Inverse Functions

Intuition

Take $f(x) = 2x + 1$ . Put in $3$ and get $7$ . Now subtract $1$ , then divide by $2$ : $7$ turns back into $3$ , two steps that exactly undo $f$ ‘s two steps. A function that does this for $f$ , always turning $f$ ‘s output back into $f$ ‘s input, is called an inverse of $f$ .

Not every function can be undone this way. Squaring turns both $3$ and $-3$ into $9$ , so the output $9$ alone does not say which input made it. Once that sign is gone, no rule can bring it back. This lesson asks exactly which functions can be undone, and it turns out the answer hinges on one word: both. An inverse has to undo $f$ in both directions, not just one, and telling those two cases apart is the whole lesson.

There is also a structural reason to care. The previous lesson looked at functions that map a set back into itself, like $f \circ f$ , and showed that composing such functions is associative and has an identity, $\operatorname{id}(x) = x$ , two of the three things a group needs. The missing third is an inverse for every element, which is why composing functions from a set to itself only makes a monoid rather than a group. This lesson supplies that missing piece for the functions that have one.

Learning goals

Verify a candidate inverse by checking that both compositions equal the identity
Explain why a function has at most one inverse
Decide whether a function is invertible by testing whether it is one-to-one
Find an inverse algebraically and state its domain and range
Justify why restricting a domain to create an inverse is a choice, not a fixed repair
Connect the horizontal line test to swapping each pair $(a, b)$ into $(b, a)$

An inverse must undo f from both sides

Give the idea its definition. A function $g$ is an inverse of $f$ when composing the two in either order returns the identity:

g \circ f = \operatorname{id} \ \text{ on } \operatorname{dom} f \qquad\text{and}\qquad f \circ g = \operatorname{id} \ \text{ on } \operatorname{dom} g.

Read both halves. The first, $g(f(x)) = x$ for every $x$ in the domain of $f$ , says $g$ undoes $f$ : feed $x$ through $f$ and $g$ brings it back. The second, $f(g(x)) = x$ for every $x$ in the domain of $g$ , says $f$ undoes $g$ the same way. Reversing has to work in both directions, and we insist on both because one alone can hold while the other fails. The example that shows this is the heart of the lesson.

Worked example 1 One side is not enough

Let $f(x) = x^2$ with domain all real numbers, and let $g(x) = \sqrt{x}$ with domain $x \ge 0$ . Test both compositions.

Start with $f \circ g$ . For any $x \ge 0$ ,

(f \circ g)(x) = f(\sqrt{x}) = (\sqrt{x})^2 = x,

so $f \circ g$ is the identity on the domain of $g$ . Stop here and you would call $g$ the inverse of $f$ . Now test the other order:

(g \circ f)(x) = g(x^2) = \sqrt{x^2} = |x|.

This equals $x$ when $x \ge 0$ , but for $x < 0$ it gives $-x$ , not $x$ . For instance $(g \circ f)(-3) = \sqrt{9} = 3 \ne -3$ . So $g \circ f$ is not the identity on the domain of $f$ ; it collapses every negative input to its positive twin. The function $g$ undoes $f$ from one side only. Squaring throws away the sign, and no square root can restore a sign that is already gone. A one-sided inverse is not an inverse, which is exactly why the definition demands both compositions.

Check your understanding

For $f(x) = x^2$ on all real numbers and $g(x) = \sqrt{x}$ on $x \ge 0$ , a student checks that $(f \circ g)(x) = x$ , concludes $g = f^{-1}$ , and adds: "inverses are unique, so no other function could also undo $f$ this way." What is wrong with the student's reasoning?

A Nothing is wrong; the uniqueness fact confirms the conclusion. B

(f \circ g)(x)

does not actually equal

x

, so the whole argument starts from a false step. C

(g \circ f)(x) = |x|

, which is not

x

when

x < 0

, so

g

never satisfies the two-sided definition and is not an inverse of

f

at all; the uniqueness fact only ever rules out a second inverse once a first one is confirmed, so it cannot rescue an

f

that has none. D The uniqueness fact only applies to linear functions, so it cannot be invoked for

f(x) = x^2

When both sides do hold, the verification is a short two-part check, and skipping either part is the most common way to get an inverse wrong.

Worked example 2 A two-sided verification

Show that $g(x) = \dfrac{x - 1}{2}$ is the inverse of $f(x) = 2x + 1$ , both on all real numbers.

