Asks when a function can be run backwards. Defines the inverse by the two-sided composition condition, shows that only one-to-one functions have one, and settles the notation trap where the superscript minus one is not a reciprocal. Finds inverse formulas by swapping and solving, reads the graph as a reflection in the line y equals x, and uses domain restriction to rescue functions that fail the horizontal line test.
Subject: Precalculus · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Precalculus · Chapter 1 — Functions
§1.7 Inverse Functions, pp. 151-167
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §1.7 Inverse Functions §1.7, pp. 151-167 — the pages these objectives are drawn from
Warm-up
Undoing is an everyday idea, and its failures are everyday too.
Discussion prompt
You are told a number was doubled and the answer was 14. Can you recover it? Now you are told a number was squared and the answer was 9. Can you recover that one?
Hint: For the second, how many numbers square to 9?
Answer:
The first is recoverable: the number was 7, and nothing else doubles to 14. Doubling can be undone.
The second is not, or not uniquely: 3 and negative 3 both square to 9, so the information given does not determine the original. Squaring destroys the sign, and no amount of cleverness recovers it afterwards.
That is exactly the one-to-one condition from Section 1.1, arriving where it matters. A rule can be undone precisely when no two inputs share an output — and this lesson is that observation turned into a definition and a method.
Concept
The inverse of a function takes an output back to the input it came from. Formally, composing the two in either order returns the input unchanged.
inverse function — For a one-to-one function f, the function that reverses it: it sends each output of f back to the input that produced it. It is written with a superscript minus one, and satisfies the composition condition in both orders.
\[ f^{-1}\bigl(f(x)\bigr)=x \quad \text{and} \quad f\bigl(f^{-1}(y)\bigr)=y \]
Both conditions are required. A rule can undo another in one direction without undoing it in the other, and such a rule is not an inverse. Checking only one composition is the standard shortcut, and it is a shortcut that lets genuine errors through.
Figure (svg): Two function machines wired in series, the second undoing the first, so that the input travels through both and comes back out unchanged
OpenStax, Precalculus, §1.7 Inverse Functions §1.7, pp. 151-154
Section
Section 1
Concept
A function has an inverse precisely when it is one-to-one. If two inputs shared an output, the inverse would be handed that output and have no way to decide which input to return.
It is worth seeing why the failure is a failure of the same kind the whole chapter is about. Inverting swaps inputs and outputs, so an output reached twice becomes an input with two outputs — and that is precisely what disqualifies a relation from being a function.
Figure (svg): Two function machines wired in series, the second undoing the first, so that the input travels through both and comes back out unchanged
OpenStax, Precalculus, §1.7 Inverse Functions §1.7, pp. 151-156
Picture it
The wire runs forward through f and back through its inverse, and the input reappears.
Figure (svg): Two function machines wired in series, the second undoing the first, so that the input travels through both and comes back out unchanged
Both compositions must return the input. A rule that undoes f in one direction only is not its inverse, and the check for that is one line of algebra each way.
Worked example
Both compositions, every time.
\[ \text{Show that } g(x)=\frac{x+5}{3} \text{ is the inverse of } f(x)=3x-5. \]
Compose one way
Why: Feed f into g.
\[ g(f(x)) = \frac{(3 x - 5) + 5}{3} \]
Simplify
Why: The fives cancel and the 3 divides out.
\[ = 3 x / 3 = x \]
Compose the other way
Why: Feed g into f.
\[ f(g(x)) = 3(\frac{x + 5}{3}) - 5 \]
Simplify
Why: The 3 cancels and the fives cancel.
\[ = x \]
Figure (svg): The solution to Worked example verify a proposed inverse shown as a ladder of expressions, one row per legal move
\[ g(f(x))=x \quad \text{and} \quad f(g(x))=x \]
Verify: check with one number as well
Why: Take x equal to 4. Then f gives 7, and g at 7 gives 12 over 3, which is 4 — back where we started. A numerical check is not a proof but it catches algebra slips instantly, and it costs one substitution each way.
OpenStax, Precalculus, §1.7 Inverse Functions §1.7, pp. 154-155
Sorting
Apply the horizontal line test to each.
Sort into buckets
Sort each function by whether it is invertible on its full domain.
