One-to-one functions and the horizontal line test, the definition of an inverse through its two cancellation equations, finding an inverse algebraically by swapping and solving, the reflection of graphs in the line y equals x, restricting a domain to force an inverse to exist, and the inverse trigonometric functions with their conventional ranges.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 1 — Functions and Graphs
Inverse Functions
Objectives
Five outcomes. The fourth explains something Section 1.3 left hanging: why the inverse sine key returns exactly one answer.
OpenStax Calculus Volume 1, §1.4 Inverse Functions §1.4, pp. 65-79 — the section these objectives are drawn from
Warm-up
Section 1.3 ended with a puzzle. The equation sine of t equals one half has infinitely many solutions, yet the inverse sine key returns exactly one.
Discussion prompt
Why can the inverse sine not simply return all of them? What would go wrong?
Hint: Ask what the definition of a function demands of anything you want to call the inverse sine.
Answer:
If the inverse sine returned every solution, it would assign many outputs to the single input one half — and that is precisely what the definition of a function forbids.
\[ \sin^{-1}\!\left(\tfrac{1}{2}\right) = \tfrac{\pi}{6} \; \text{only, not also } \tfrac{5\pi}{6} \]
So before sine can have an inverse function at all, its domain must be cut down until it hits each output only once. This section is about when that is possible, how it is done, and what it costs.
Concept
A function can be undone precisely when no two inputs share an output. Such a function is called one-to-one, and only then does reversing every input-output pair produce something that is itself a function.
one-to-one function — A function for which no two different inputs produce the same output. Equivalently, whenever the outputs at two inputs agree, the inputs must have been equal.
\[ f(x_1) = f(x_2) \;\Longrightarrow\; x_1 = x_2 \]
Section 1.1's definition already allowed two inputs to share an output — that was the deliberate asymmetry. Being one-to-one is the extra condition that closes it, and it is exactly what an inverse needs.
Figure (svg): A one-to-one function passing the horizontal line test beside a parabola failing it
OpenStax Calculus Volume 1, §1.4 Inverse Functions §1.4, pp. 65-67
Section
Section 1
Concept
The vertical line test asks whether a curve is a function at all: does any input have two outputs? The horizontal line test asks whether that function is one-to-one: does any output come from two inputs? A curve can pass the first and fail the second.
horizontal line test — A function is one-to-one exactly when every horizontal line meets its graph at most once. A horizontal line meeting the graph twice exhibits two inputs sharing an output.
\[ \text{one-to-one} \iff \text{every horizontal line meets the graph at most once} \]
A function that is strictly increasing throughout, or strictly decreasing throughout, is automatically one-to-one — because a later input always gives a strictly larger output, so no two can agree. That observation will become a theorem once Chapter 4 can detect monotonicity from the derivative.
Figure (svg): A one-to-one function passing the horizontal line test beside a parabola failing it
OpenStax Calculus Volume 1, §1.4 Inverse Functions §1.4, pp. 66-68 — one-to-one functions and the horizontal line test
Picture it
A cubic and a parabola, each met by a horizontal line.
Figure (svg): A one-to-one function passing the horizontal line test beside a parabola failing it
The parabola is a perfectly good function — it passes the vertical line test everywhere. It simply cannot be undone, because being told the output is 4 leaves you unable to say whether the input was 2 or negative 2.
Worked example
Example 1.29. Assume the outputs agree and see what follows.
\[ \text{Is } f(x) = x^3 + 4 \text{ one-to-one? Is } g(x) = x^2 \text{?} \]
Assume the outputs agree at two inputs
Why: This is the definition's hypothesis.
\[ a ^{3} + 4 = b ^{3} + 4 \]
Simplify and see whether the inputs must agree
Why: Subtracting 4 and taking cube roots, which is unambiguous for real numbers.
\[ a ^{3} = b ^{3},\text{ so } a = b \]
Conclude for the cubic
Why: The inputs were forced to be equal, so it is one-to-one.
Run the same argument on the square
Why: Squares agreeing does NOT force the inputs to agree.
\[ a ^{2} = b ^{2}\text{ gives } a = +- b \]
Produce a concrete counterexample
Why: One pair of distinct inputs sharing an output settles it.
\[ g(2) = g(-2) = 4 \]
Figure (svg): The solution to Worked example testing one-to-one algebraically shown as a ladder of expressions, one row per legal move
\[ f \text{ is one-to-one}; \qquad g \text{ is not, since } g(2) = g(-2) \]
Verify: check both against the horizontal line test
Why: The cubic is increasing everywhere, so a horizontal line at any height crosses it exactly once — consistent with the algebra. The parabola is met twice by every horizontal line above the vertex, and the pair 2 and negative 2 is one such crossing. The odd power being reversible and the even power not is the same distinction that made odd roots defined everywhere and even roots not.
OpenStax Calculus Volume 1, §1.4 Inverse Functions §1.4, pp. 67-68
Sorting
Apply the horizontal line test to each.
Sort into buckets
Sort each function on its natural domain.
