Inverse functions come back constantly in calculus. e^x and \ln x undo each other, and \sin x and \arcsin x undo each other. How fast an inverse function changes (a whole section, three chapters from now) is built entirely on the relationship refreshed here. It's worth having completely automatic before it's one more thing to think about mid-proof.
An inverse function is a rewind button. If f turns "3" into "9" by squaring it, f^{-1} has to turn "9" back into "3" by taking its square root. It can't turn "9" into "-3", even though (-3)^2 is also 9. A rewind button can only have one place to land. That's the whole reason for restricting the domain first. y=x^2 has two different inputs landing on the same output. There's no single, unambiguous way to rewind it until you agree in advance which half of the picture you're rewinding.
An inverse function, written f^{-1}, is a second function that exactly undoes f. If f sends a to b, then f^{-1} sends b back to a, so f^{-1}(f(x)) = x for every x in f's domain. Not every function has one. An inverse exists only when f is one-to-one, meaning no two different inputs ever produce the same output. You test that graphically with the horizontal line test. If any horizontal line crosses the graph more than once, two different x-values share that output, and no single rule could correctly "rewind" that value back to just one of them. Graphically, y = f^{-1}(x) is the reflection of y = f(x) across the line y = x: every point (a, b) on f becomes the point (b, a) on f^{-1}. When f fails the horizontal line test, the domain has to be restricted to a piece where it does pass before an inverse can be built at all. That restriction is exactly what happens to build \arcsin, \arccos, and \arctan out of ordinary \sin, \cos, and \tan, none of which are one-to-one on their full domain.
Formula
Inverse Function
also known as: the undo-button relationship
Recalled from College Algebra · Composition and inverse functions — first taught there.
If f is one-to-one, its inverse f^{-1} is defined by f^{-1}(y) = x \iff f(x) = y, so that f^{-1}(f(x)) = x for all x in the domain of f, and f(f^{-1}(x)) = x for all x in the domain of f^{-1}.
Theorem
Horizontal Line Test
Recalled from College Algebra · Composition and inverse functions — first taught there.
A function f is one-to-one if and only if no horizontal line intersects its graph more than once.
The horizontal line y = k meets the graph of f exactly at the points (x, k) with f(x) = k — one point for every input that produces output k.
If f is one-to-one, then f(a) = f(b) forces a = b, so no two distinct inputs share an output: every horizontal line meets the graph at most once. Conversely, suppose some horizontal line y = k met the graph twice, at (a, k) and (b, k) with a \neq b. Then f(a) = k = f(b) with a \neq b — two different inputs sharing an output, which is exactly the negation of one-to-one. So f fails to be one-to-one exactly when some horizontal line crosses its graph more than once.
Drag the point — watch its mirror
Drag the point along y=x^2 restricted to x \ge 0 (so it's one-to-one) — watch its mirror trace y=\sqrt{x} across the line y=x.
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Odd power vs. even power
Sort each power function as one-to-one or not — an odd exponent passes the horizontal line test everywhere; an even exponent always fails it once you look at both sides of the y-axis.
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One-to-one, or not?
Slide a — for a \le 0 the curve climbs the whole way through, so it's one-to-one everywhere. Once a > 0 it grows a hump: a horizontal line through the middle of that hump crosses the curve three times, so the horizontal line test fails right there.
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True, or false?
Sort each claim about inverse functions into True or False.
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Where this goes next
How fast an inverse function changes
Once that machinery exists, there's an elegant shortcut for how fast f^{-1} changes, built entirely from how fast f changes. You never need to find a formula for the inverse function first. It works because the mirror-across-y=x picture refreshed here does real work: reflecting a graph reflects its slopes too, in a specific, computable way.
Check your understanding (counts toward your progress)
If (4, 9) is a point on the graph of f, which point must be on the graph of f^{-1}?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
This is the original point on f itself -- f^{-1}'s graph is the coordinates SWAPPED, not repeated.
Reflecting across y=x swaps the coordinates; it doesn't negate either one.
Reflecting across y=x swaps both coordinates together; it doesn't negate just one of them.
Why
The graph of f^{-1} is the reflection of f's graph across y=x, which swaps every point's coordinates -- (4,9) on f becomes (9,4) on f^{-1}.
For f(x)=-8x+2, find f^{-1}(-70).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
f^{-1}(x)=\dfrac{x-(2)}{-8}, so f^{-1}(-70)=\dfrac{-70-(2)}{-8}=9.
Correct answer
Also accepted:
Why
Solve y=-8x+2 for x in terms of y, then substitute y=-70.