Check $g \circ f$ first, substituting $f(x)$ into $g$ :

(g \circ f)(x) = g(2x + 1) = \frac{(2x + 1) - 1}{2} = \frac{2x}{2} = x.

Now check $f \circ g$ , substituting $g(x)$ into $f$ :

(f \circ g)(x) = f\!\left(\frac{x - 1}{2}\right) = 2 \cdot \frac{x - 1}{2} + 1 = (x - 1) + 1 = x.

Both compositions return $x$ for every real number, and both functions have all of $\mathbb{R}$ as their domain, so $g = f^{-1}$ . Each line undid the other: the first canceled “times two then plus one” by “minus one then over two,” and the second ran that cancellation in the opposite order.

There is only one inverse

Because the definition is two-sided, a function cannot have two different inverses. Here is why: suppose $g$ and $h$ both undid $f$ from both directions. Take any output $b$ of $f$ . Since $h$ is an inverse, $f(h(b)) = b$ ; feed that into $g$ , and since $g$ undoes $f$ , $g(f(h(b))) = h(b)$ . But the left side is also $g(b)$ , because $f(h(b)) = b$ . So $g(b) = h(b)$ for every $b$ : $g$ and $h$ agree everywhere, which makes them the same function. Because there is at most one inverse, we may speak of the inverse of $f$ and give it the name $f^{-1}$ .

Check your understanding

In the uniqueness argument, $g$ and $h$ both undo $f$ from both sides. For an output $b$ of $f$ , the single expression $g(f(h(b)))$ is read two ways: inside-out as $g(f(h(b))) = g(b)$ (using $f(h(b)) = b$ , since $h$ undoes $f$ ), and regrouped as $(g \circ f)(h(b)) = h(b)$ (using $g \circ f = \operatorname{id}$ , since $g$ undoes $f$ ). What justifies reading the same expression both ways?

A Composition is associative, so

g(f(h(b)))

and

(g \circ f)(h(b))

name the exact same value; the two readings just group the same three applications differently B

g

and

h

are already known to be equal, so any grouping gives the same answer C Composition is commutative, so

g

f

, and

h

can be applied in any order D

f

is its own inverse, so

f(h(b))

and

h(b)

are interchangeable

That name carries a warning you first met at the end of the previous lesson. The superscript in $f^{-1}$ marks the inverse for composition, and it never means the reciprocal $\dfrac{1}{f}$ . For $f(x) = x + 3$ the inverse is $f^{-1}(x) = x - 3$ , while the reciprocal $\bigl(f(x)\bigr)^{-1} = \dfrac{1}{x + 3}$ is a completely different function. Keep $f^{-1}$ for undoing and $\dfrac{1}{f}$ for dividing.

Which functions can be inverted

Not every function has an inverse; Worked Example 1 already met one that does not. The question is which do, and the answer is the cleanest statement in the chapter, provided we say one thing about the codomain out loud and then never forget it.

Throughout the rest of this lesson, take a function’s codomain to be its range, the set of values it actually outputs. A function is one-to-one when different inputs always give different outputs, equivalently when $f(a) = f(b)$ forces $a = b$ . With the codomain fixed as the range, invertibility and one-to-oneness are the same condition.

Here is why. Draw an arrow from each input to the output it produces. If $f$ is one-to-one, no two arrows point at the same output, so every output in the range has exactly one arrow pointing at it. Reverse every arrow and you get a new set of arrows, one leaving each output and landing back on the input it came from, and that is a function: it is $f^{-1}$ . If instead two arrows had pointed at the same output, reversing them would leave that output with two arrows going out to two different inputs, which is not a function at all, since a function may only give one answer per input. So the reversed arrows form a function exactly when the original arrows never collided, which is exactly one-to-one.

The codomain hypothesis is not decoration, and dropping it is the classic way this statement turns false. If the codomain is left larger than the range, some declared output is never actually produced, and no $g$ can satisfy $f \circ g = \operatorname{id}$ there. That is because $f(g(b))$ is always a genuine output of $f$ , and it can never equal a value that $f$ misses. With a larger codomain the correct condition is that $f$ be both one-to-one and onto, that is a bijection. Taking the codomain to be the range is exactly what makes “onto” automatic, and that leaves one-to-one as the only thing to check. Whenever this lesson says “invertible” it means “one-to-one, as a map onto its range.”