Sine is the worst case on the list: every horizontal line between negative 1 and 1 meets it infinitely often. That is why its restriction has to be so severe, and why the convention for which branch to keep matters so much.
Worked example
Checkpoint 1.29. The same rule can be one-to-one or not, depending on where it lives.
\[ \text{Is } f(x) = x^2 \text{ one-to-one on } [0, \infty)? \text{ On } [-1, 3]? \]
Test the first domain
Why: On the non-negative inputs, squares agreeing forces the inputs to agree.
\[ a, b \ge 0\text{ and } a ^{2} = b ^{2}\text{ gives } a = b \]
Conclude for the first
Why: No two distinct inputs share an output.
\[ \text{one-to-one on } [0, \infty] \]
Test the second domain
Why: Look for two inputs in the interval with the same square.
\[ -1\text{ and } 1\text{ are both in } [-1, 3] \]
Produce the counterexample
Why: Their squares agree.
\[ f(-1) = f(1) = 1 \]
Figure (svg): The solution to Worked example one-to-one on a restricted domain shown as a ladder of expressions, one row per legal move
\[ \text{one-to-one on } [0,\infty); \quad \text{not on } [-1, 3] \]
Verify: draw the horizontal line on each piece
Why: On the non-negative half the parabola rises steadily, so every horizontal line meets it once. On the interval from negative 1 to 3 the vertex is interior, so lines at heights between 0 and 1 meet it twice. Being one-to-one is a property of the function TOGETHER with its domain, not of the formula alone — which is exactly what makes restriction a usable technique.
OpenStax Calculus Volume 1, §1.4 Inverse Functions §1.4, pp. 68-68
Trap
\[ f(x) = x^2 \]
Apply the vertical line test and conclude an inverse exists
Why: The student checks that it is a function and stops there.
\[ \text{passes the vertical line test} \;\Longrightarrow\; f^{-1} \text{ exists} \quad \text{(wrong)} \]
Being a function is necessary but nowhere near sufficient. The parabola is a function and has no inverse.
\[ \text{vertical line test} \Rightarrow \text{it is a function} \]
\[ \text{horizontal line test} \Rightarrow \text{it can be inverted} \]
Ask the two questions separately
Why: One asks whether any INPUT has two outputs; the other whether any OUTPUT has two inputs.
They are genuinely independent questions about the same picture, and the words 'vertical' and 'horizontal' are the only thing distinguishing them — which is why saying the question out loud is worth more than remembering which line goes with which name.
Fill the middle
Testing whether the cubic from the worked example is one-to-one.
Fill in the blanks
a^3 + 4 = b^3 + 4 \;\Longrightarrow\; a^3 = b^3 \;\Longrightarrow\; a = b
Why: Every real number has exactly one real cube root, so equal cubes force equal inputs. Contrast the square, where equal squares only force the inputs to be equal up to a sign — and that ambiguity is exactly what destroys invertibility.
Two truths and a lie
All three are about one-to-one functions.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The vertical line test only certifies that the curve is a function; the horizontal line test is what certifies invertibility. The parabola passes the first and fails the second, and it is the standard counterexample precisely because the two tests are so easily confused.
Prediction
Commit before reasoning.
Predict first
Which of these guarantees a function is one-to-one on its whole domain?
Correct: Being strictly increasing throughout, or strictly decreasing throughout.
\[ x_1 < x_2 \;\Longrightarrow\; f(x_1) < f(x_2) \;\Longrightarrow\; f(x_1) \ne f(x_2) \]
Why: Strict monotonicity means the outputs are always moving one way, so no output can be revisited. Polynomials, continuity and a full domain guarantee nothing: x squared is a continuous polynomial defined on all reals and is not one-to-one. This criterion becomes genuinely powerful in Chapter 4, where the sign of the derivative detects monotonicity — a function whose derivative is always positive is one-to-one, and so is invertible, without ever drawing the graph.
Section
Section 2
Concept
For a one-to-one function, the inverse is the function that reverses every input-output pair. Its defining property is stated by two equations: composing in either order returns the input untouched.
inverse function — For a one-to-one function f with domain D and range R, the inverse is the function with domain R and range D that sends each output of f back to the input it came from. It is written f with a superscript negative one, which does not mean a reciprocal.
\[ f^{-1}(f(x)) = x \;\text{ on } D, \qquad f(f^{-1}(y)) = y \;\text{ on } R \]
The notation is genuinely unfortunate. The superscript negative one on a function means the inverse, while on a number it means the reciprocal. The inverse of the cubing function is the cube root, not one over the cube.
Figure (svg): The two cancellation equations drawn as a round trip through both machines
OpenStax Calculus Volume 1, §1.4 Inverse Functions §1.4, pp. 68-71 — the definition of an inverse function
Picture it
Two machines in series, in either order.
Figure (svg): The two cancellation equations drawn as a round trip through both machines
Both equations are required. A function can undo another on part of its range while failing elsewhere, and only checking both directions catches that. It is the reason the domain and range swap: the inverse's domain is the original's range.
