Applies §1.7's inverse machinery to the families of this chapter. Restricts a quadratic or higher power to a branch on which it is one-to-one, inverts it to produce a radical function, and inverts rational functions by collecting and factoring. Solves radical equations, where squaring both sides can create solutions the original never had.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 3 — Polynomial and Rational Functions
§3.8 Inverses and Radical Functions, pp. 421-437
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §3.8 Inverses and Radical Functions §3.8, pp. 421-437 — the pages these objectives are drawn from
Warm-up
Section 1.7 said an inverse exists exactly when the function is one-to-one. Apply that test to the chapter so far.
Discussion prompt
Can a quadratic be inverted? A cubic? A rational function like one over x?
Hint: Run the horizontal line test on each.
Answer:
A quadratic cannot be inverted as it stands: every horizontal line above its vertex cuts the parabola twice, so two inputs share an output.
A cubic can, provided it is increasing throughout — the basic cubing rule is one-to-one, so its inverse is the cube root with no restriction needed.
The reciprocal can too: no two inputs share an output. So the chapter contains both kinds, and the question for each is the same one from §1.7 — with the difference that here the restriction, when needed, genuinely changes the answer.
Concept
Most functions in this chapter fail the horizontal line test. Restricting the domain to a stretch on which the function is one-to-one makes an inverse exist, and the branch chosen decides which inverse you get.
\[ f(x)=x^2 \text{ on } x\ge 0 \;\Longrightarrow\; f^{-1}(x)=\sqrt{x}; \qquad \text{on } x\le 0 \;\Longrightarrow\; f^{-1}(x)=-\sqrt{x} \]
The two answers are equally correct and they are different functions. Which one a problem wants is decided by the restriction it states, so the restriction is part of the question rather than a preliminary to it — and an answer given without naming its branch is incomplete.
Figure (svg): A parabola with its two halves shaded differently, and the two different inverses each half produces, drawn as the upper and lower halves of a sideways parabola
OpenStax, Precalculus, §3.8 Inverses and Radical Functions §3.8, pp. 421-426
Section
Section 1
Concept
A function is one-to-one on any interval where it is strictly increasing or strictly decreasing. Restricting to such an interval makes an inverse exist.
That last point is why the loss is smaller than it looks. Restricting the squaring rule to the nonnegative inputs halves its domain and keeps every one of its outputs — the discarded half was producing duplicates rather than new values.
Figure (svg): A parabola with its two halves shaded differently, and the two different inverses each half produces, drawn as the upper and lower halves of a sideways parabola
OpenStax, Precalculus, §3.8 Inverses and Radical Functions §3.8, pp. 421-427
Picture it
The same parabola cut at its vertex, and the two inverses the halves produce.
Figure (svg): A parabola with its two halves shaded differently, and the two different inverses each half produces, drawn as the upper and lower halves of a sideways parabola
The two inverses together form the sideways parabola, which is not a function. Choosing a branch is exactly what makes one of the two halves selectable.
Worked example
The vertex is where the direction changes, so it is where to cut.
\[ \text{Restrict } f(x)=x^2-6x+5 \text{ so that an inverse exists.} \]
Find the vertex
Why: Negative b over 2a.
\[ x = 3 \]
Note the direction on each side
Why: Decreasing left of 3, increasing right.
\[ \text{turns at } 3 \]
Choose a branch
Why: Either works; take the right one.
\[ \text{restrict to } x \ge 3 \]
Confirm it is one-to-one
Why: Strictly increasing throughout.
Figure (svg): A parabola with its two halves shaded differently, and the two different inverses each half produces, drawn as the upper and lower halves of a sideways parabola
\[ \text{restrict to } x\ge 3 \]
Verify: check the range is preserved
Why: The vertex's output is 9 minus 18 plus 5, which is negative 4, and on the right branch the outputs run from negative 4 upward — the same range as the full parabola. Nothing was lost but the duplicates. The left branch would have been equally valid and would give a different inverse.
OpenStax, Precalculus, §3.8 Inverses and Radical Functions §3.8, pp. 422-424
Sorting
For the squaring rule, which restrictions make it one-to-one?