Worked example
The horizontal line test, applied before any algebra is attempted.
\[ \text{Does } f(x)=x^2-4 \text{ have an inverse on its full domain?} \]
Look for two inputs sharing an output
Why: Try a number and its negative.
\[ f(3) = f(-3) = 5 \]
Apply the horizontal line test
Why: A line at height 5 cuts the parabola twice.
Say what would go wrong
Why: The inverse at 5 would have two candidate outputs.
\[ 3\text{ or } -3,\text{ no way to choose} \]
Conclude
Why: No inverse exists on the full domain.
Figure (svg): The squaring rule shown failing the horizontal line test on its full domain, and then restricted to the nonnegative inputs where it passes, with the resulting square root inverse drawn beside it
\[ \text{No: } f(3)=f(-3)=5, \text{ so } f \text{ is not one-to-one.} \]
Verify: note that this is not the end of the story
Why: The function is perfectly respectable and merely cannot be reversed as it stands. Restricting the domain to the inputs at or above zero makes it one-to-one, and the inverse is then the square root of the quantity y plus 4. The failure is a licence to restrict, not a dead end.
OpenStax, Precalculus, §1.7 Inverse Functions §1.7, pp. 155-156
Trap
\[ f(x)=x^2, \; g(x)=\sqrt{x}: \quad f(g(x)) = (\sqrt{x})^2 = x \;\Longrightarrow\; \text{inverses} \]
Compose in one direction and observe the identity
Why: The square of the square root simplifies to x, so the two rules appear to cancel.
They are declared inverses on the strength of that single composition.
The other direction fails. Composing the other way gives the square root of x squared, which is the absolute value of x, not x. At the input negative 3 it returns 3.
So the squaring rule is not invertible on its full domain, exactly as the horizontal line test predicted. The one composition that worked did so only on the nonnegative inputs, where the square root's own domain had already restricted things.
Check both directions. The two-sided condition is not redundancy — it is what rules out precisely this case, and this case is the standard example rather than an exotic one.
Prediction
A function sends both 2 and 5 to the output 9.
Predict first
What happens when you try to invert it?
Correct: The inverse would have to send 9 to two different outputs.
Why: Inverting swaps the roles, so the output 9 becomes an input, and it inherits both 2 and 5 as its partners. One input with two outputs is exactly what disqualifies a relation from being a function, so the proposed inverse is not one. Note that it is not undefined at 9 — the problem is the opposite, that it is over-defined there.
Two truths and a lie
Two of these are true of inverses and one is false.
Eliminate the wrong options
One of these claims is wrong.
Survives elimination: B
Why: B is the false claim. Only one-to-one functions have inverses, and most functions are not one-to-one — the squaring rule and the absolute value are two of the six toolkit functions and neither qualifies without restriction.
Socratic
The definition requires the composition to give the identity in both orders.
Discussion prompt
Give an example where one direction works and the other does not, and explain what that shows.
Hint: The squaring rule and the square root are the standard case.
Answer:
Squaring the square root of x gives x, for every x where the square root is defined. So that direction works.
Taking the square root of x squared gives the absolute value of x, which is not x when x is negative. So that direction fails, and the failure is exactly on the inputs the first composition never had to consider.
What this shows is that the two compositions test different things. One asks whether the outer rule recovers what the inner produced; the other asks the same question with the roles swapped, and the two can disagree whenever the domains do not match up. The two-sided condition is precisely what forces the domains to be compatible.
Section
Section 2
Concept
Written on a function's name, the superscript minus one means the inverse function. Written on a quantity, it means the reciprocal. The two are different, and the position decides which is meant.
\[ f^{-1}(x) = \text{the inverse}, \qquad \bigl(f(x)\bigr)^{-1} = \frac{1}{f(x)} \]
The notation is genuinely unfortunate and it is not going away, so the defence is to test. Take the doubling rule: its inverse is halving, while its reciprocal is one over two x. At the input 4 those give 2 and one eighth. Nothing about them is close.
Figure (svg): A warning card contrasting the inverse function notation with the reciprocal, showing that the superscript minus one on a function name does not mean one over the function
OpenStax, Precalculus, §1.7 Inverse Functions §1.7, pp. 156-158
Picture it
Where the superscript sits is the entire difference.
Figure (svg): A warning card contrasting the inverse function notation with the reciprocal, showing that the superscript minus one on a function name does not mean one over the function
When a question uses the notation ambiguously, evaluate both readings at one input. They will differ, and the context will make clear which was meant.