Worked example
Example 1.30. Both compositions, every time.
\[ \text{Verify that } g(x) = \frac{x+4}{3} \text{ is the inverse of } f(x) = 3x - 4. \]
Compose one way
Why: Substitute g into f.
\[ f(g(x)) = 3(\frac{x + 4}{3}) - 4 \]
Simplify
Why: The 3s cancel and the constants do too.
\[ = x + 4 - 4 = x \]
Compose the other way
Why: Substitute f into g.
\[ g(f(x)) = \frac{(3 x - 4) + 4}{3} \]
Simplify
Why: The constants cancel and the 3s do too.
\[ = 3 x / 3 = x \]
Conclude only after both
Why: Each equation holds on the appropriate set.
Figure (svg): The solution to Worked example verifying a proposed inverse shown as a ladder of expressions, one row per legal move
\[ f(g(x)) = x \;\text{ and }\; g(f(x)) = x \]
Verify: trace one number through both machines
Why: Take the input 5: f of 5 is 11, and g of 11 is 15 over 3, which is 5 again. Going the other way, g of 5 is 3, and f of 3 is 9 minus 4, which is 5. The round trip returns the starting number in both orders, which is what the two equations assert. A single numerical trace is a fast confidence check, though only the algebra proves it for every input.
OpenStax Calculus Volume 1, §1.4 Inverse Functions §1.4, pp. 70-70
Matching
Ask what undoes the operation, in reverse order.
Match the pairs
Why: The last is its own inverse, which is perfectly allowed — its graph is already symmetric in the line y equals x, so reflecting changes nothing. The third needs its domain restriction stated, since the bare rule would accept inputs the inverse must refuse.
Worked example
Checkpoint 1.30. The inverse inherits its domain from the original's range.
\[ \text{For } f(x) = \sqrt{x-2} \text{ with domain } [2,\infty), \text{ give the domain and range of } f^{-1}. \]
Find the range of the original
Why: A square root outputs zero or more, and every such value is attained.
\[ \text{range of } f = [0, \infty] \]
The inverse's domain is that range
Why: The inverse accepts exactly what the original produced.
\[ \text{domain of inverse } = [0, \infty] \]
The inverse's range is the original's domain
Why: The inverse produces exactly what the original accepted.
\[ \text{range of inverse } = [2, \infty] \]
Find the rule by swapping and solving
Why: Square both sides and add 2.
Figure (svg): The solution to Worked example domain and range swap shown as a ladder of expressions, one row per legal move
\[ f^{-1}(x) = x^2 + 2, \quad D = [0,\infty), \; R = [2,\infty) \]
Verify: notice that the stated domain is doing real work
Why: The rule x squared plus 2 is defined for every real number, but the inverse's domain is only the non-negative half. Without that restriction the composition would fail: at x equal to negative 3 the rule gives 11, and f of 11 is 3, not negative 3. The domain is part of the answer, not an afterthought — which is exactly why the restricted parabola and the square root are inverses only when the restriction is carried along.
OpenStax Calculus Volume 1, §1.4 Inverse Functions §1.4, pp. 71-71
Error analysis
A student is asked for the inverse of the cubing function.
Annotate
On: \( f(x) = x^3 \;\Longrightarrow\; f^{-1}(x) = \frac{1}{x^3} \)
The reliable defence is never to trust the notation but to run a cancellation check. If composing the two functions does not return the input, whatever you wrote is not the inverse, whatever it is called.
Fill the middle
Verifying the linear pair from the worked example.
Fill in the blanks
f(g(x)) = 3\!\left(\fracx___\right) - 4 = x + 4 - 4 = ___
Why: The composition collapses to x, which is one of the two cancellation equations. The other direction must be checked as well before declaring them inverses, because a function can undo another in one direction only.
Ranking
Confirming that a proposed g really is the inverse of f.
Put in order
Why: Step a is skipped most often and is the one that saves wasted effort: if f is not one-to-one, no g can satisfy both equations and there is nothing to verify. Step e is what turns a formula into a correct answer, since the restricted parabola and the square root are inverses only with their domains attached.
Prediction
Commit before reasoning.
Predict first
If g(f(x)) equals x for every x in the domain of f, must g be the inverse of f?
Correct: Not necessarily. Both equations are required.
\[ g(f(x)) = x \text{ on } D_f \quad \text{and} \quad f(g(y)) = y \text{ on } R_f \]
Why: A function can undo another on one side while failing on the other, typically because its domain is larger than the original's range. Take f as the square root on the non-negative numbers and g as squaring on ALL reals: g of f of x is x for every non-negative x, but f of g of negative 3 is 3, not negative 3. The second equation fails, and g is not the inverse until its domain is cut down. This is precisely why the previous worked example insisted on stating the inverse's domain.
Section
Section 3
Concept
To find an inverse algebraically, write the rule as an equation in x and y, swap the two letters, and solve the result for y. The swap is the mathematical content; the solving is ordinary algebra.
reflection in the line y equals x — The graph of an inverse is obtained from the graph of the original by reflecting it in the diagonal line y equals x, because reversing an input-output pair swaps the two coordinates of every point.
\[ (a, b) \text{ on } f \iff (b, a) \text{ on } f^{-1} \]
Swapping first and solving second is safer than solving first and renaming afterwards, because after the swap every subsequent step is about the new function and there is no bookkeeping left to forget.
