Functions and algebra · Eqora guide
Inverse functions in math: undo, don't reciprocate
Recover an input by reversing the operations. Check uniqueness, swap the domain and range, and verify the whole round trip.
You are given f(x) = 3x − 6 and asked for f⁻¹(x). Does the raised minus one tell you to write 1/(3x − 6)? No: inverse functions in math recover the input that produced an output. Here the rule triples an input, then subtracts six. To return, add six to the output, then divide by three. The inverse is f⁻¹(x) = (x + 6)/3, not the reciprocal of the original expression.
Reversing arithmetic is only part of the job. A valid inverse function must return exactly one permitted input for each output in the original range. Squaring can lose the sign of an input; a restricted domain can restore uniqueness. We will build the inverse, explain the notation, decide when a quadratic needs a restriction and test both directions. You can complete every step with paper and a pencil, without relying on a solver.
Eqora publishes this guide as a learning resource, not a guarantee of correct answers or grades and not a replacement for your own examination work. Check the full problem, its notation, domains and assumptions. The sail-loft photographs illustrate undoing a process and choosing a path. They are not graphs or numerical evidence; use the stated mathematical relationships for every calculation.
Inverse functions in math return the original input
Start with a concrete trip through f. An input of five gives f(5) = 3 × 5 − 6 = 9. The inverse takes that output nine as its own input and returns five: f⁻¹(9) = (9 + 6)/3 = 5. The roles of the quantities change, although their values stay the same. A point (5, 9) on the original graph therefore becomes (9, 5) on the inverse graph.
The University of Connecticut definition describes these round trips as f⁻¹(f(x)) = x and f(f⁻¹(y)) = y, on the appropriate domains. The inner function acts first. The first trip begins with an original input, while the second begins with an original output. Writing different letters for the two starting roles helps you see why their allowed sets need not be identical.
Keep the notation attached to its purpose. The expression f⁻¹(x) names the inverse function evaluated at x. The expression 1/f(x) instead divides one by the value f(x). For our example, f⁻¹(9) is five, whereas 1/f(9) = 1/21. These quantities solve different questions. A negative exponent on an ordinary number means reciprocal; the superscript in the name of an inverse function follows a different convention.
Undo the last operation first
For f(x) = 3x − 6, the forward order is multiply by three, then subtract six. Undo subtraction first by adding six; undo multiplication next by dividing by three. Subtracting six and then multiplying by three is a different chain. Compare (x + 6)/3 with x/3 + 6: the latter adds six after division and therefore fails to undo the original subtraction at the correct stage.
The algebra makes the order visible. Write y = 3x − 6, add six to both sides to get y + 6 = 3x, and divide both sides by three to get x = (y + 6)/3. At this stage the original input x is expressed in terms of the original output y. Rename that output variable x to write f⁻¹(x) = (x + 6)/3. Renaming is not a second arithmetic operation.
You may instead swap x and y at the start, then solve x = 3y − 6 for y. Either route gives the same rule. Choose the order that lets you track the roles clearly; do not mix the two halfway through. Parentheses matter: the whole quantity x + 6 is divided by three. Before simplifying, say the reverse operations aloud and compare them with your written expression.
Check whether the output identifies one input
Consider q(x) = x² on all real numbers. Both q(3) and q(−3) equal nine. If you receive nine, you cannot know which of those two inputs to return. The expression ±√x describes two candidates for a positive input, not one output from an inverse function. Solving one equation can legitimately produce two answers; defining one inverse function requires a unique answer at each allowed input.
The property you need is one-to-one, also called injective: different permitted inputs produce different outputs. A graph passes the horizontal-line test when no horizontal line meets it more than once. The vertical-line test asks a different question: whether each input has one output in the original rule. A parabola passes the vertical test but fails the horizontal test when both sides are included.
Use a collision as an efficient counterexample. Two distinct allowed inputs with equal outputs prove that a function is not one-to-one. One successful pair of inverse calculations does not prove that it is one-to-one everywhere. A strictly increasing or strictly decreasing function on an interval is one-to-one there, but you still need to state the interval. Never silently replace an unrestricted problem with the branch you prefer.
The domain and range exchange jobs
The original domain contains allowed inputs; the range contains outputs actually reached. An inverse starts from those reached outputs and returns the corresponding original inputs. Thus its domain is the original range, and its range is the original domain. Distinguish the range from a larger declared target set: an inverse on that entire target also requires every target value to be reached, the surjectivity part of bijectivity.
For f(x) = 3x − 6 with 0 ≤ x ≤ 4, the range is −6 ≤ y ≤ 6. The inverse rule remains (x + 6)/3, but its domain is now [−6, 6] and its range is [0, 4]. Although the expression can be evaluated at twelve, twelve is outside this inverse's domain. A formula's largest algebraically possible domain does not override an explicit restriction in the problem.