Worked example 3 Finding an inverse, then proving it

Find the inverse of $f(x) = x^3 + 1$ on all real numbers, and verify it on both sides.

Write $y = x^3 + 1$ and solve for $x$ :

y - 1 = x^3 \quad\Longrightarrow\quad x = \sqrt[3]{y - 1}.

Renaming the input, the candidate is $f^{-1}(x) = \sqrt[3]{x - 1}$ . Solving only produces a candidate, so verify both orders before trusting it:

(f^{-1} \circ f)(x) = \sqrt[3]{(x^3 + 1) - 1} = \sqrt[3]{x^3} = x,

(f \circ f^{-1})(x) = \left(\sqrt[3]{x - 1}\right)^3 + 1 = (x - 1) + 1 = x.

Both give $x$ across all of $\mathbb{R}$ , so $f^{-1}(x) = \sqrt[3]{x - 1}$ is confirmed. Notice that this check alone is enough: it proves $f$ has an inverse and pins down exactly what it is, all in one step, with no separate one-to-one argument needed first.

Worked example 4 Inverting a linear-fractional rule

Find the inverse of $f(x) = \dfrac{x}{x - 2}$ on $x \ne 2$ , and state the domain of the inverse.

Write $y = \dfrac{x}{x - 2}$ and solve for $x$ . Clear the denominator, then gather every term containing $x$ on one side:

y(x - 2) = x \quad\Longrightarrow\quad yx - 2y = x \quad\Longrightarrow\quad yx - x = 2y.

Now factor $x$ out of the left side and divide:

x(y - 1) = 2y \quad\Longrightarrow\quad x = \frac{2y}{y - 1}.

So $f^{-1}(x) = \dfrac{2x}{x - 1}$ . The move that carries any linear-fractional rule is exactly this one, collect the $x$ terms, factor, then divide. Check both orders before trusting the candidate:

(f \circ f^{-1})(x) = f\!\left(\frac{2x}{x - 1}\right) = \frac{\dfrac{2x}{x-1}}{\dfrac{2x}{x-1} - 2} = \frac{\dfrac{2x}{x-1}}{\dfrac{2}{x-1}} = \frac{2x}{2} = x,

(f^{-1} \circ f)(x) = f^{-1}\!\left(\frac{x}{x-2}\right) = \frac{2 \cdot \dfrac{x}{x-2}}{\dfrac{x}{x-2} - 1} = \frac{\dfrac{2x}{x-2}}{\dfrac{2}{x-2}} = \frac{2x}{2} = x.

Both collapse to $x$ . For the domain, the inputs of $f^{-1}$ are the values $f$ actually produces, and $f$ never outputs $1$ , since $\frac{x}{x - 2} = 1$ forces $x = x - 2$ , which is impossible. So $\operatorname{dom} f^{-1}$ excludes $1$ , matching the denominator of the formula $\frac{2x}{x - 1}$ .

Check your understanding

Consider $f(x) = \dfrac{3x}{x + 1}$ on $x \ne -1$ . Which statement correctly decides whether $f$ has an inverse and, if so, finds it?

f

has no inverse because its rule involves a fraction. B

f

is one-to-one because solving

y = \dfrac{3x}{x+1}

for

x

produces a single expression with no

\pm

choice, and that expression is

f^{-1}(x) = \dfrac{x}{3 - x}

, with domain

x \ne 3

. C

f

is one-to-one, so

f^{-1}(x) = \dfrac{x+1}{3x}

, the reciprocal of

f(x)

, with domain

x \ne 0

. D

f

is one-to-one, and solving for

x

just gives back the same rule,

f^{-1}(x) = \dfrac{3x}{x+1}

, with domain

x \ne -1

Swapping pairs is what inverting really does

Take the three pairs $(1, 4), (2, 7), (3, 10)$ , the input-output pairs of some function $f$ . Swap every pair’s coordinates and you get $(4, 1), (7, 2), (10, 3)$ : the same information, read backward, output first. Every input of $f$ becomes an output, and every output becomes an input. That swapped set of pairs is exactly $f^{-1}$ , when it is a function at all.

Now try it on a function that is not one-to-one: $(-2, 4)$ and $(2, 4)$ . Swapping gives $(4, -2)$ and $(4, 2)$ , two different pairs that start with the same first number, $4$ . A function may only give one answer per input, so this swapped set is not a function; there is no way to know whether $4$ should go back to $-2$ or to $2$ . A repeated output is exactly what breaks the swap.