Figure (svg): The three-step algebraic recipe for finding an inverse, worked on one example
OpenStax Calculus Volume 1, §1.4 Inverse Functions §1.4, pp. 71-74 — finding inverse functions and their graphs
Picture it
The squaring function on the non-negative inputs, with the square root.
Figure (svg): A function and its inverse reflected in the line y equals x, with a matched pair of points
The point (2, 4) on the parabola becomes (4, 2) on the root, and the segment joining them crosses the diagonal at a right angle. Any point where a function meets the diagonal is fixed by the reflection, so the two graphs meet there.
Worked example
Example 1.31. Three steps, in order.
\[ \text{Find the inverse of } f(x) = 3x - 4. \]
Write the rule as an equation
Why: Replace the function notation with y.
\[ y = 3 x - 4 \]
Swap x and y
Why: This is the operation that reverses every pair.
\[ x = 3 y - 4 \]
Solve for y
Why: Add 4 to both sides.
\[ x + 4 = 3 y \]
Finish and rename
Why: Divide by 3 and write it as the inverse.
Figure (svg): The three-step algebraic recipe for finding an inverse, worked on one example
\[ f^{-1}(x) = \frac{x+4}{3} \]
Verify: run a cancellation check and a graph check
Why: Composing gives x in both directions, as the earlier example showed. Geometrically, f has slope 3 and the inverse has slope one third — reciprocal slopes, which is what reflecting a line in the diagonal must do. Both intercepts also swap in the expected way. When the reciprocal-slope relation fails for a linear function, the algebra has gone wrong somewhere.
OpenStax Calculus Volume 1, §1.4 Inverse Functions §1.4, pp. 72-72
Ranking
Given a one-to-one rule, produce its inverse.
Put in order
Why: Step a is what makes the rest legitimate, and step d's branch choice is decided entirely by the restriction chosen in step a. Doing the swap before solving keeps every later line a statement about the inverse, with no renaming left to remember.
Worked example
Example 1.33. No inverse exists until the domain is cut down.
\[ \text{Restrict } f(x) = (x-1)^2 \text{ so that an inverse exists, and find it.} \]
Locate the vertex, where the two halves meet
Why: The squared quantity is zero there.
\[ \text{vertex at } x = 1 \]
Keep one side of it
Why: Choosing the right-hand branch makes the function increasing.
\[ \text{restrict to } x \ge 1 \]
Swap and solve, taking the root that matches
Why: The non-negative root corresponds to the branch kept.
\[ x = (y - 1) ^{2}\text{ gives } y = 1 + \sqrt{x} \]
State the domain of the inverse
Why: It is the range of the restricted original.
\[ D = [0, \infty] \]
Figure (svg): The solution to Worked example an inverse needing a restriction shown as a ladder of expressions, one row per legal move
\[ f^{-1}(x) = 1 + \sqrt{x}, \quad D = [0,\infty), \; R = [1,\infty) \]
Verify: check that the other branch gives a different, equally valid answer
Why: Restricting instead to x at most 1 gives the inverse 1 minus the square root of x, with range from negative infinity to 1. Both are correct inverses of different restrictions of the same rule. That is the honest situation: the restriction is a CHOICE, and the inverse you get depends on it. Testing the first at x equal to 4 gives 3, and f of 3 is 4 — consistent, and 3 does lie in the branch that was kept.
OpenStax Calculus Volume 1, §1.4 Inverse Functions §1.4, pp. 74-75
Trap
\[ f(x) = (x-1)^2 \text{ restricted to } x \le 1 \]
Swap, solve, and take the positive square root by habit
Why: The student writes the root with a plus sign automatically.
\[ f^{-1}(x) = 1 + \sqrt{x} \quad \text{(wrong branch)} \]
Test it: f inverse of 4 gives 3, but 3 is not in the restricted domain, which was inputs at most 1. The answer lands outside the set it was supposed to return to.
\[ f^{-1}(x) = 1 - \sqrt{x}, \quad R = (-\infty, 1] \]
Choose the sign that lands in the domain you kept
Why: The inverse's range must be the restricted original's domain, and that requirement picks the sign.
The check is quick and decisive: f inverse of 4 is now negative 1, which is at most 1, and f of negative 1 is 4. The general rule is that the inverse's RANGE must equal the restricted domain, so compute one value and see where it lands.
Fill the middle
Finding the inverse of the linear rule from the worked example.
Fill in the blanks
x = 3y - 4 \;\Longrightarrow\; y = (x+4)/3
Why: Adding 4 and dividing by 3 isolates y. Note the reciprocal slope: the original had slope 3 and the inverse has one third, which is what reflection in the diagonal does to any line.
Prediction
Commit before reasoning.
Predict first
If the graphs of f and its inverse intersect, where must the intersection lie?