Sort into buckets
Sort each restriction.
Worked example
Some of the chapter's functions are already one-to-one.
\[ \text{Does } f(x)=x^3+2 \text{ need a restriction?} \]
Identify the parent
Why: The cubing rule, shifted up.
Check for turning points
Why: The cubing rule never turns.
Apply the horizontal line test
Why: Increasing everywhere.
Conclude
Why: No restriction is needed.
Figure (svg): The solution to Worked example a function needing no restriction shown as a ladder of expressions, one row per legal move
\[ \text{No restriction needed: } f \text{ is increasing everywhere.} \]
Verify: contrast with an even power
Why: The fourth power would need a restriction for exactly the reason the square does, since an even power sends a number and its negative to the same output. Odd powers preserve sign and therefore never repeat a value, which is §3.3's parity distinction deciding an inverse question.
OpenStax, Precalculus, §3.8 Inverses and Radical Functions §3.8, pp. 424-426
Trap
\[ f(x)=x^2 \text{ on } -2\le x\le 2 \;\Longrightarrow\; \text{now invertible} \]
Restrict the domain to a bounded interval
Why: The domain is now finite, which is taken to be enough.
The function is declared one-to-one on that interval.
The interval still straddles the vertex, so it still contains both 1 and negative 1, which share the output 1. Restricting alone is not enough — the restriction has to remove the duplication.
A working restriction keeps inputs on one side of the turning point: from 0 to 2, or from negative 2 to 0.
Check the restricted function is monotone, not merely that the domain got smaller. The condition is one-to-one, and a bounded interval containing a turning point never satisfies it.
Prediction
Consider the power functions from §3.3.
Predict first
Which need a restriction before they can be inverted?
Correct: The even powers only.
Why: An even power sends a number and its negative to the same output, so it fails the horizontal line test and needs a restriction. An odd power preserves sign and is increasing throughout, so it is one-to-one already and its inverse — an odd root — needs no restriction at all.
Faded example
Restrict the function x squared plus 8x plus 3 to a branch.
Fill in the blanks
\text-4 x = -\frac______ = ___, \text___ x \ge ___ \text___ x \le ___
Why: The vertex formula gives negative 4, which is where the parabola turns. Either branch works, and each gives a different inverse differing in the sign of the root. Stating which branch was chosen is part of the answer.
Socratic
Restricting halves the domain and keeps every output.
Discussion prompt
Explain why restricting a parabola to one branch loses no outputs.
Hint: How many inputs produced each output before the restriction?
Answer:
Before restricting, every output above the vertex was produced twice, once from each side. The two inputs were different but the output was identical.
Discarding one side removes one of the two routes to each output, but not the output itself — it is still reached from the surviving side. So the range is unchanged.
That is why the restriction is a cheap operation despite discarding half the domain. It removes redundancy rather than information, and it is exactly what makes the square root able to return every nonnegative value despite being the inverse of only half a parabola.
Section
Section 2
Concept
Inverting a restricted quadratic follows §1.7's swap-and-solve, with one extra decision: taking a square root introduces a sign, and the restriction says which one to keep.
Completing the square is not an optional tidying step here. In expanded form the variable appears in two terms and cannot be isolated at all; vertex form puts it in one place, and that is precisely what makes solving for it possible.
Figure (svg): A parabola with its two halves shaded differently, and the two different inverses each half produces, drawn as the upper and lower halves of a sideways parabola
OpenStax, Precalculus, §3.8 Inverses and Radical Functions §3.8, pp. 427-432
Picture it
The colour of the branch on the left matches the sign of the root on the right.
Figure (svg): A parabola with its two halves shaded differently, and the two different inverses each half produces, drawn as the upper and lower halves of a sideways parabola
Keeping the right-hand branch gives the positive root; keeping the left-hand one gives the negative root. Both are inverses, of different functions.