Worked example
The gap between them is the point.
\[ \text{For } f(x)=2x, \text{ find } f^{-1}(4) \text{ and } \bigl(f(4)\bigr)^{-1}. \]
Find the inverse rule
Why: Doubling is undone by halving.
\[ f\text{ inverse } (x) = \frac{x}{2} \]
Evaluate the inverse at 4
Why: Halve it.
\[ f\text{ inverse } (4) = 2 \]
Evaluate f at 4
Why: Double it.
\[ f(4) = 8 \]
Take the reciprocal of that
Why: One over the output.
\[ \frac{1}{8} \]
Figure (svg): The solution to Worked example compute both readings shown as a ladder of expressions, one row per legal move
\[ f^{-1}(4)=2, \qquad \bigl(f(4)\bigr)^{-1}=\tfrac{1}{8} \]
Verify: check what each answer means
Why: The 2 answers the question 'what input doubles to 4', which is what an inverse is for. The one eighth answers 'what is one over the output at 4', which is arithmetic on a number. The two answer completely different questions, and no reading of the notation makes them agree here.
OpenStax, Precalculus, §1.7 Inverse Functions §1.7, pp. 157-157
Discrimination
The position of the superscript decides.
Sort into buckets
Sort each expression by what it denotes.
Worked example
Deciding which is meant from what the expression is doing.
\[ \text{In the equation } f^{-1}(7)=2, \text{ what has been claimed?} \]
Locate the superscript
Why: It is on the function's name, not on an output.
Translate the claim
Why: The inverse sends 7 to 2.
\[ \text{inverse of } 7\text{ is } 2 \]
Say the same thing forwards
Why: So f must send 2 to 7.
\[ f(2) = 7 \]
State what it does not say
Why: It says nothing about one over anything.
Figure (svg): The solution to Worked example read the notation in context shown as a ladder of expressions, one row per legal move
\[ f^{-1}(7)=2 \;\Longleftrightarrow\; f(2)=7 \]
Verify: check the equivalence both ways
Why: If f sends 2 to 7, then the rule that undoes f must send 7 back to 2, which is the claim. The two statements carry identical information written in opposite directions, and being able to flip between them quickly is most of what inverse problems require.
OpenStax, Precalculus, §1.7 Inverse Functions §1.7, pp. 158-158
Error analysis
A student is asked for the inverse of the tripling rule.
Annotate
On: \( f(x)=3x \;\Longrightarrow\; f^{-1}(x)=\frac{1}{3x} \)
Test any proposed inverse by composing it with the original. The reciprocal will fail that test immediately, which is a faster check than trying to remember which reading the notation intends.
Prediction
For most functions the inverse and the reciprocal are different.
Predict first
Is there a function for which they coincide?
Correct: Yes, the reciprocal rule itself.
Why: The rule that sends x to one over x is its own inverse, since applying it twice returns x, and its inverse also happens to be its reciprocal by construction. It is a genuine coincidence rather than a pattern, and it is worth knowing precisely because it is the one case where the confusion would go undetected.
Translation
Each statement says something specific about a function or its inverse.
Match the pairs
Why: The first two carry identical information in opposite directions, which is the defining relationship. The third is arithmetic on an output and says nothing about inverses. The fourth is a different claim from the second, and mixing those two up is the commonest slip once the notation itself is understood.
Counterexample
A classmate claims the inverse of a function is always one over it.
Discussion prompt
Give a function where this fails badly, and say why the belief is tempting.
Hint: Try the rule that adds 5.
Answer:
Adding 5. Its inverse is subtracting 5, while one over the output is one over the quantity x plus 5. At the input 3, the inverse gives negative 2 and the reciprocal gives one eighth — no resemblance at all.
The belief is tempting because the notation is genuinely the same symbol used for reciprocals of numbers, where the minus one exponent does mean one over. Nothing in the symbol itself signals the change of meaning.
The defence is procedural rather than mnemonic: compose your candidate with the original and see whether you get x back. The reciprocal essentially never survives that test, so it is caught in one line whenever the temptation arises.
Section
Section 3
Concept
To find an inverse's formula, write the function as an equation in x and y, swap the two letters, and solve for y. The swap is what reverses the roles of input and output.