This connects to a claim from the first lesson of this chapter, Relations and Functions: swapping the two variables reflects the whole picture across $y = x$ , and the horizontal line test is the vertical line test asked about the swapped picture. Both claims are now proven at once. A function $f$ is, underneath, its set of pairs $(a, f(a))$ . Swapping every pair gives the inverse relation, and that swapped set is a function of its first coordinate exactly when no two of $f$ ‘s pairs share a second coordinate, that is, exactly when $f$ is one-to-one, which on a graph is the horizontal line test. So the swapped relation is a function precisely when $f$ passes the horizontal line test, and only then is there an inverse function $f^{-1}$ .

Being a function was never a property of a curve alone; it was a property of a curve together with a choice of input. Inverting changes that choice, feeding in what used to come out. So “can $f$ be undone” and “does $f$ pass the horizontal line test” are the same question asked two ways.

Restricting the domain is a real choice

Worked Example 1 showed that $f(x) = x^2$ on all real numbers has no inverse, because it fails the horizontal line test: $f(-2) = f(2) = 4$ . The standard repair is to shrink the domain until the rule becomes one-to-one. It is worth being honest about what that move is. You are not fixing $f$ ; by the equality test of the Function Notation lesson, a rule together with a smaller domain is a different function. Restricting the domain replaces the function that had no inverse with a neighboring one that does.

Worked example 5 Two honest restrictions of $x^2$

The rule $x^2$ becomes one-to-one on the inputs $x \ge 0$ , and also on the inputs $x \le 0$ . Find the inverse in each case.

On the domain $x \ge 0$ , the outputs are all the $y \ge 0$ , and the input that produced a given $y$ is the nonnegative square root:

f(x) = x^2 \ (x \ge 0) \quad\Longrightarrow\quad f^{-1}(x) = \sqrt{x}.

Check both orders. The easy side is $(f \circ f^{-1})(x) = (\sqrt{x})^2 = x$ for $x \ge 0$ , the domain of $f^{-1}$ . The awkward side is $(f^{-1} \circ f)(x) = \sqrt{x^2} = |x| = x$ , which holds because $x \ge 0$ .

On the domain $x \le 0$ , the same outputs $y \ge 0$ now arrive from negative inputs, so the inverse is the negative root:

f(x) = x^2 \ (x \le 0) \quad\Longrightarrow\quad f^{-1}(x) = -\sqrt{x}.

Check both orders again. $(f \circ f^{-1})(x) = (-\sqrt{x})^2 = x$ for $x \ge 0$ , the domain of $f^{-1}$ . And $(f^{-1} \circ f)(x) = -\sqrt{x^2} = -|x| = -(-x) = x$ , using $|x| = -x$ for $x \le 0$ .

Both restrictions are legitimate, and they give different inverses, $\sqrt{x}$ and $-\sqrt{x}$ . So there is no such thing as the restriction that makes $x^2$ invertible; there is a choice, and the choice is part of the answer.

Check your understanding

Take the pairs $(0, 0), (1, 1), (2, 4), (3, 9)$ from $f(x) = x^2$ restricted to $x \ge 0$ . A classmate says: (1) 'swapping every pair to $(b, a)$ gives a function here because no two of these pairs share a second coordinate,' and (2) ' $x \ge 0$ is the one restriction that fixes $x^2$ , since it worked.' Which claim, if either, is correct?

A Both are correct. B Only claim (1) is correct: the swap gives a function because no output repeats, which is exactly the horizontal line test, but claim (2) is wrong, since

x \le 0

restricts

x^2

to a different, equally valid one-to-one rule with inverse

-\sqrt{x}

. C Only claim (2) is correct:

x \ge 0

is the one true fix, and the pair swap has nothing to do with why it works. D Neither is correct;

x^2

can never be inverted, no matter how the domain is restricted.

Domain and range trade places

Swapping every pair $(a, b)$ into $(b, a)$ swaps the two coordinates’ roles: the inputs of $f^{-1}$ are the outputs of $f$ , and the outputs of $f^{-1}$ are the inputs of $f$ . In symbols, $\operatorname{dom} f^{-1} = \operatorname{range} f$ and $\operatorname{range} f^{-1} = \operatorname{dom} f$ .