The Stuttgart mathematics course explains the exchange of domain and range and the graph's reflection across y = x. Reflection swaps coordinates, not signs: (0, −6) becomes (−6, 0), and (4, 6) becomes (6, 4). If you need to organize coordinates first, our guide to reading equation graphs explains the distinction between an input and its plotted output. Preserve included endpoints when you swap the two intervals.
Choose the quadratic branch from the stated domain
Let h(x) = (x − 2)² + 1, restricted to x ≥ 2. Its smallest output is one, and it increases from that point, so its range is [1, ∞). Solve y − 1 = (x − 2)². Because the domain says x − 2 ≥ 0, take x − 2 = √(y − 1), giving h⁻¹(x) = 2 + √(x − 1), with x ≥ 1. The positive root is justified by the domain, not by a general rule that inverses always use plus.
If the original restriction were x ≤ 2 instead, the same original formula would have inverse h⁻¹(x) = 2 − √(x − 1), again with x ≥ 1. Now x − 2 is nonpositive, so the minus branch returns the permitted original input. At an inverse input of ten, the plus branch returns five and the minus branch returns minus one. Both square to nine after subtracting two, but each belongs to a different original domain.
Without a restriction, the full parabola has no inverse function on its range. If the exercise asks you to choose a suitable restriction, either complete side is possible; state your choice and its inverse together. If it supplies a restriction, follow it. Replacing ± with a single sign without explaining why may create a tidy-looking answer that solves a different problem.
Verify both compositions, including the restrictions
For the linear example, f⁻¹(f(x)) = (3x − 6 + 6)/3 = x, and f(f⁻¹(x)) = 3((x + 6)/3) − 6 = x. Each identity holds on the appropriate starting set. With restricted domains, first check that the inner output is allowed as an input to the outer function. Formal cancellation alone cannot make an undefined composition valid.
For the quadratic with x ≥ 2, h⁻¹(h(x)) = 2 + √((x − 2)²) = 2 + |x − 2| = x, precisely because x − 2 ≥ 0. In the other direction, h(h⁻¹(y)) = (√(y − 1))² + 1 = y for y ≥ 1. The absolute value in the first calculation is essential. In general, √(a²) = |a|, not automatically a.
A substitution check catches arithmetic mistakes quickly: h(5) = 10 and h⁻¹(10) = 5. But it supplements the domain argument rather than replacing it. Our independent answer-checking guide shows how to compare a candidate with its original conditions. Try a boundary as well: h(2) = 1 and h⁻¹(1) = 2. Endpoints often reveal an omitted equality sign or the wrong square-root branch.
A rational rule has two exclusions to track
Try r(x) = (2x + 1)/(x − 3), with x ≠ 3. Set y(x − 3) = 2x + 1, collect terms to obtain x(y − 2) = 3y + 1, and solve x = (3y + 1)/(y − 2). Therefore r⁻¹(x) = (3x + 1)/(x − 2), with x ≠ 2. Do not lose the original exclusion three while discovering the inverse exclusion two.
Why can the original never output two? Setting (2x + 1)/(x − 3) = 2 leads to 2x + 1 = 2x − 6, a contradiction. Conversely, every y ≠ 2 gives the displayed preimage, and that preimage cannot be three: 3y + 1 = 3(y − 2) would imply 1 = −6. This proves the range is all real numbers except two and justifies the inverse's domain.
Check r(4) = 9 and r⁻¹(9) = 4, then verify the symbolic compositions on their permitted sets. Cross-multiplication assumes the denominator is nonzero; it does not authorize dividing by an excluded value later. Keep exclusions next to each line of work. This example also shows why swapping the numbers in a formula is not a reliable substitute for solving the input-output equation.
Build and test a fresh inverse on paper
Work independently with p(x) = 5 − 2x on −1 ≤ x ≤ 3. Before reading further, find the range, inverse rule and inverse domain. Then test both endpoints and one interior value. This decreasing example deliberately changes the sign and interval; using the earlier three-times rule from memory will not help. You need the relationship between input and output.
The endpoints give p(−1) = 7 and p(3) = −1. Since the rule is decreasing, its range is [−1, 7], written in increasing interval order. Solve y = 5 − 2x to get x = (5 − y)/2. Thus p⁻¹(x) = (5 − x)/2 on [−1, 7], with range [−1, 3]. At the interior point one, p(1) = 3 and p⁻¹(3) = 1.
Finish by explaining why the inverse is unique, which set supplies its inputs and how its operations undo the original order. If you made an error, identify whether it was notation, arithmetic, uniqueness or the domain. That diagnosis gives your next practice attempt a precise purpose. An inverse is complete when its rule and permitted sets work together, not merely when the letters have been exchanged.