The swap is not a trick. A point on the original graph has coordinates given by an input and its output; on the inverse those roles are exchanged, so the point has the coordinates swapped. Doing that to the equation is doing it to every point at once.
Figure (svg): A card showing that the domain of a function becomes the range of its inverse and the range becomes the domain, with an arrow crossing between the two columns
OpenStax, Precalculus, §1.7 Inverse Functions §1.7, pp. 158-163
Picture it
Reversing the arrows exchanges which set is the input and which the output.
Figure (svg): A card showing that the domain of a function becomes the range of its inverse and the range becomes the domain, with an arrow crossing between the two columns
This is why a restricted range on the original becomes a restricted domain on the inverse, and it is how the inverse's domain is found without any extra work.
Worked example
Swap, then undo the operations in reverse order.
\[ \text{Find the inverse of } f(x)=4x-9. \]
Write it as an equation
Why: Replace the function notation with y.
\[ y = 4 x - 9 \]
Swap the letters
Why: This performs the inversion.
\[ x = 4 y - 9 \]
Undo the subtraction
Why: Add 9 to both sides.
\[ x + 9 = 4 y \]
Undo the multiplication
Why: Divide by 4.
\[ y = \frac{x + 9}{4} \]
Figure (svg): The solution to Worked example a linear rule shown as a ladder of expressions, one row per legal move
\[ f^{-1}(x)=\frac{x+9}{4} \]
Verify: compose both ways
Why: Feeding f into the candidate gives 4x minus 9, plus 9, all over 4, which is x. Feeding the candidate into f gives 4 times the quantity x plus 9 over 4, minus 9, which is also x. Both directions return the input, so this really is the inverse.
OpenStax, Precalculus, §1.7 Inverse Functions §1.7, pp. 159-160
Ranking
For finding an inverse formula.
Put in order
Why: Writing it as an equation comes first so there are two letters to swap. The swap performs the inversion, and solving then puts the new rule in usable form. Stating the domain is last because it comes from the original function's range rather than from anything the algebra produced.
Worked example
The original's range becomes the inverse's domain, and it has to be stated.
\[ \text{Find the inverse of } f(x)=\sqrt{x-2}+1. \]
Swap the letters
Why: Starting from y equals the rule.
\[ x = \sqrt{y - 2} + 1 \]
Isolate the radical
Why: Subtract 1 from both sides.
\[ x - 1 = \sqrt{y - 2} \]
Square both sides
Why: Undo the square root.
\[ (x - 1) ^{2} = y - 2 \]
Solve for y and state the domain
Why: Add 2; the original's range was y at or above 1.
\[ y = (x - 1) ^{2} + 2, x \ge 1 \]
Figure (svg): The solution to Worked example an inverse whose domain is restricted shown as a ladder of expressions, one row per legal move
\[ f^{-1}(x)=(x-1)^2+2, \qquad x \ge 1 \]
Verify: check why the restriction is needed
Why: Without it the formula is a full parabola, which is not one-to-one and cannot be anyone's inverse. The original's outputs were all at or above 1, so those are the only inputs the inverse should ever receive. Squaring both sides is the step that quietly enlarged the domain, and stating the restriction is what repairs it.
OpenStax, Precalculus, §1.7 Inverse Functions §1.7, pp. 161-163
Trap
\[ f(x)=\sqrt{x-2}+1 \;\Longrightarrow\; f^{-1}(x)=(x-1)^2+2 \quad \text{for all } x \]
Swap and solve, then report the resulting formula
Why: The algebra is carried out correctly and the parabola is the honest result of it.
The formula is given with no restriction, so it is read as defined for every real input.
Squaring both sides enlarged the domain. The original function only ever produced outputs at or above 1, so those are the only inputs its inverse can legitimately receive.
Without the restriction the formula is a full parabola, which fails the horizontal line test and therefore cannot be the inverse of anything.
The inverse's domain is the original's range. Read it off the original rather than off the new formula, and state it alongside the answer — an inverse given without its domain is only half an answer.
Faded example
Invert the rule that halves its input and then adds 3.
Fill in the blanks
x = \tfrac23+3 \;\Longrightarrow\; x - 3 = \tfrac______ \;\Longrightarrow\; y = ___(x-___)
Why: After swapping, the operations are undone in reverse order: subtract the 3 first, then multiply by 2 to undo the halving. The inverse is twice the quantity x minus 3. Note the reversal — the original halved and then added, so the inverse subtracts and then doubles.