Correct: On the line y equals x — those are the points the reflection leaves alone.
\[ (a,b) = (b,a) \;\Longrightarrow\; a = b \]
Why: Reflecting in the diagonal sends the point (a, b) to (b, a). For a point to lie on both graphs it must be its own reflection, which forces a to equal b. So intersections sit on the diagonal. They need not be at the origin, and they certainly can occur: the reciprocal function is its own inverse and meets the diagonal at 1 and negative 1. A useful consequence is that solving f of x equals x finds those meeting points without ever computing the inverse.
Sorting
Ask whether the natural domain is already one-to-one.
Sort into buckets
Sort each function.
Even powers always need restricting and odd powers never do, which is the same parity distinction that decided even and odd symmetry in Section 1.1 and even and odd roots in Section 1.2. Cosine needs the most severe restriction of all, and the conventional choice is the interval from 0 to pi.
Section
Section 4
Concept
Each trigonometric function is restricted to one branch on which it is monotone, and the inverse of that branch is the standard inverse trigonometric function. The chosen branch is a convention, but it is a universal one.
principal values — The conventional restricted ranges of the inverse trigonometric functions: inverse sine and inverse tangent return values between negative pi over two and pi over two, while inverse cosine returns values between zero and pi.
\[ \sin^{-1}: [-1,1] \to \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right], \qquad \cos^{-1}: [-1,1] \to [0, \pi] \]
The two ranges differ, and the difference matters. Inverse sine returns a value in the right half of the circle, which may be negative; inverse cosine returns a value in the upper half, which is never negative. Using one range for the other is the commonest error here.
Figure (svg): The sine curve with its conventional restricted domain highlighted, and the inverse sine beside it
OpenStax Calculus Volume 1, §1.4 Inverse Functions §1.4, pp. 75-79 — inverse trigonometric functions
Picture it
Sine cut down to a single increasing piece, and the inverse that produces.
Figure (svg): The sine curve with its conventional restricted domain highlighted, and the inverse sine beside it
The restricted branch runs from negative pi over 2 to pi over 2, which is exactly the range of the inverse. Reflecting that branch in the diagonal gives the inverse sine graph beside it, with domain and range swapped.
Worked example
Example 1.35. Respect the range, always.
\[ \text{Evaluate } \sin^{-1}\!\left(-\tfrac{1}{2}\right) \text{ and } \cos^{-1}\!\left(-\tfrac{1}{2}\right). \]
Recall the inverse sine's range
Why: Between negative pi over 2 and pi over 2.
\[ \text{answer in } [-\frac{\pi}{2}, \frac{\pi}{2}] \]
Find the angle in that range with sine negative one half
Why: Negative pi over 6 works and lies in the range.
\[ \arcsin(-\frac{1}{2}) = -\frac{\pi}{6} \]
Recall the inverse cosine's range
Why: Between 0 and pi, so never negative.
\[ \text{answer in } [0, \pi] \]
Find the angle in that range with cosine negative one half
Why: Two pi over 3 lies in the second quadrant, inside the range.
\[ \arccos(-\frac{1}{2}) = 2 \pi / 3 \]
Figure (svg): The solution to Worked example evaluating inverse trigonometric functions shown as a ladder of expressions, one row per legal move
\[ \sin^{-1}\!\left(-\tfrac{1}{2}\right) = -\tfrac{\pi}{6}, \qquad \cos^{-1}\!\left(-\tfrac{1}{2}\right) = \tfrac{2\pi}{3} \]
Verify: check each answer lies in its own range
Why: Negative pi over 6 is inside the interval from negative pi over 2 to pi over 2, so the first is legitimate. Two pi over 3 is inside the interval from 0 to pi, so the second is too. Note how differently the same input is treated: the inverse sine went negative and the inverse cosine did not, purely because the conventional ranges differ. Answering 4 pi over 3 for the second would have the right cosine but sits outside the range, so it is wrong.
OpenStax Calculus Volume 1, §1.4 Inverse Functions §1.4, pp. 77-78
Matching
Respect each function's own range.
Match the pairs
Why: The second and third share an input of negative one half and give completely different answers, because their ranges differ. Inverse sine may return a negative angle; inverse cosine never does. That single distinction accounts for most errors on this topic.
Worked example
Example 1.36. Draw a triangle rather than computing an angle.
\[ \text{Simplify } \cos\!\left(\sin^{-1} x\right) \text{ for } -1 \le x \le 1. \]
Name the inner angle
Why: Let theta be the angle whose sine is x.
\[ \theta = \arcsin x,\text{ so } \sin \theta = x \]
Use the Pythagorean identity
Why: Cosine squared is one minus sine squared.
\[ \cos ^{2} \theta = 1 - x ^{2} \]
Take the root, choosing the sign from the range
Why: Theta lies between negative pi over 2 and pi over 2, where cosine is never negative.
\[ \cos \theta = +\sqrt{1 - x ^{2}} \]
State the result
Why: No inverse trigonometric function remains.
\[ \cos(\arcsin x) = \sqrt{1 - x ^{2}} \]
Figure (svg): The solution to Worked example a composition with an inverse shown as a ladder of expressions, one row per legal move
\[ \cos\!\left(\sin^{-1} x\right) = \sqrt{1 - x^2} \]
Verify: test the endpoints and one interior value
Why: At x equal to 0: arcsin 0 is 0 and cosine of 0 is 1, matching the root of 1. At x equal to 1: arcsin 1 is pi over 2 and its cosine is 0, matching the root of 0. At x equal to negative one half the formula gives root of three quarters, about 0.866, and cosine of negative pi over 6 is indeed that. The positive sign was forced by the range — this is where the convention does real work, and choosing the negative root would fail every one of these checks.
