Inverse Functions

Intuition

The previous lesson left one thread hanging on purpose. Composition of functions is associative and has an identity, the function $\operatorname{id}(x) = x$ that changes nothing, but it does not hand you an inverse for every function. That missing inverse made the whole collection of functions a monoid rather than a group. This lesson supplies the missing piece for the functions that admit it. You already know the mechanics from an earlier course, swap $x$ and $y$ and solve, and this lesson does not repeat them. It asks what an inverse really is, and the answer is structural: $f^{-1}$ is the inverse element of $f$ for composition, the function that undoes $f$ from both directions. That one word, both, is the whole lesson. It tells apart a function that truly reverses $f$ from one that only appears to, and it is what makes the inverse unique. That word is also what finally answers the question the first lesson of this chapter asked and could not yet settle: which rules can be run backward at all.

Learning goals

Require both compositions to be the identity
Prove the inverse unique from associativity
Tie invertibility to being one-to-one onto the range
Swap the ordered pairs, which the horizontal line test measures
Trade domain for range between the two functions
Restrict a domain knowing the choice is not unique

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 $(f \circ g)(x) = x$ and concludes $g = f^{-1}$ . Why is the conclusion wrong?

A It is correct; one identity is enough. B

(f \circ g)(x)

does not actually equal

x

. C

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

, which is not

x

when

x < 0

, so

g

undoes

f

from only one side. D

\sqrt{x}

is not a function, so it cannot be an inverse.

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 cancelled “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. This is the payoff of the previous lesson’s hardest theorem, associativity, and it is what earns us the right to write “the” inverse and give it a name.

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.

Because there is at most one, we may speak of the inverse of $f$ and give it the name $f^{-1}$ . 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.

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.

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.

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.

First, the function is one-to-one: if $a^3 + 1 = b^3 + 1$ then $a^3 = b^3$ , and

a^3 - b^3 = (a - b)(a^2 + ab + b^2) = 0

forces $a = b$ , because the second factor is positive except at $a = b = 0$ . So an inverse exists. To find it, 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 $f^{-1}(x) = \sqrt[3]{x - 1}$ . Verify both orders:

(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. Solving for $x$ produced a candidate; the two-sided check is what turned the candidate into a proof.

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. 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

Taking each function's codomain to be its range, which one has an inverse?

f(x) = x^2

with domain all real numbers B

f(x) = |x|

with domain all real numbers C

f(x) = x^3

with domain all real numbers D

f(x) = x^4

with domain all real numbers

The question Chapter one asked

The first lesson of this chapter, Relations and Functions, opened with a claim that can now be cashed in. It said that whether a set of points is a function depends on which variable you feed in. It added: the horizontal line test is nothing but the vertical line test asked about the other variable, and swapping the two variables reflects the whole picture across $y = x$ . It promised that this swap is exactly what forming an inverse would mean. Here is that promise paid.

A function $f$ is, underneath, its set of input-output pairs $(a, f(a))$ . Swapping the two coordinates of every pair produces a new set of points,

\{\, (f(a),\, a) : a \in A \,\},

the inverse relation: the same pairings read backward, output first. That swapped relation is a function of its first coordinate exactly when no two of its pairs share that first coordinate. That in turn means no two pairs of $f$ share the same second coordinate, which means no two inputs of $f$ produce the same output. That last phrase is one-to-one, and on a graph it is the horizontal line test applied to $f$ . So the swapped relation is a function precisely when $f$ is one-to-one, and only then is there an inverse function $f^{-1}$ .

This is the chapter closing on itself. Being a function was never a property of a curve; it was a property of a curve together with a choice of input. Inverting is the act of changing that choice, feeding in what used to come out. The vertical line test on the swapped picture is the horizontal line test on the original. So the two questions, “can $f$ be undone” and “does $f$ pass the horizontal line test”, were the same question in two costumes all along.

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 neighbouring 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 the awkward side: $(f^{-1} \circ f)(x) = \sqrt{x^2} = |x| = x$ , 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: $(f^{-1} \circ f)(x) = -\sqrt{x^2} = -|x| = -(-x) = x$ , 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. This is the same honesty the transformations and composition lessons practised when they refused to call a non-unique decomposition “the” decomposition.