Prediction
A function has domain the interval from 0 to 5 and range the interval from 2 to 12.
Predict first
What are the domain and range of its inverse?
Correct: Domain 2 to 12; range 0 to 5.
Why: Inverting swaps the two sets: what the original produced is what the inverse accepts, and what the original accepted is what the inverse produces. This is the fastest way to state an inverse's domain, and it requires no work on the new formula at all.
Step zero
You are asked to find the inverse of a quadratic function.
Discussion prompt
Before doing any algebra, what must you check, and what will you probably have to do?
Hint: Does a quadratic pass the horizontal line test?
Answer:
Check whether it is one-to-one, and a quadratic on its full domain is not: its parabola fails the horizontal line test at every height above or below the vertex.
So you will have to restrict the domain first, conventionally to one side of the vertex, and say which side you chose. Both choices are legitimate and they give different inverses.
Only then is the swap-and-solve worth starting. Completing the square is usually the next move, since it puts x in one place and makes solving for y possible — which is the same reason §1.5 needed it.
Section
Section 4
Concept
Swapping a point's coordinates reflects it across the line y equals x. Since inverting swaps every point's coordinates, the two graphs are mirror images in that line.
That last point is the cleanest explanation of why one-to-one is the right condition. Reflecting turns horizontal lines into vertical ones, so a graph passing the horizontal line test reflects into one passing the vertical line test — which is exactly the condition for the reflection to be a function.
Figure (svg): A function and its inverse drawn on the same axes as mirror images across the dashed line y equals x, with a pair of corresponding points marked showing that their coordinates are swapped
OpenStax, Precalculus, §1.7 Inverse Functions §1.7, pp. 163-167
Picture it
The dashed diagonal is the mirror, and the two marked points are a reflected pair.
Figure (svg): A function and its inverse drawn on the same axes as mirror images across the dashed line y equals x, with a pair of corresponding points marked showing that their coordinates are swapped
Fold the page along the dashed line and the two curves land on each other. The square root and the restricted squaring rule are the standard example of this pairing.
Worked example
No formula needed — reflect a few points and join them.
\[ \text{Given the graph of } f, \text{ sketch } f^{-1}. \]
Draw the line y equals x
Why: This is the mirror.
Pick several points on the original
Why: Corners and intercepts are the useful ones.
Swap each point's coordinates
Why: A point at (4, 2) becomes (2, 4).
Join the reflected points
Why: Keeping the same overall shape, mirrored.
Figure (svg): A function and its inverse drawn on the same axes as mirror images across the dashed line y equals x, with a pair of corresponding points marked showing that their coordinates are swapped
\[ \text{reflect every point } (a,b) \longmapsto (b,a) \]
Verify: check a point that lies on the mirror
Why: Any point where the original crosses the line y equals x has equal coordinates, so swapping leaves it alone — it is a fixed point of the reflection and lies on both graphs. Finding one is a quick confirmation that the mirror line was drawn correctly.
OpenStax, Precalculus, §1.7 Inverse Functions §1.7, pp. 164-165
Matching
Reflecting in the diagonal swaps the coordinates.
Match the pairs
Why: Every pair is simply swapped, signs and all. The third is on the mirror line itself, so swapping leaves it where it is — such points lie on both the function and its inverse. The last one shows that negative coordinates are swapped like any others, with no sign change.
Worked example
One graph answers questions about both functions.
\[ \text{If } f(2)=7, \text{ what is } f^{-1}(7), \text{ and where is that point?} \]
Read the given statement
Why: The input 2 produces the output 7.
\[ f: 2 \to 7 \]
Reverse it
Why: The inverse sends that output back.
State the value
Why: The inverse at 7 is 2.
\[ f\text{ inverse } (7) = 2 \]
Locate the point
Why: Coordinates swapped from the original's.
\[ \text{point } (7, 2) \]
Figure (svg): The solution to Worked example read values off the two graphs shown as a ladder of expressions, one row per legal move
\[ f^{-1}(7)=2, \text{ at the point } (7,\,2) \]
Verify: check against the reflection
Why: The original passes through the point with coordinates 2 and 7. Reflecting in the diagonal swaps them to 7 and 2, which is where the inverse's point sits. Every question about an inverse's values can be answered this way, without ever finding its formula.
