(x) = (5 − x)/3. Verify: f(f⁻¹(x)) = 5 − 3 · (5 − x)/3 = 5 − (5 − x) = x, and f⁻¹(f(x)) = (5 − (5 − 3x))/3 = 3x/3 = x. Both compositions return x. Numerical check: f(4) = 5 − 12 = −7 and f⁻¹(−7) = (5 + 7)/3 = 4, the original input.

Worked domain restriction: Make q(x) = (x + 2)² invertible and find the inverse. The parabola has vertex (−2, 0) and repeats every positive output, for example q(0) = 4 = q(−4). Restrict to x ≥ −2, the right arm, where outputs never repeat. Swap: x = (y + 2)², and because y + 2 ≥ 0 on the restricted domain, y + 2 = √x and y = √x − 2. So q⁻¹(x) = √x − 2 for x ≥ 0. Check by composition: q(q⁻¹(x)) = (√x − 2 + 2)² = (√x)² = x for x ≥ 0, and q⁻¹(q(x)) = √((x + 2)²) − 2 = (x + 2) − 2 = x, where the square root of the square is x + 2 rather than |x + 2| precisely because x ≥ −2. The restriction is what makes the last step legal.

Representation task: On axes from 0 to 10 in both directions, plot f(x) = x² for x ≥ 0 using x = 0, 1, 2, 3, plot f⁻¹(x) = √x using x = 0, 1, 4, 9, and draw the line y = x. Mark (2, 4) on f and (4, 2) on f⁻¹ and draw the segment joining them, noting that y = x cuts it in half at a right angle. Then make a two-row table of f and, under it, the table of f⁻¹ obtained by swapping the rows, and write one sentence explaining why the left arm of the parabola had to be removed before the inverse could be a function.

Independent performance: Choose three functions: a linear function, a cubic or rational function that has an inverse, and a quadratic that does not. For the first two, find the inverse algebraically, verify by composition in both orders, and confirm with one numerical pair. For the quadratic, show with a horizontal line and two inputs that it has no inverse, restrict the domain to one side of the vertex, find the inverse of the restricted function, and state its domain. Then build a table of six values of one of your invertible functions, hand your partner the table with both columns but without the formula, and ask them to read two values of the inverse from it and one from a sketch of the graph. Success criteria: every inverse is written as a function of x with its domain; every verification shows both compositions simplifying to x; the restricted domain is stated before the root is chosen; and the informal "find the input from the output" reasoning is shown for at least one value.