Worked example
Complete the square, swap, solve, choose the sign.
\[ \text{Invert } f(x)=x^2-6x+5 \text{ on } x\ge 3. \]
Complete the square
Why: Half of negative 6 is negative 3.
\[ (x - 3) ^{2} - 4 \]
Swap the letters
Why: Starting from y equals the rule.
\[ x = (y - 3) ^{2} - 4 \]
Solve for the bracket
Why: Add 4 and take a root.
\[ y - 3 = +- \sqrt{x + 4} \]
Choose the sign from the restriction
Why: Inputs at or above 3 means y at or above 3.
Figure (svg): The solution to Worked example invert a restricted quadratic shown as a ladder of expressions, one row per legal move
\[ f^{-1}(x)=3+\sqrt{x+4}, \qquad x\ge -4 \]
Verify: check a point both ways
Why: The original at 5 gives 25 minus 30 plus 5, which is zero. The inverse at zero gives 3 plus the root of 4, which is 5 — back where we started. The domain of the inverse is x at or above negative 4, which is the original's range, confirming the swap.
OpenStax, Precalculus, §3.8 Inverses and Radical Functions §3.8, pp. 428-430
Prediction
A parabola with vertex at x equal to 2 is restricted to inputs at or below 2.
Predict first
Which root does its inverse use?
Correct: The negative root, so the inverse's outputs stay at or below 2.
Why: The inverse's outputs are the original's inputs, which were restricted to at or below 2. Subtracting a nonnegative root from 2 gives values at or below 2, so the negative root is the one that produces the right outputs. Checking what the inverse must output is the reliable way to pick the sign.
Worked example
The same function, the other restriction, a different inverse.
\[ \text{Invert the same } f \text{ on } x\le 3. \]
Reach the same point
Why: The algebra to the root is identical.
\[ y - 3 = +- \sqrt{x + 4} \]
Read the restriction
Why: Inputs at or below 3.
\[ y \le 3 \]
Choose the sign that gives that
Why: The bracket must be at or below zero.
Write the inverse
Why: Three minus the root.
\[ 3 - \sqrt{x + 4} \]
Figure (svg): A parabola with its two halves shaded differently, and the two different inverses each half produces, drawn as the upper and lower halves of a sideways parabola
\[ f^{-1}(x)=3-\sqrt{x+4}, \qquad x\ge -4 \]
Verify: check a point on this branch
Why: The original at 1 gives 1 minus 6 plus 5, which is zero. The inverse at zero gives 3 minus 2, which is 1 — the correct input for this branch. Note that the previous inverse returned 5 at the same input, which is the other branch's answer. Both are correct inverses of different restricted functions.
OpenStax, Precalculus, §3.8 Inverses and Radical Functions §3.8, pp. 430-432
Error analysis
A student inverts a restricted quadratic.
Annotate
On: \( f^{-1}(x)=3\pm\sqrt{x+4} \)
The plus-or-minus is the sideways parabola, which is not a function. Resolving it is the last step of the inversion, not an optional refinement, and the restriction is what resolves it.
Faded example
Invert the rule that adds 2 to the cube of x, which needs no restriction.
Fill in the blanks
x = y^3 + 2 \;\Longrightarrow\; y^3 = x - 2 \;\Longrightarrow\; y = \sqrt[3]___}}
Why: Subtracting 2 and taking a cube root inverts it, and no sign question arises because an odd root has exactly one real value. This is why odd powers are easier to invert than even ones: there is no branch to choose.
Sorting
Only even roots introduce a plus or minus.
Sort into buckets
Sort each function by whether inverting it requires choosing a sign.
Step zero
You are asked to invert a quadratic given in expanded form.
Discussion prompt
What must be done before the swap-and-solve can work, and why?
Hint: How many places does the variable appear in expanded form?
Answer:
Complete the square. In expanded form the variable appears in two terms, and there is no way to isolate it — every rearrangement leaves it in two places.
Vertex form puts it in exactly one place, inside a single bracket, and that is what makes solving for it possible: undo the outside operations, then take a root.
The vertex also tells you where to restrict, so completing the square does double duty. It is the same technique from §3.2 doing the same job in a different question, which is worth noticing — the form is useful because the variable appears once, and that is useful for more than one reason.
Section
Section 3
Concept
Inverting a rational function needs one step the earlier cases did not: after clearing the denominator, the new variable appears in two terms and must be collected and factored out.