OpenStax, Precalculus, §1.7 Inverse Functions §1.7, pp. 165-166
Error analysis
A student sketches an inverse by reflecting across an axis.
Annotate
On: \( f(2)=7 \;\Longrightarrow\; f^{-1} \text{ passes through } (-2,7) \)
The mirror for an inverse is always the diagonal. Drawing that line before starting the sketch is the single most effective way to avoid this.
Prediction
The original graph passes the horizontal line test.
Predict first
What does its reflection in the diagonal pass?
Correct: The vertical line test, so it is a function.
Why: Reflecting in the diagonal turns horizontal lines into vertical ones, so a graph meeting every horizontal line at most once reflects into one meeting every vertical line at most once. That is precisely the condition for the reflection to be a function, which is the cleanest explanation of why one-to-one is the requirement for an inverse to exist.
Faded example
A function passes through the points with coordinates 1 and 4, and 3 and 10.
Fill in the blanks
f^1(4) = 3, \qquad f^___(10) = ___
Why: Each output of the original becomes an input of the inverse, returning the input it came from. So 4 goes back to 1 and 10 goes back to 3. No formula is needed at any stage — the two points alone determine these two values of the inverse.
Explain it to yourself
The mirror line for an inverse is always y equals x.
Discussion prompt
Explain why swapping a point's coordinates is the same as reflecting it in that particular line.
Hint: What is special about the points on that line?
Answer:
The line y equals x consists of exactly the points whose two coordinates are equal, so swapping does nothing to them. A reflection leaves its mirror fixed, and this operation leaves precisely that line fixed — which already identifies it as the mirror.
For any other point, swapping moves it to the other side of that line, and it moves it to the position the same distance away measured perpendicular to the line. That is what a reflection is.
So the diagonal is not a convention chosen for convenience. It is the only line whose reflection has the effect of swapping coordinates, which is what inverting a function does to every point of its graph.
Section
Section 5
Concept
A function that fails the horizontal line test can often be made one-to-one by discarding part of its domain. The inverse then exists, but only for the piece that was kept.
This is where the square root comes from. Squaring is not invertible, but squaring restricted to the nonnegative inputs is, and its inverse is what everyone calls the square root. The restriction is a convention that was agreed long ago and is now invisible, which is why the negative root has to be written explicitly when it is wanted.
Figure (svg): The squaring rule shown failing the horizontal line test on its full domain, and then restricted to the nonnegative inputs where it passes, with the resulting square root inverse drawn beside it
OpenStax, Precalculus, §1.7 Inverse Functions §1.7, pp. 155-163
Picture it
The same parabola, before and after the restriction.
Figure (svg): The squaring rule shown failing the horizontal line test on its full domain, and then restricted to the nonnegative inputs where it passes, with the resulting square root inverse drawn beside it
Nothing about the outputs changed: every nonnegative value is still produced. What was discarded is the duplicate route to each of them, which is exactly what was blocking the inverse.
Worked example
State the restriction, then do the usual swap and solve.
\[ \text{Find an inverse for } f(x)=x^2+3 \text{ on a suitable domain.} \]
Note the failure and choose a side
Why: The parabola turns at x equal to 0.
\[ \text{restrict to } x \ge 0 \]
Swap the letters
Why: Starting from y equals the rule.
\[ x = y ^{2} + 3 \]
Solve for y
Why: Subtract 3, then take a root.
\[ y = \sqrt{x - 3} \]
Keep only the sign matching the restriction
Why: The restriction kept nonnegative inputs.
Figure (svg): The solution to Worked example restrict and invert shown as a ladder of expressions, one row per legal move
\[ f^{-1}(x)=\sqrt{x-3}, \qquad x \ge 3 \]
Verify: check the domain and one value
Why: The restricted function's outputs run from 3 upward, so the inverse's domain is x at or above 3, which matches the radicand's requirement exactly. Testing: f at 2 is 7, and the inverse at 7 is the square root of 4, which is 2. Had the restriction been to the nonpositive inputs instead, the inverse would have been the NEGATIVE square root — a different and equally valid answer.
OpenStax, Precalculus, §1.7 Inverse Functions §1.7, pp. 161-163
Sorting
A restriction works when the function is one-to-one on what remains.
Sort into buckets
Sort each restriction of the squaring rule.