Practice: Finding and checking inverses (3 problems · doesn’t affect your progress)
For g(x) = x^3+2, find g^{-1}(29).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Solve x^3+2=29: x^3=27, so x=3 -- meaning g^{-1}(29)=3.
Correct answer
Also accepted:
Why
Set x^3+2 equal to 29 and solve for x -- take a cube root at the end.
Compare f(x)=x^2 and g(x)=x^3: which one needs its domain restricted before it can have an inverse function?
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This has the two swapped -- x^3 passes the horizontal line test everywhere already, while x^2 does not.
x^3 doesn't need a restriction -- every horizontal line crosses an odd-power curve like x^3 exactly once.
x^2 fails the horizontal line test on its full domain -- it does need a restriction, unlike x^3.
Why
x^2 sends both 2 and -2 to 4, failing the horizontal line test -- it needs a restricted domain (like x\ge0) first. x^3 is one-to-one over every real number already, since odd powers never repeat an output.
Check each function against the horizontal line test over all real numbers, without restricting anything first.
Only f(x)=x^2 needs a restriction -- g(x)=x^3 is already one-to-one on its own
Which pair of functions are genuine inverses of each other?
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Check f(g(x)): \dfrac{(3x-1)-1}{3}=\dfrac{3x-2}{3}, which isn't x -- this pair fails the inverse check.
These two just shift by opposite constants without undoing each other's scaling correctly: f(g(x))=2(2x-1)+1=4x-1\neq x.
f(g(x))=(x+1)+1=x+2\neq x -- a function can't be its own inverse unless composing it with itself actually returns x.
Why
f(g(x))=\dfrac{(3x+1)-1}{3}=\dfrac{3x}{3}=x, and g(f(x))=3\cdot\dfrac{x-1}{3}+1=(x-1)+1=x -- both directions confirm this pair genuinely undoes each other.
Compute f(g(x)) for each pair and check whether it simplifies all the way down to plain x.
f(x)=\dfrac{x-1}{3} and g(x)=3x+1
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
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Slide a horizontal line up and down across the pictured V-shaped graph. Does it pass the horizontal line test?
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Straightness only helps if there's a SINGLE line — here two different straight arms meet at the bottom, and both arms reach the same heights.
Having one lowest point doesn't prevent two DIFFERENT inputs from reaching the same height above it — which is exactly what happens on either arm.
Steepness changes how wide the V looks, but both arms still climb to reach every height above the vertex — the collision survives regardless.
Why
This is f(x)=|x|. Any horizontal line above the vertex crosses both the left arm and the right arm — two different inputs (e.g. x=3 and x=-3) sharing the same output 3. Not one-to-one, so absolute value has no inverse function over its full domain.
Pick a height above the vertex and check whether both arms of the V reach it.
From College Algebra
The point (-2, 7) lies on the graph of a one-to-one function f. What is f^{-1}(7)?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Inverse points are mirror images: (-2,7) on f becomes (7,-2) on f^{-1}. So f^{-1}(7) = -2.
Correct answer
Also accepted:
Why
Recall that the graph of f^{-1} swaps the coordinates of every point on f's graph.
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
Which function is odd?
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Both terms have even powers (the constant is x^0), making this even, not odd.
The x^3 part is odd but the +1 is even — the mixture is neither. Check: f(-1) = 0 but -f(1) = -2.
|-x| = |x|, so absolute value gives equal outputs for opposite inputs — that's even.
Why
Test f(x) = x^3 - 4x: f(-x) = (-x)^3 - 4(-x) = -x^3 + 4x = -f(x). Exact negative, so it's odd. Shortcut: every term has an odd power of x. Mixing odd-power terms with even-power terms (like +1) destroys the symmetry.
Test each function by substituting -x and comparing the result to the original.
From College Algebra
Classify h(x) = \dfrac{x}{x^2 + 1}.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Even would need h(-x) = h(x) — but the numerator's sign flips while the denominator doesn't, so the whole fraction flips sign.
Check algebraically: h(-x) = \frac{-x}{x^2+1} = -h(x) — an exact negative, which is the definition of odd, not "neither."
Symmetry is a property of the whole formula, not restricted to half the domain — the sign-flip works for every x, positive or negative.
Why
h(-x) = \frac{-x}{(-x)^2+1} = \frac{-x}{x^2+1} = -h(x). The denominator (even powers only) is unaffected by the sign flip, while the lone x on top flips — odd.
Substitute -x into the formula and simplify to compare it with the original.