The collecting step is what makes this case look harder than it is. Without it the variable is stuck on both sides and no amount of rearranging isolates it; with it, one factoring and one division finish the job.
Figure (svg): The steps of inverting a rational function, showing the swap, the clearing of the denominator, the collection of the y terms, and the factoring that isolates y
OpenStax, Precalculus, §3.8 Inverses and Radical Functions §3.8, pp. 432-435
Picture it
Steps three and four are the ones the earlier cases did not need.
Figure (svg): The steps of inverting a rational function, showing the swap, the clearing of the denominator, the collection of the y terms, and the factoring that isolates y
Collecting and factoring is the whole difficulty. Everything before it is the standard swap and everything after it is one division.
Worked example
Clear, expand, collect, factor, divide.
\[ \text{Invert } f(x)=\frac{x+1}{x-2}. \]
Swap and clear the denominator
Why: Multiply both sides up.
\[ x(y - 2) = y + 1 \]
Expand
Why: Distributing the x.
\[ x y - 2 x = y + 1 \]
Collect the y terms
Why: All on the left, the rest on the right.
\[ x y - y = 2 x + 1 \]
Factor and divide
Why: Take y out and divide by the bracket.
\[ y = \frac{2 x + 1}{x - 1} \]
Figure (svg): The steps of inverting a rational function, showing the swap, the clearing of the denominator, the collection of the y terms, and the factoring that isolates y
\[ f^{-1}(x)=\frac{2x+1}{x-1} \]
Verify: compose one way
Why: Feeding the inverse into the original: the numerator becomes 2x plus 1 over x minus 1, plus 1, which is 3x over x minus 1; the denominator becomes 2x plus 1 over x minus 1, minus 2, which is 3 over x minus 1. Dividing gives x, as required. Note the asymptotes swapped: the original had a vertical asymptote at 2 and a horizontal one at 1, and the inverse has them at 1 and 2.
OpenStax, Precalculus, §3.8 Inverses and Radical Functions §3.8, pp. 433-434
Ranking
For a rational function.
Put in order
Why: The swap performs the inversion and clearing removes the fraction. Expanding then reveals where the variable is, collecting brings the occurrences together, and factoring reduces them to one so that division finishes it. Attempting to divide before collecting is what leaves the variable on both sides.
Worked example
The reflection swaps horizontal and vertical.
\[ \text{A function has a vertical asymptote at } x=4 \text{ and a horizontal one at } y=3. \text{ Its inverse?} \]
Recall what inverting does
Why: It reflects in the line y equals x.
See what a vertical line becomes
Why: Reflected, it is horizontal.
Apply it to the vertical asymptote
Why: At input 4 becomes at output 4.
\[ \text{horizontal at } y = 4 \]
Apply it to the horizontal one
Why: At output 3 becomes at input 3.
\[ \text{vertical at } x = 3 \]
Figure (svg): The solution to Worked example read the inverse's asymptotes for free shown as a ladder of expressions, one row per legal move
\[ \text{VA } x=3, \quad \text{HA } y=4 \]
Verify: check against the worked example
Why: In the previous example the original had a vertical asymptote at 2 and a horizontal one at 1, and the inverse came out with them at 1 and 2 — swapped, exactly as this predicts. The rule saves computing them from the new formula, and it is a useful check that the inversion was performed correctly.
OpenStax, Precalculus, §3.8 Inverses and Radical Functions §3.8, pp. 434-435
Trap
\[ xy-2x = y+1 \;\Longrightarrow\; y = \frac{xy-2x-1}{1} \]
Rearrange to put y alone on the left
Why: The other terms are moved across in the usual way.
The result still contains y on the right-hand side.
Nothing has been isolated, because y still appears on both sides. Moving terms one at a time cannot help when the variable is in two places.
The move is to collect all the y terms on one side first, then factor y out of them. That converts two occurrences into one, which is what makes division possible.
Whenever a variable appears twice, collect and factor before dividing. It is the same principle that made completing the square necessary for quadratics: get the variable into one place, and only then undo the operations around it.