For the $x^2$ example just above, that swap is easy to miss, because $x \ge 0$ happened to be both the domain of $f$ and the range of $\sqrt{x}$ . Here is a case where domain and range actually look different. Let $f(x) = x + 3$ on $0 \le x \le 5$ . Its outputs run from $f(0) = 3$ to $f(5) = 8$ , so its range is $3 \le y \le 8$ . The inverse is $f^{-1}(x) = x - 3$ , and by the swap its domain is $3 \le x \le 8$ , the range $f$ just had, while its own range is $0 \le y \le 5$ , the domain $f$ just had. Inverting does not merely flip a formula; it swaps the two sets a function lives between, and the domain is genuine data, carried along with the rule.

Check your understanding

Let $f(x) = 2x - 1$ on the domain $1 \le x \le 4$ , so its range is $1 \le y \le 7$ . What are the domain AND range of $f^{-1}$ ?

A Domain

1 \le x \le 7

, range

1 \le y \le 4

B Domain

1 \le x \le 4

, range

1 \le y \le 7

(same as

f

) C Domain

1 \le x \le 7

, range

0 \le y \le 4

D Domain

0 \le x \le 4

, range

1 \le y \le 7

Reading the inverse off the graph

The coordinate swap has a picture. A point $(a, b)$ lies on the graph of $f$ exactly when $b = f(a)$ , and that is exactly when $(b, a)$ lies on the graph of $f^{-1}$ . So the graph of $f^{-1}$ is the set of points $(b, a)$ as $(a, b)$ runs over the graph of $f$ . The map that sends every point $(a, b)$ to $(b, a)$ is reflection in the line $y = x$ : it fixes that line and exchanges the horizontal and vertical directions. Therefore the graph of $f^{-1}$ is the mirror image of the graph of $f$ across $y = x$ . That is not by decree but because inverting is swapping coordinates, and swapping coordinates is that reflection.

The rule x squared on the inputs x greater than or equal to zero, labeled x squared, and its inverse the square root, labeled square root of x. Each is the mirror image of the other across the dashed line y = x, and the two curves meet on that line at the origin and at the point (1, 1).

Reflection across $y = x$ also preserves a graph’s direction: if $f$ is increasing, so is $f^{-1}$ , and if $f$ is decreasing, so is $f^{-1}$ . Here is why: “increasing” means that whenever the input goes up, the output goes up too, so an input-step and its matching output-step always point the same way. Swapping coordinates swaps which one you call the input, but it cannot flip just one of the two steps, so they still point the same way after the swap. A rising curve reflects to a rising curve, and a falling one to a falling one.

That mirror is a reliable way to picture an inverse, with one caution. It is tempting to think the two graphs can only meet on the line $y = x$ , since that is where the mirror itself sits. That is not true in general: the reflection carries the two graphs into each other, so wherever they cross, the crossings mirror each other too, but a set can mirror itself without lying entirely on the mirror line. What is always true is narrower: if $f(a) = a$ for some input $a$ , then $a$ is a fixed point, $(a, a)$ sits on $y = x$ , and since $f^{-1}(a) = a$ too, that point lies on both graphs. So every fixed point is a common point on $y = x$ , but a common point off $y = x$ can still happen, as the next example shows.

Worked example 6 Where $f(x) = -x^3$ meets its inverse

The function $f(x) = -x^3$ on all real numbers is one-to-one, and its inverse is $f^{-1}(x) = -\sqrt[3]{x}$ , the negative of the real cube root. Find every point the two graphs share.

A shared point is an $x$ with $f(x) = f^{-1}(x)$ , so set the two rules equal:

-x^3 = -\sqrt[3]{x} \quad\Longrightarrow\quad x^3 = \sqrt[3]{x}.

Cube both sides to clear the root, writing $\sqrt[3]{x}$ as $x^{1/3}$ :

x^9 = x \quad\Longrightarrow\quad x^9 - x = 0 \quad\Longrightarrow\quad x(x^8 - 1) = 0.

The solutions are $x = 0$ and $x^8 = 1$ , that is $x = 1$ and $x = -1$ . Read off the meeting points: at $x = 0$ the graphs meet at $(0, 0)$ , which lies on $y = x$ ; at $x = 1$ they meet at $(1, -1)$ ; and at $x = -1$ they meet at $(-1, 1)$ . Check the last two on both graphs: $f(1) = -1$ and $f^{-1}(1) = -\sqrt[3]{1} = -1$ , so $(1, -1)$ sits on each; and $f(-1) = 1$ with $f^{-1}(-1) = -\sqrt[3]{-1} = 1$ , so $(-1, 1)$ does too. Two of the three meeting points, $(1, -1)$ and $(-1, 1)$ , are not on $y = x$ : the claim that intersections can happen only on $y = x$ is false in general, even though it happens to be true whenever $f$ is increasing, since $f(x) = -x^3$ is decreasing.