Worked example
A convention so familiar that it is easy to forget it is one.
\[ \text{Explain why } \sqrt{9}=3 \text{ and not } -3. \]
Note that both square to 9
Why: Nothing arithmetic distinguishes them.
\[ 3 ^{2} = (-3) ^{2} = 9 \]
Recall that a function returns one output
Why: The root symbol names a function.
Identify the restriction
Why: Squaring was restricted to nonnegative inputs.
\[ \text{restricted to } x \ge 0 \]
Read off the consequence
Why: The inverse returns only values from that set.
\[ \text{returns } 3 \]
Figure (svg): The solution to Worked example where the square root comes from shown as a ladder of expressions, one row per legal move
\[ \sqrt{\;\;} \text{ inverts } x^2 \text{ on } x \ge 0, \text{ so it returns the nonnegative root} \]
Verify: notice where the other root went
Why: It is still there, and it is why solving an equation like x squared equals 9 requires writing plus or minus explicitly. The equation has two solutions; the root symbol names only one of them. Conflating the two is a persistent source of lost solutions, and this is where the distinction originates.
OpenStax, Precalculus, §1.7 Inverse Functions §1.7, pp. 156-157
Trap
\[ f(x)=x^2 \;\Longrightarrow\; f^{-1}(x)=\sqrt{x} \]
Give the square root as the inverse of squaring
Why: It is the standard pairing and the composition works in one direction.
No restriction is mentioned, so the claim reads as being about the squaring rule on its full domain.
As stated it is false. The squaring rule on its full domain has no inverse at all, because it fails the horizontal line test.
The correct statement names the restriction: the square root inverts squaring on the nonnegative inputs. Check the other composition to see why it matters — the square root of x squared is the absolute value of x, which is not x when x is negative.
State the restriction as part of the answer. An inverse of a non-one-to-one function is only defined relative to a choice, and leaving the choice unstated makes the answer ambiguous at best.
Prediction
The squaring rule is restricted to the inputs at or below zero instead of at or above.
Predict first
What is its inverse?
Correct: The negative square root.
Why: The restricted function's inputs were all at or below zero, so its inverse must return values at or below zero, which means the negative root. The restriction is one-to-one and perfectly valid, and it gives a genuinely different inverse from the conventional choice — which is why the choice has to be stated.
Elimination
Several conditions are proposed for a restriction to produce an inverse. Rule out the ones that are genuinely required.
Eliminate the wrong options
One of these is NOT required.
Survives elimination: B
Why: B is not required, and in fact a restriction containing the vertex of a parabola in its interior is precisely what fails — it keeps inputs on both sides and preserves the duplication. The useful restrictions stop at the vertex or avoid it entirely.
Real world
The trigonometric functions in Chapter 6 need exactly this treatment, and much more aggressively.
Discussion prompt
The sine function repeats forever, so it is very far from one-to-one. What must be done before it can have an inverse, and what does that cost?
Hint: How much of its domain can be kept?
Answer:
It must be restricted to a single stretch on which it is one-to-one — conventionally the interval from negative one half turn to one half turn measured in quarter-turns, where it rises steadily from negative 1 to 1.
The cost is severe: almost the entire domain is discarded. But the range is fully preserved, exactly as with the squaring rule, so the inverse can still return every value the original ever produced.
The consequence is one students meet constantly: the inverse sine of a value returns one angle, while the equation it came from usually has infinitely many solutions. That gap between the function's single output and the equation's many solutions is the same gap as between the square root and the plus-or-minus, and it originates here.
Comparison
Fill the blanks from memory. Every row is the same swap, seen differently.
Comparison matrix
| the function f | its inverse | |
|---|---|---|
| sends | inputs to outputs | outputs back to inputs |
| domain | the legal inputs | the range of f |
| range | the attainable outputs | the domain of f |
| a point on its graph | (a, b) | (b, a) |
| graph | the original curve | its reflection in the line y = x |
| test it must pass | the horizontal line test | the vertical line test |
The last row is the cleanest statement of why one-to-one is required: the reflection turns one test into the other, so the original passing the horizontal test is exactly what makes the reflection a function.
Pattern
Six steps, and the first and last are the ones that get skipped.
Step 5 is where an otherwise correct answer loses its marks, particularly after squaring both sides, which always enlarges the domain of what it produces.