Faded example
From the equation xy plus 3x equal to 2y minus 1, isolate y.
Fill in the blanks
xy - 2y = -1 - 3x \;\Longrightarrow\; y(x - ___) = -1-3x
Why: Subtracting 2y from both sides and moving 3x across collects the y terms on the left, and factoring y out gives a single occurrence. Dividing by the bracket then isolates y. The step that mattered was recognising that both y terms had to move to the same side before anything could be factored.
Prediction
A rational function has a horizontal asymptote at y equal to 5.
Predict first
What does its inverse have?
Correct: A vertical asymptote at x equal to 5.
Why: Inverting reflects the graph in the line y equals x, which turns horizontal lines into vertical ones and swaps the coordinate. So a horizontal asymptote at height 5 becomes a vertical asymptote at input 5. This is a fast check that an inversion was carried out correctly.
Explain it to yourself
Inverting a linear rule needs no collecting; inverting a rational one does.
Discussion prompt
Explain why the rational case needs a step the linear case does not.
Hint: After clearing the denominator, how many times does the new variable appear?
Answer:
In the linear case, after the swap the new variable appears once, so undoing the operations around it isolates it directly.
In the rational case it appears in both the numerator and the denominator. Clearing the denominator multiplies it out, and the variable ends up in two terms — one from each place it started.
Collecting and factoring turns those two occurrences back into one, at which point the linear method applies. So the extra step is not a new technique; it is the preparation that makes the old technique usable, in the same way completing the square was for quadratics.
Section
Section 4
Concept
The inverse's domain is the original's range and its range is the original's domain. Reading them off is faster and safer than deriving them from the new formula.
The fourth bullet is the practical reason to prefer reading them off. Solving for the new variable often involves squaring or clearing a denominator, both of which enlarge what the formula accepts, so the formula's own domain is typically larger than the inverse's actual one.
Figure (svg): A function and its inverse drawn as reflections in the line y equals x, with the domain of one shaded to match the range of the other
OpenStax, Precalculus, §3.8 Inverses and Radical Functions §3.8, pp. 429-433
Picture it
The two curves are mirror images, and so are their domains and ranges.
Figure (svg): A function and its inverse drawn as reflections in the line y equals x, with the domain of one shaded to match the range of the other
The shaded band along one axis for the original becomes the band along the other axis for the inverse. Reading them off costs nothing and derives nothing.
Worked example
Find the original's domain and range, then swap.
\[ \text{For } f(x)=(x-3)^2-4 \text{ on } x\ge 3, \text{ give the inverse's domain and range.} \]
State the original's domain
Why: The restriction.
\[ x \ge 3 \]
State the original's range
Why: Outputs from the vertex upward.
\[ y \ge - 4 \]
Swap for the inverse's domain
Why: The original's range.
\[ \text{domain } x \ge - 4 \]
Swap for its range
Why: The original's domain.
\[ \text{range } y \ge 3 \]
Figure (svg): A function and its inverse drawn as reflections in the line y equals x, with the domain of one shaded to match the range of the other
\[ \text{domain } [-4,\infty), \quad \text{range } [3,\infty) \]
Verify: check against the formula
Why: The inverse is 3 plus the root of x plus 4. Its radicand needs x at or above negative 4, matching the domain. And since the root is nonnegative, its outputs start at 3, matching the range. Here the formula agrees, but the swap was faster and it does not depend on the formula being right.
OpenStax, Precalculus, §3.8 Inverses and Radical Functions §3.8, pp. 430-431
Faded example
A function has domain from 2 to 9 and range from negative 1 to 5.
Fill in the blanks
\text-1 [9, 5], \qquad \text___ [2, ___]
Why: The inverse's domain is the original's range, from negative 1 to 5, and its range is the original's domain, from 2 to 9. No formula is needed at any point, which is exactly the advantage of the swap.