The decreasing function f sending x to minus x cubed and its inverse, the function sending x to minus the cube root of x. The two curves are mirror images across the dashed line y = x, yet they cross at (-1, 1) and (1, -1), marked with dots, neither of which lies on that line.

Common mistakes

Checking only one composition. $g \circ f = \operatorname{id}$ does not force $f \circ g = \operatorname{id}$ . For $f(x) = x^2$ and $g(x) = \sqrt{x}$ , $f \circ g$ is the identity on $x \ge 0$ but $g \circ f = |x|$ , so $g$ is not an inverse. Always verify both sides.
Stating “invertible exactly when one-to-one” with no codomain. The clean biconditional needs the codomain taken to be the range. If the codomain is larger, you also need $f$ onto, that is a bijection; one-to-one alone is not enough.
Reading $f^{-1}$ as $\dfrac{1}{f}$ . The superscript marks the compositional inverse, never the reciprocal. For $f(x) = x + 3$ , $f^{-1}(x) = x - 3$ , while $\bigl(f(x)\bigr)^{-1} = \dfrac{1}{x + 3}$ is a different function.
Believing the graphs meet only on $y = x$ . They can cross off it: $f(x) = -x^3$ meets its inverse at $(1, -1)$ and $(-1, 1)$ . The one-way statement that does hold is that if $f(a) = a$ , then $(a, a)$ lies on both graphs and on $y = x$ . The reverse fails, as $(1, -1)$ shows, so a shared point need not sit on $y = x$ .
Treating a domain restriction as unique. The rule $x^2$ is made one-to-one on $x \ge 0$ (inverse $\sqrt{x}$ ) or on $x \le 0$ (inverse $-\sqrt{x}$ ); both are valid, so a restriction is a choice, not “the” restriction.
Trusting a simplified formula over the domain. $(g \circ f)(x)$ may reduce to $x$ yet fail to be the identity, because the composition can carry a smaller domain than $\mathbb{R}$ . For instance $(\sqrt{x})^2$ reduces to $x$ , but it is the identity only on $x \ge 0$ , not on all of $\mathbb{R}$ ; the inverse must match domains, not just formulas.

Key takeaways

$g$ is the inverse of $f$ when both $g \circ f = \operatorname{id}$ on $\operatorname{dom} f$ and $f \circ g = \operatorname{id}$ on $\operatorname{dom} g$ . A one-sided inverse is not an inverse.
The inverse is unique: two functions that both undid $f$ from both directions would have to agree everywhere, which is why we may write $f^{-1}$ ; the superscript means the compositional inverse, never $\dfrac{1}{f}$ .
With a function’s codomain taken to be its range, $f$ has an inverse exactly when it is one-to-one; with a larger codomain the condition is one-to-one and onto, a bijection.
Inverting swaps the ordered pairs $(a, b) \mapsto (b, a)$ , so the swapped relation is a function precisely when $f$ passes the horizontal line test.
Domain and range trade places: $\operatorname{dom} f^{-1} = \operatorname{range} f$ and $\operatorname{range} f^{-1} = \operatorname{dom} f$ .
The graph of $f^{-1}$ is the reflection of the graph of $f$ across $y = x$ , and the two can meet off that line; only $f(a) = a$ forces a common point onto $y = x$ .
Restricting the domain to gain an inverse produces a different function, and the restriction is a choice (both $x \ge 0$ and $x \le 0$ work for $x^2$ ), not a unique repair.

Practice

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More resources (optional)

Other explanations of this lesson, if you want a second take.

3.7 Inverse Functions (OpenStax)
Definition of Inverse and Inverting from a Graph (Purplemath)
Is the Inverse Also a Function? (Purplemath)
Finding the Inverse of a Function (Purplemath)

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Why an inverse can never have a rival

The main lesson gives the reason in plain words. Here is the formal version, which pins down exactly where the previous lesson’s associative law is needed.