Domain and range trade places

The pair-swap picture makes one more fact plain, and it is worth stating as a small theorem. When you reverse every pair $(a, b)$ into $(b, a)$ , the first and second coordinates switch roles. So the inputs of $f^{-1}$ are the outputs of $f$ , and the outputs of $f^{-1}$ are the inputs of $f$ .

$\operatorname{dom} f^{-1} = \operatorname{range} f$ and $\operatorname{range} f^{-1} = \operatorname{dom} f$ #

By the definition of the inverse, $f^{-1}$ carries an output of $f$ back to the input it came from, so its allowed inputs are exactly the values $f$ actually produces. That set is the range of $f$ , giving $\operatorname{dom} f^{-1} = \operatorname{range} f$ .

For the other equation, $f^{-1}$ returns inputs of $f$ , and it returns every one of them, since each $a \in \operatorname{dom} f$ is handed back from $b = f(a)$ by the relation $f^{-1}(f(a)) = a$ . So the outputs of $f^{-1}$ are precisely $\operatorname{dom} f$ , giving $\operatorname{range} f^{-1} = \operatorname{dom} f$ .

This is the structural version of a fact the earlier lessons kept circling: the domain is genuine data. Inverting does not merely flip a formula; it swaps the two sets a function lives between. In the example just above, $f(x) = x^2$ on $x \ge 0$ has range $y \ge 0$ , and its inverse $\sqrt{x}$ has domain $x \ge 0$ and range $y \ge 0$ , the two sets traded exactly as the theorem says.

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, drawn as the steeper curve, and its inverse the square root, drawn as the shallower curve. 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, and this is worth pinning down because it is easy to assume without proof. If $f$ is increasing, so is $f^{-1}$ . To see it, 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.

That mirror is a reliable way to picture an inverse, and it is also where a tempting falsehood creeps in, the kind this chapter has spent six lessons hunting. It is often said that the graphs of $f$ and $f^{-1}$ can meet only on the line $y = x$ . That is false. The reflection carries the two graphs into each other, so their set of common points is symmetric across $y = x$ . But a set can be symmetric across a line without lying on it. The extreme case is a function that is its own inverse: $f(x) = \dfrac{1}{x}$ satisfies $f(f(x)) = x$ , so the graph of $f$ and the graph of $f^{-1}$ are the very same curve. The two graphs therefore meet at every one of its points, including $\left(2, \tfrac{1}{2}\right)$ , which is nowhere near $y = x$ . A non-degenerate example makes the point without that coincidence.

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 that line is simply wrong.

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.

There is a true statement nearby, and keeping it apart from the false one is the whole discipline of this chapter. If $f(a) = a$ , then $a$ is a fixed point: the point $(a, a)$ lies on $y = x$ , and since $f^{-1}(a) = a$ as well, it lies on both graphs. So every fixed point is a common point sitting on the line $y = x$ . That is the one direction that holds. Its converse, that every common point is such a fixed point, is the claim Worked Example 6 just refuted. For a strictly increasing $f$ the two directions do coincide, and then every intersection really does land on $y = x$ , which is why the false version sounds right. But increasing is a hypothesis, not a general truth, and $f(x) = -x^3$ is decreasing, so it is free to break the pattern.

Check your understanding

Which statement about the graphs of a one-to-one function $f$ and its inverse $f^{-1}$ is correct?

A They can intersect only on the line

y = x

. B Being reflections across

y = x

, they can never intersect. C They can intersect off the line

y = x

; for

f(x) = -x^3

they meet at

(1, -1)

and

(-1, 1)

. D They can intersect only at the origin.

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 $\mathbb{R}$ ; 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, by associativity ( $g = g \circ (f \circ h) = (g \circ f) \circ h = h$ ), 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, which answers Chapter one’s opening question.
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.

Inverse Functions

Learning goals

An inverse must undo f from both sides

There is only one inverse

Which functions can be inverted

The question Chapter one asked

Restricting the domain is a real choice

Domain and range trade places

Reading the inverse off the graph

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)