Worked example
The reason to read off rather than derive.
\[ \text{Invert } f(x)=x^2 \text{ on } x\ge 0 \text{ and compare the two domains.} \]
Find the inverse
Why: The positive root.
\[ \sqrt{x} \]
Read the domain by swapping
Why: The original's range was y at or above 0.
\[ \text{domain } x \ge 0 \]
Note what would happen for the other branch
Why: The inverse there is the negative root.
\[ -\sqrt{x} \]
Compare with a squared formula
Why: Squaring in an intermediate step can enlarge things.
Figure (svg): The solution to Worked example when the formula's domain is too large shown as a ladder of expressions, one row per legal move
\[ \text{domain } [0,\infty) \]
Verify: see the general danger
Why: In problems where solving requires squaring both sides, the resulting formula often accepts inputs the inverse should refuse — squaring destroys a sign restriction that was doing real work. Reading the domain off the original's range sidesteps that entirely, because the original's range was determined before any squaring happened.
OpenStax, Precalculus, §3.8 Inverses and Radical Functions §3.8, pp. 431-433
Error analysis
A student states the domain of an inverse from its formula.
Annotate
On: \( f^{-1}(x)=3+\sqrt{x+4}: \quad \text{domain} = \text{all reals where the radicand is nonnegative} \)
Read the domain off the original's range. The formula's own restrictions are a consequence rather than the definition, and they can be looser after an irreversible step.
Prediction
A quadratic is restricted to inputs at or above its vertex before inverting.
Predict first
Where does that restriction show up in the inverse?
Correct: As the inverse's range.
Why: The original's domain becomes the inverse's range, so the restriction reappears as a limit on the inverse's outputs. That is why the sign of the root had to be chosen: the positive root produces outputs at or above the vertex and the negative root produces outputs at or below it, and the restriction decides which is wanted.
Discrimination
The swap is the whole rule.
Sort into buckets
Sort each statement about a function and its inverse.
Counterexample
A classmate says the inverse's domain can always be found from the inverse's own formula.
Discussion prompt
Describe a case where that gives the wrong answer.
Hint: What happens when solving required squaring both sides?
Answer:
Take the squaring rule restricted to inputs at or below zero. Its inverse is the negative square root, whose formula accepts every nonnegative input — which happens to be right here.
But consider a case where the algebra squares both sides in an intermediate step. Squaring destroys a sign condition, so the resulting formula accepts inputs for which the original equation had no solution, and reading its domain gives too much.
The safe route is always to read the inverse's domain off the original's range, which was determined before any irreversible operation. That is the definition rather than a consequence, so it cannot be enlarged by anything the algebra does afterwards.
Section
Section 5
Concept
Solving an equation containing a radical usually requires squaring both sides, which is not a reversible step. Every answer must be checked in the original equation.
The reason squaring adds solutions is that it destroys sign information: a number and its negative square to the same thing, so an equation that was false because of a sign becomes true after squaring. The check is not a formality — it is the only thing that detects this.
Figure (svg): A warning card showing that squaring both sides of an equation can introduce solutions, with a worked instance where a false solution appears and is caught by checking
OpenStax, Precalculus, §3.8 Inverses and Radical Functions §3.8, pp. 435-437
Picture it
An equation with no solutions acquires one when both sides are squared.
Figure (svg): A warning card showing that squaring both sides of an equation can introduce solutions, with a worked instance where a false solution appears and is caught by checking
The squaring step lost the reason there was no solution, which was a sign. Only substituting back into the original recovers it.
Worked example
Isolate, square, solve, check.
\[ \text{Solve } \sqrt{x+7}=x-5. \]
The radical is already isolated
Why: Nothing added to it.
Square both sides
Why: Removing the radical.
\[ x + 7 = x ^{2} - 10 x + 25 \]
Rearrange and solve
Why: A quadratic against zero.
\[ x ^{2} - 11 x + 18 = 0 \]
Factor and read the roots
Why: Nine and two.
\[ x = 9\text{ or } x = 2 \]
Figure (svg): A warning card showing that squaring both sides of an equation can introduce solutions, with a worked instance where a false solution appears and is caught by checking
\[ x = 9 \]
Verify: check both in the ORIGINAL
Why: At 9: the root of 16 is 4, and 9 minus 5 is 4 — it works. At 2: the root of 9 is 3, and 2 minus 5 is negative 3 — the two sides differ in sign, so 2 fails. Squaring made negative 3 and positive 3 indistinguishable, which is exactly how the extraneous solution got in.