The inverse is unique#

Suppose $g$ and $h$ are both inverses of $f$ . Write $A$ for the domain of $f$ and $B$ for its range, so $f$ sends $A$ to $B$ while $g$ and $h$ send $B$ back to $A$ . Being inverses means $g \circ f = \operatorname{id}_A$ and $f \circ g = \operatorname{id}_B$ , and likewise $h \circ f = \operatorname{id}_A$ and $f \circ h = \operatorname{id}_B$ .

Now run a single chain and justify each step:

g = g \circ \operatorname{id}_B = g \circ (f \circ h) = (g \circ f) \circ h = \operatorname{id}_A \circ h = h.

The first equality composes $g$ with the identity on $B$ , its own domain, which changes nothing. The second replaces $\operatorname{id}_B$ by $f \circ h$ , equal to it because $h$ is an inverse of $f$ . The third is the associative law from the previous lesson, and it is the one place the whole argument rests on it. The fourth uses $g \circ f = \operatorname{id}_A$ , and the last drops the identity on $A$ . So $g = h$ : any two inverses of $f$ are the same function. Notice that associativity is not a decoration here. Without it the middle step would fail, and the phrase “the inverse” would not even be well defined.

Proving invertible means one-to-one, in both directions

The main lesson explains this with arrows. Here is the same claim proved both ways.

With codomain taken as the range, $f$ is invertible exactly when it is one-to-one#

Take $f$ with domain $A$ and codomain its range $B = f(A)$ , so every element of $B$ equals $f(a)$ for at least one $a \in A$ .

Suppose first that $f$ is one-to-one. Each $b \in B$ is $f(a)$ for at least one $a$ , because $B$ is the range, and for at most one $a$ , because $f$ is one-to-one, so for exactly one. Define $g(b)$ to be that unique $a$ . Then for every $a \in A$ we get $g(f(a)) = a$ , since $a$ is the one input that $f$ sends to $f(a)$ , so $g \circ f = \operatorname{id}_A$ . Likewise, for every $b \in B$ we get $f(g(b)) = b$ , since $g(b)$ was chosen as an input that $f$ sends to $b$ , so $f \circ g = \operatorname{id}_B$ . Both sides hold, so $g$ is a two-sided inverse.

Conversely, suppose $f$ has an inverse $g$ . If $f(a_1) = f(a_2)$ , apply $g$ to both sides. Because $g \circ f = \operatorname{id}_A$ ,

a_1 = g(f(a_1)) = g(f(a_2)) = a_2,

so equal outputs force equal inputs and $f$ is one-to-one. That settles both directions.

Why reflecting across y = x preserves direction

The main lesson states this fact. Here is a proof that it holds for every increasing function, not just the ones a picture happens to show.

If $f$ is increasing, so is $f^{-1}$ #

Take two values $u < v$ in the range of $f$ and pull them back through $f$ , writing $u = f(a)$ and $v = f(b)$ , so $a = f^{-1}(u)$ and $b = f^{-1}(v)$ . If $a \ge b$ held, the increasing $f$ would give $f(a) \ge f(b)$ , that is $u \ge v$ , contradicting $u < v$ ; so $a < b$ , which says $f^{-1}(u) < f^{-1}(v)$ . A rising graph reflects to a rising graph, and by the same argument a falling one reflects to a falling one.

A bit of history (optional)

Mathematics sometimes gets a definition right by accident, and notices only years later.

Heinrich Weber, a German mathematician, wrote down axioms for a group in 1882. He asked for three things. Combining two elements must give a third element of the system. The combining must be associative. And you may cancel from either side: from $ab = ac$ it follows that $b = c$ , and from $ba = ca$ it also follows that $b = c$ . Nowhere did he ask that inverses exist. His definition was correct anyway, because every group he had in mind was finite. On a finite set, cancellation quietly delivers the inverses for nothing.

The reason is a counting argument. Combining with a fixed element $a$ is itself a map from the set to itself, sending each $x$ to $ax$ , and cancellation says that map never sends two inputs to one output. On a finite set, a map from a set to itself that never sends two inputs to one output must reach every element: its outputs are all different, so there are as many of them as there are elements, and nothing is left over to be missed. A one-sided inverse for a map of a finite set to itself is therefore two-sided already. The gap this lesson warns about cannot open there.

The functions here live where that argument collapses. On the whole real line, squaring sends two inputs to one output and still leaves outputs unused. That is why the square root undoes it from one side and fails from the other. Weber came to see this himself. His textbook of 1895 says that the infinite case must ask for inverses outright. That demand is the pair of compositions this lesson made you check by hand.