OpenStax Algebra and Trigonometry 2e, §3.7 Inverse Functions §3.7
Check
The notation names an inverse, not a reciprocal.
Check your understanding
If f(5) = 12, what is f inverse of 12?
Answer: A
Why: The inverse sends each output back to the input it came from, so the output 12 goes back to 5. Equivalently, the point at 5 and 12 on the original reflects to the point at 12 and 5 on the inverse.
Check
Swap and solve.
Check your understanding
What is the inverse of the rule f(x) = 5x + 2?
Answer: A
Why: Swapping gives x equal to 5y plus 2. Subtracting 2 and dividing by 5 gives y as x minus 2, all over 5. Checking: at the input 3 the original gives 17, and the inverse at 17 gives 15 over 5, which is 3.
Check
The mirror is the diagonal.
Check your understanding
The graph of a function passes through the point with coordinates 2 and 9. Which point is on its inverse?
Answer: A
Why: Inverting swaps a point's coordinates, which is the effect of reflecting it in the line y equals x. So the point at 2 and 9 becomes the point at 9 and 2.
Real world
Encoding and decoding are inverse functions, and the one-to-one requirement is a real design constraint.
Discussion prompt
A system compresses files and later restores them. What does the one-to-one condition mean here, and what happens when it fails?
Hint: What would it mean for two different files to compress to the same thing?
Answer:
One-to-one means no two different files compress to the same output. If two did, the decompressor would be handed that output and have no way to know which original to return — exactly the situation that stops an inverse existing.
When it fails, the compression is lossy: the original cannot be recovered, only something close to it. Image and audio formats do this deliberately, discarding information to save space, and they are explicitly not invertible.
Lossless formats are designed to be one-to-one precisely so the inverse exists. So the distinction between lossy and lossless compression is, in this vocabulary, exactly the distinction between a function that fails the horizontal line test and one that passes it.
Commit first
State your confidence along with your answer.
Predict first
A function is invertible. What must be true of its graph?
Correct: No horizontal line crosses it more than once.
Why: That is the horizontal line test, which is exactly the one-to-one condition. The vertical line test is required of every function, invertible or not, so it does not distinguish. Symmetry about the diagonal would make a function its own inverse, which is a much stronger and rarer property, and passing through the origin is irrelevant.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate why the squaring rule has no inverse but the square root exists anyway, without contradicting yourself.
Hint: Which squaring rule does the square root actually invert?
Answer:
The squaring rule on its full domain has no inverse, because 3 and negative 3 both go to 9 and nothing could decide which to return.
The square root inverts a different function: squaring restricted to the nonnegative inputs. That restricted rule is one-to-one, so it has an inverse, and the root symbol names it.
There is no contradiction because the two are not the same function — they have different domains. A good explanation makes the restriction visible, because it is normally invisible: everyone writes the square root without ever mentioning the convention that makes it well defined, and that silence is what makes the question confusing in the first place.
Exit ticket
One honest answer, so the next chapter can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The second costs more marks than its difficulty warrants and is worth fixing immediately. The fourth returns in Chapter 4 for logarithms and again in Chapter 6 for the inverse trigonometric functions, so time spent on it now is repaid twice.
Connect it up
One page, drawn from memory, closes the chapter.
Draw it
Draw a function and its inverse as mirror images in the line y equals x, marking one pair of reflected points with their swapped coordinates. Beside the picture, write the two-sided composition condition. Then write the chain: fails the horizontal line test, so restrict the domain, so the inverse exists — and give the squaring rule as your worked instance of it, naming the restriction explicitly.
If your page connects the horizontal line test to the existence of the inverse and the reflection to the swapping of coordinates, you have the whole of Chapter 1 in one diagram, since those two facts need every earlier section to state.
Recap
Five things, closing a chapter whose sections all meet in this one.
| if you remember one thing | it should be this |
|---|---|
| about existence | one-to-one is exactly the condition, no more and no less |
| about notation | the superscript minus one on a name is not a reciprocal |
| about the method | swap the letters, then solve, then state the domain |
| about the graph | reflect in the line y equals x, not in either axis |
Chapter 2 leaves general functions behind and studies one family in depth — the linear ones — where every question in this chapter has a clean and complete answer.
OpenStax, Precalculus, §1.7 Inverse Functions §1.7, pp. 151-167 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Precalculus — $55/session, free consultation.