OpenStax Calculus Volume 1, §1.4 Inverse Functions §1.4, pp. 78-79
Error analysis
A student evaluates the inverse sine.
Annotate
On: \( \sin^{-1}\!\left(\tfrac{1}{2}\right) = \tfrac{5\pi}{6}, \quad \text{since } \sin\tfrac{5\pi}{6} = \tfrac{1}{2} \)
This is the distinction between SOLVING an equation and EVALUATING an inverse. Solving gives every angle with that sine; evaluating the inverse gives the one in the conventional range. Section 1.3's trap was forgetting the extra solutions, and this one is including them where they do not belong.
Fill the middle
The composition from the worked example, using the Pythagorean identity.
Fill in the blanks
\cos\!\left(\sin^x^2 x\right) = \sqrt___}}
Why: If theta is the angle whose sine is x, then cosine squared theta is 1 minus x squared. The positive root is taken because the inverse sine's range lies where cosine is non-negative — the convention deciding the sign.
Two truths and a lie
All three are about inverse trigonometric functions.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and it is the subtler of the two cancellation directions. The equation holds only when t is already inside the restricted range. At t equal to 5 pi over 6, the sine is one half and the inverse sine of that is pi over 6, not 5 pi over 6. The composition returns the representative of t within the principal range, which is why the two cancellation equations have different domains attached.
Prediction
Commit before reasoning.
Predict first
Why is inverse cosine's range taken as 0 to pi rather than negative pi over 2 to pi over 2, as for sine?
Correct: Because cosine is even, so on the interval from negative pi over 2 to pi over 2 it takes each value twice.
\[ \cos(-t) = \cos t \;\Longrightarrow\; \text{not one-to-one on } \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right] \]
Why: Cosine of negative t equals cosine of t, so that symmetric interval fails the horizontal line test badly — cosine of pi over 3 and of negative pi over 3 are both one half. The interval from 0 to pi is where cosine is strictly decreasing, so it is one-to-one there and every value from negative 1 to 1 is attained exactly once. The choice is forced by cosine's symmetry, not by taste, and it is why the two inverse functions have genuinely different ranges.
Section
Section 5
Concept
Any function can be made one-to-one by shrinking its domain to a piece on which it is monotone. Doing so always works, but it discards inputs, and different choices of piece give genuinely different inverses.
restricted domain — A subset of a function's domain chosen so that the function is one-to-one on it. The inverse of the restricted function has that subset as its range, so the choice of restriction determines which inverse you get.
\[ f(x) = x^2 \text{ on } [0,\infty) \;\Longrightarrow\; f^{-1}(x) = \sqrt{x} \]
Every familiar inverse in mathematics is the result of such a choice. The square root is the inverse of the right half of the parabola, and the decision to prefer the positive root rather than the negative one is exactly this convention, made so long ago that it now looks like a fact.
Figure (svg): The parabola with its left half greyed out, showing the restriction that makes an inverse possible
OpenStax Calculus Volume 1, §1.4 Inverse Functions §1.4, pp. 73-76 — restricting domains
Picture it
The same rule, before and after the restriction.
Figure (svg): The parabola with its left half greyed out, showing the restriction that makes an inverse possible
Nothing about the formula changed; only the set of allowed inputs did. That is enough to turn a function with no inverse into one with a perfectly good inverse, at the cost of the inputs thrown away.
Worked example
The choice is real, and both answers are correct.
\[ \text{Invert } f(x) = x^2 - 4 \text{ on } [0,\infty) \text{ and again on } (-\infty, 0]. \]
Swap and solve, leaving the sign open
Why: Add 4 and take roots.
\[ x = y ^{2} - 4\text{ gives } y = +- \sqrt{x + 4} \]
For the right-hand restriction, take the positive root
Why: The inverse's range must be the non-negative inputs kept.
\[ y = +\sqrt{x + 4} \]
For the left-hand restriction, take the negative root
Why: Now the range must be the non-positive inputs.
\[ y = -\sqrt{x + 4} \]
State the shared domain
Why: Both inverses accept the original's range.
\[ D = [-4, \infty] \]
Figure (svg): The solution to Worked example two restrictions, two inverses shown as a ladder of expressions, one row per legal move
\[ f^{-1}(x) = \pm\sqrt{x+4}, \text{ the sign fixed by the branch kept} \]
Verify: test one input against each branch
Why: Take x equal to 5. The first inverse gives 3, and f of 3 is 9 minus 4, which is 5, with 3 in the non-negative half. The second gives negative 3, and f of negative 3 is also 5, with negative 3 in the non-positive half. Both round trips succeed, on their own branches. That two different correct answers exist is the honest content here — the inverse is a function of the restriction as much as of the rule.