OpenStax, Precalculus, §3.8 Inverses and Radical Functions §3.8, pp. 435-436
Prediction
Solving a radical equation gives two candidates, and the right-hand side of the original was x minus 4.
Predict first
Which kind of candidate is likely to be extraneous?
Correct: One that makes the right-hand side negative.
Why: A principal square root is never negative, so the original equation cannot hold when the other side is negative. Squaring destroyed that constraint, which is how such a candidate got in. Checking the sign of the non-radical side is a fast way to spot extraneous answers before substituting fully.
Worked example
Squaring an unisolated radical does not remove it.
\[ \text{Solve } \sqrt{2x+3}+4 = 9. \]
Isolate the radical first
Why: Subtract 4 from both sides.
\[ \sqrt{2 x + 3} = 5 \]
Square both sides
Why: Now the radical is alone.
\[ 2 x + 3 = 25 \]
Solve
Why: Subtract 3 and halve.
\[ x = 11 \]
Check in the original
Why: The root of 25 is 5, plus 4 is 9.
Figure (svg): The solution to Worked example isolate before squaring shown as a ladder of expressions, one row per legal move
\[ x = 11 \]
Verify: see what squaring too early would have done
Why: Squaring the original directly gives 2x plus 3, plus 8 times the root, plus 16 — the radical survives, multiplied by 8, and the equation is worse than before. Isolating first is not a tidiness preference; squaring only removes a radical that stands alone on its side.
OpenStax, Precalculus, §3.8 Inverses and Radical Functions §3.8, pp. 436-437
Trap
\[ \sqrt{x+7}=x-5 \;\Longrightarrow\; x=9 \text{ or } x=2, \text{ both reported} \]
Square, solve the quadratic, and report both roots
Why: The algebra is correct throughout and two roots are found.
Both values are given as solutions of the original equation.
Only 9 works. At 2 the left side is 3 and the right side is negative 3, so the original equation is false there.
The value 2 solves the squared equation, which is a different equation. Squaring made the two sides' signs indistinguishable, so it admitted a solution the original refused.
Every answer must be checked in the original. This is the one place in the course where a correct chain of algebra can produce a wrong answer, and checking is the only defence.
Faded example
Prepare the equation with the root of x minus 1, plus 3, equal to 7.
Fill in the blanks
\sqrt3 = 7 - 4 = ___
Why: Subtracting 3 isolates the radical, giving 4 on the right. Squaring then gives x minus 1 equal to 16, so x is 17. Checking: the root of 16 is 4, plus 3 is 7 — correct. Squaring before isolating would have left the radical in the equation.
Sorting
Reversible steps cannot; irreversible ones can.
Sort into buckets
Sort each operation.
Explain it to yourself
Most algebraic steps never require checking afterwards.
Discussion prompt
Explain what makes squaring different from adding or dividing.
Hint: Is squaring one-to-one?
Answer:
Adding and dividing by a nonzero constant are reversible: you can undo them exactly, so the new equation has precisely the same solutions as the old one.
Squaring is not one-to-one — it sends a number and its negative to the same output. So two things that were different before squaring become equal after, and an equation that was false because its two sides differed in sign becomes true.
This is §1.7's condition doing real work outside its own section: an operation is safe to apply to both sides exactly when it is one-to-one. Squaring fails that test, so it can add solutions, and checking is the only way to remove them again.
Comparison
Fill the blanks from memory. Three families, three amounts of extra work.
Comparison matrix
| linear | restricted quadratic | rational | |
|---|---|---|---|
| restriction needed | no | yes, at the vertex | usually no |
| preparation | none | complete the square | clear the denominator |
| the hard step | none | choosing the sign of the root | collecting and factoring y |
| the result is | linear | a radical function | another rational function |
In every case the method is §1.7's swap and solve. What differs is the preparation needed to get the variable into one place so that solving is possible at all.
Pattern
Six steps, and the first two are the ones §1.7 did not need.