OpenStax Calculus Volume 1, §1.4 Inverse Functions §1.4, pp. 74-75
Matching
The branch kept decides the sign.
Match the pairs
Why: In every row the inverse's range is exactly the interval that was kept — that is the general rule, and it is what fixes the sign or the branch. The first two rows show two correct inverses of the same formula, differing only in the restriction chosen.
Worked example
The same technique, applied where it matters most.
\[ \text{Which restriction of } \cos x \text{ makes an inverse possible, and what is its range?} \]
Find an interval on which cosine is monotone
Why: It falls steadily from 1 to negative 1.
\[ \text{cosine decreases on } [0, \pi] \]
Check every output is attained exactly once
Why: The range from negative 1 to 1 is covered without repetition.
\[ \text{one-to-one on } [0, \pi] \]
The inverse's domain is that range
Why: It accepts what cosine produced.
\[ D = [-1, 1] \]
The inverse's range is the restricted domain
Why: It returns what cosine accepted.
\[ R = [0, \pi] \]
Figure (svg): The solution to Worked example restricting a trigonometric function shown as a ladder of expressions, one row per legal move
\[ \cos^{-1} : [-1, 1] \to [0, \pi] \]
Verify: confirm the interval is the largest that works
Why: Extending to any interval larger than 0 to pi would include a turning point of cosine and immediately repeat a value: adding a little past pi makes cosine start rising again, so it would take some value twice. Adding a little below 0 does the same by evenness. So this interval is maximal, which is one reason the convention settled there rather than on some other decreasing stretch such as 2 pi to 3 pi.
OpenStax Calculus Volume 1, §1.4 Inverse Functions §1.4, pp. 76-77
Trap
\[ x^2 = 9 \;\Longrightarrow\; x = \sqrt{9} = 3 \]
Apply the square root as though it undid squaring everywhere
Why: The student treats root and square as exact opposites.
\[ x = 3 \quad \text{(incomplete)} \]
The value negative 3 also squares to 9 and has been silently discarded.
\[ x^2 = 9 \;\Longrightarrow\; x = \pm 3 \]
Remember the root inverts only the restricted half
Why: The square root is the inverse of squaring on the non-negative inputs, so it can only ever return one of the two solutions.
The general fact is that the root of x squared is the ABSOLUTE VALUE of x, not x. When solving an equation you must restore the sign the restriction removed — which is the same shape of error as reporting one family of solutions to a trigonometric equation.
\[ \sqrt{x^2} = |x|, \quad \text{not } x \]
Fill the middle
Simplifying an expression that catches almost everyone.
Fill in the blanks
\sqrt|x| = ___
Why: At x equal to negative 3 the left side is the root of 9, which is 3, not negative 3. So the root of x squared is the absolute value of x. The square root only inverts squaring on the non-negative half, and the absolute value is exactly what records that.
Prediction
Commit before reasoning.
Predict first
What is given up by restricting a domain to force an inverse to exist?
Correct: The discarded inputs — and with them the completeness of any solution set the inverse produces.
\[ \text{restriction} \Rightarrow \text{inverse exists}; \quad \text{but solving needs the discarded branch back} \]
Why: The restricted function still has the same formula and the same range; what is lost is half the domain. The practical consequence appears whenever you solve: taking a square root or an inverse sine returns only the representative from the branch that was kept, so the other solutions must be restored by hand. That is why solving x squared equals 9 needs the plus-or-minus, and why solving sine of t equals one half needs a second family. The two errors are the same error.
Sorting
One asks for every input; the other asks for a single principal value.
Sort into buckets
Sort each task by how many answers it has.
The pairs a-b and c-d are the same underlying question asked two ways, and they have different answers. Being clear about which one is being asked is the single most useful habit from this section.
Comparison
Fill the blanks. Confusing these two is the most expensive error in the section.
Comparison matrix
| Test | Question it asks | What passing it means |
|---|---|---|
| Vertical line | does any input have two outputs? | the curve is a function |
| Horizontal line | does any output have two inputs? | the function is one-to-one, so it has an inverse |
| Both | neither failure occurs | the pairs can be reversed to give a function |
| Neither | it is not even a function | nothing can be done until it is split into functions |
The middle two rows are the whole section. A curve can be a perfectly good function and still have no inverse, and the parabola is the example to keep in mind.
Pattern
Given a function and asked for its inverse.
Step two's record-keeping is what makes step three's branch choice automatic. Skipping it is why the plus-or-minus so often gets resolved by habit rather than by reasoning.
Stewart, Calculus: Early Transcendentals 8e, §1.5 Inverse Functions and Logarithms §1.5, pp. 55-67
Check
One-to-one. Ask about shared outputs.
Check your understanding
Which of these is one-to-one on its natural domain?
Answer: A
Why: The cubic is increasing everywhere, so every horizontal line meets it exactly once.
Check
Finding an inverse. Swap, then solve.
Check your understanding
Find the inverse of f(x) = 3x - 4.
Answer: A
Why: Swapping gives x = 3y - 4; adding 4 and dividing by 3 gives (x + 4)/3.
Check
Inverse trigonometry. Respect the range.
Check your understanding
Evaluate arccos(-1/2).
Answer: A
Why: Inverse cosine returns values in [0, pi], and 2 pi / 3 is the angle there with cosine negative one half.
Real world
A conversion between temperature scales is given by F equals nine fifths of C plus thirty-two, and a shipping firm charges a flat fee plus a rate per kilogram.
Discussion prompt
Find the inverse of the temperature conversion and say what it is for. Then explain why a charge structure that gives the same price for two different weights has no inverse, and what that means in practice.
Hint: An inverse answers the reversed question, and it exists exactly when the reversed question has one answer.
Answer:
\[ F = \tfrac{9}{5}C + 32 \;\Longrightarrow\; C = \tfrac{5}{9}(F - 32) \]
The inverse converts the other way. It exists because the conversion is linear with non-zero slope, hence strictly increasing, hence one-to-one — every Fahrenheit reading came from exactly one Celsius reading.
The charge structure is different. If the firm rounds weights up to the nearest kilogram, then a 2.1 kg and a 2.9 kg parcel both cost the same. The cost function is a staircase, it fails the horizontal line test, and it has no inverse.
In practice that means the reversed question — given the price, what did the parcel weigh? — has no single answer, and no clever algebra will supply one. The information was destroyed when the weight was rounded. This is the everyday form of the point about restriction: an inverse recovers the input only when nothing was thrown away.
\[ \text{one-to-one} \iff \text{the input can be recovered from the output} \]
Commit first
Answer, then rate your confidence honestly.
Predict first
What is the root of x squared, simplified?
Correct: The absolute value of x.
\[ \sqrt{(-3)^2} = \sqrt{9} = 3 = |-3| \]
Why: Test it at a negative input: the root of the square of negative 3 is the root of 9, which is 3 — not negative 3. The square root symbol always denotes the non-negative root, because it is the inverse of squaring restricted to the non-negative inputs. Writing plus or minus is wrong for a different reason: the expression is a single number, not two. The plus-or-minus belongs when SOLVING x squared equals 9, which is a different question from simplifying this expression.
Explain it
They can find inverses algebraically but do not see why the parabola needs a restriction when the algebra seems to work fine.
Discussion prompt
In four sentences or fewer, explain why squaring has no inverse until its domain is cut down.
Hint: Ask them what the input was, given that the output is 9.
Answer:
Ask them: the output is 9, so what was the input? They cannot say — it might have been 3 or negative 3, and nothing in the output distinguishes them. An inverse has to answer that question with a single number, and here there is no single answer to give.
Cutting the domain to the non-negative inputs throws one of the two candidates away, and then the question does have one answer. That is what the square root is: not the undoing of squaring in general, but the undoing of squaring on the half where the ambiguity was removed.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For one-to-one, say the horizontal question out loud: does any output come from two inputs? For branches, remember that the inverse's range must equal the domain you kept, then test one value. For the trigonometric ranges, hold onto the two pictures — inverse sine lives in the right half of the circle, inverse cosine in the upper half. For solving versus evaluating, ask whether the question wants every input or one principal value. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Draw one set of axes with the line y equals x dashed across it. On those axes sketch the cubing function and its inverse the cube root, and mark a matched pair of points showing the coordinate swap. Beside it, draw a second set of axes with the parabola, mark a horizontal line meeting it twice, and write one sentence saying which output has two inputs. Then grey out the left half, redraw the surviving half with the square root reflected in the diagonal, and state the domain and range of both. Below, find the inverse of 3x minus 4 in full, showing the swap on its own line, and verify it with both cancellation equations. At the bottom, draw sine over three periods, shade the branch from negative pi over 2 to pi over 2, and beside it write the value of the inverse sine of one half and the complete solution set of sine t equals one half, with one sentence on why those two answers differ. In a margin, write what the root of x squared simplifies to, and why.
If your cube root graph is not the mirror image of your cubic in the dashed diagonal, check a single point: whatever pair (a, b) sits on one, the pair (b, a) must sit on the other. That one check is faster than redrawing.
Recap
Five things, and the last resolves the puzzle Section 1.3 left open.
| If you see | Then |
|---|---|
| A horizontal line meeting twice | No inverse without a restriction |
| A strictly monotone function | It is one-to-one, so an inverse exists |
| A superscript negative one on a function | It means the inverse, not the reciprocal |
| A restricted domain | The inverse's range must equal it |
| The root of a square | It is the absolute value |
| An inverse trig evaluation | One answer, inside the conventional range |
| A trig equation to solve | Every answer, restoring the discarded branches |
Section 1.5 takes the last family of Chapter 1: exponential and logarithmic functions, which are inverses of each other in exactly the sense defined here — and whose calculus in Chapter 3 turns out to be the simplest of all.
OpenStax Calculus Volume 1, §1.4 Inverse Functions §1.4, pp. 65-79 — everything on these slides traces back here
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