Step 5 is where the marks are and where the shortcut is. Deriving the domain from the new formula is slower and can be wrong; swapping is instant and is the definition.
OpenStax Algebra and Trigonometry 2e, §5.7 Inverses and Radical Functions §5.7
Check
Which branch was kept?
Check your understanding
The squaring rule restricted to inputs at or below zero has which inverse?
Answer: A
Why: The inverse's outputs must be the original's inputs, which were at or below zero, so the inverse must return non-positive values. The negative square root does exactly that.
Check
Swap the sets.
Check your understanding
A function has domain x at or above 2 and range y at or above 7. What is its inverse's domain?
Answer: A
Why: The inverse's domain is the original's range, so it is the inputs at or above 7. The inverse's range is the original's domain, at or above 2.
Check
Squaring is not reversible.
Check your understanding
Solving a radical equation gives x = 3 and x = 8. Substituting shows x = 3 fails. What is x = 3 called?
Answer: A
Why: It solves the squared equation but not the original, which is what extraneous means. Squaring destroyed sign information and admitted it, and checking in the original is what identifies and removes it.
Real world
Any formula you need to run backwards is an inverse problem, and most need a branch chosen.
Discussion prompt
A projectile's height is a quadratic in time, and you want the time at which a given height is reached. What does the branch choice mean physically?
Hint: How many times does a thrown ball pass a given height?
Answer:
A thrown ball passes any height below its peak twice — once going up and once coming down. That is exactly the failure of the horizontal line test, in physical form.
Restricting to the rising branch answers 'when does it first reach that height'; restricting to the falling branch answers 'when does it pass that height on the way down'. Both are meaningful questions with different answers.
So the branch choice is not a mathematical formality — it is which question you are asking. A problem that asks for a single time has to say which, and the plus-or-minus in the quadratic formula is precisely the two answers offered together.
Commit first
State your confidence along with your answer.
Predict first
Why must every solution of a radical equation be checked in the original?
Correct: Because squaring is not reversible and can add solutions.
Why: Squaring sends a number and its negative to the same place, so an equation false because of a sign becomes true after squaring. The resulting equation genuinely has solutions the original does not, and only substituting back distinguishes them. This is a property of the step rather than a risk of miscalculation.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate why restricting a parabola's domain produces two different correct inverses, and why that is not a contradiction.
Hint: Are the two restricted functions the same function?
Answer:
The two restricted functions are different functions. They have the same formula but different domains, and §1.1 says a function is its rule together with its domain.
Different functions are entitled to different inverses, so there is no contradiction. The positive root inverts the right-hand branch and the negative root inverts the left-hand one, and each is the unique inverse of its own function.
What would be a contradiction is one function having two inverses, and that does not happen. A good explanation makes clear that the restriction is part of the function's identity, which is why an answer that omits it is genuinely incomplete rather than merely untidy.
Exit ticket
One honest answer, so the next lesson 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 is where sign errors cluster, and they are hard to spot because the wrong answer is a perfectly good function. The fourth is the only place in the course where flawless algebra produces a wrong answer, which makes the checking habit worth building deliberately.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Draw a parabola and mark its vertex. Shade the two branches in different colours, and beside the picture write the two different inverses each branch produces, noting which sign of the root belongs to which. Then write the inverse's domain and range for one of them, obtained by swapping the original's rather than from the formula.
If your two inverses differ only in the sign of the root, and each one's range matches the branch it came from, you have the fact that makes the restriction part of the answer.
Recap
Five things, and the first is what §1.7 could only describe in the abstract.
| if you remember one thing | it should be this |
|---|---|
| about restriction | the branch you keep decides which inverse you get |
| about the sign | the inverse's outputs are the original's inputs; that picks the sign |
| about domains | swap, do not derive |
| about squaring | it can create solutions, so checking is compulsory |
Section 3.9 closes the chapter with the modelling counterpart: direct, inverse and joint variation, which are power functions met through the situations that produce them.
OpenStax, Precalculus, §3.8 Inverses and Radical Functions §3.8, pp. 421-437 — everything on these slides traces back here
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