Inverse Functions: Free Response

Substitute each candidate into $f$ and settle the sign inside the bars on the domain $x \ge 2$.

For $g$, the quantity $\dfrac{x - 2}{5}$ is nonnegative there, so the bars come off unchanged:

$5 \left\lvert \frac{x - 2}{5} \right\rvert + 2 = (x - 2) + 2 = x$

For $h$, the quantity $\dfrac{2 - x}{5}$ is nonpositive there, so the bars flip its sign before the multiplication happens:

$5 \left\lvert \frac{2 - x}{5} \right\rvert + 2 = (x - 2) + 2 = x$

Both compositions return their input for every $x \ge 2$, so on that side each of $g$ and $h$ does everything the identity does.

Substitute $f(x) = 5\lvert x \rvert + 2$ into each candidate. The $+2$ cancels at once in both cases:

$(g \circ f)(x) = \frac{(5\lvert x \rvert + 2) - 2}{5} = \lvert x \rvert$

$(h \circ f)(x) = \frac{2 - (5\lvert x \rvert + 2)}{5} = -\lvert x \rvert$

Evaluate at the two inputs. At $x = -4$ the first gives $\lvert -4 \rvert = 4$, which is not $-4$, while the second gives $-\lvert -4 \rvert = -4$, which is the input. At $x = 4$ the first gives $4$, the input, while the second gives $-4$, which is not.

Each candidate therefore returns the input on one half of the number line and the wrong sign on the other, so neither composition is the identity on all real numbers. Part A is untouched by this: both candidates still undo $f$ perfectly in the first order.

Look for two inputs of $f$ that share an output. Opposite inputs are the natural place, because the bars discard the sign:

$f(-4) = 5 \cdot 4 + 2 = 22 = f(4)$

Suppose some rule $k$ satisfied $k(f(x)) = x$ for every real $x$. Taking $x = -4$ forces $k(22) = -4$, and taking $x = 4$ forces $k(22) = 4$. One input, two required outputs: no function does that, so no such $k$ exists.

Nothing about $g$ or $h$ was used, so the obstruction belongs to $f$ alone. Two different inputs share the output $22$, which is to say the horizontal line at that height meets the graph of $f$ twice and $f$ is not one-to-one. That is why part A settles nothing on its own. Several rules can undo $f$ in one order while no rule undoes it in the other, and the definition of an inverse asks for both compositions precisely to keep those rules out.

Start at $g$ and insert the identity on $B$, the set $g$ takes its inputs from, which changes nothing. Then replace that identity by $f \circ h$, equal to it because $h$ undoes $f$ from the right:

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

Now regroup the triple composition and finish:

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

The regrouping is the associative law, and it is the one link in the chain that uses it; delete associativity and the argument stops there with no route to $h$. The next link uses the other hypothesis, $g \circ f = \operatorname{id}_A$, and the last drops the identity on $A$, the set $h$ delivers its outputs to. Reading the chain end to end gives $g = h$: a rule undoing $f$ from the left and a rule undoing it from the right cannot be two different rules.

Name the two right undoers $h_1$ and $h_2$, with $h_1(b) \ne h_2(b)$ at some input $b$. Suppose the function also had a left undoer $g$. Part A compares one left undoer with one right undoer and says they coincide, and nothing stops running that comparison twice, once against each:

$g = h_1 \quad \text{and} \quad g = h_2$

Two rules equal to the same rule are equal to each other, so $h_1 = h_2$ and in particular $h_1(b) = h_2(b)$. That contradicts the assumption that they disagree at $b$. The only assumption available to withdraw is the one that was supposed: there is no left undoer.

So one-sided undoers can multiply, but only while the other side stays empty. The moment a function has an undoer on each side, part A collapses every one of them into a single rule, and that rule is what earns the name $f^{-1}$.

Audit the chain link by link and list what each one consumes: the identity on $B$ and on $A$, the hypothesis $f \circ h = \operatorname{id}_B$, the hypothesis $g \circ f = \operatorname{id}_A$, and the regrouping

$g \circ (f \circ h) = (g \circ f) \circ h$

No link inspects the rule $f$ follows, and no link asks whether two inputs of $f$ can share an output. That is exactly why the conclusion is available for every $f$ at once, and it is a conclusion about counting rather than about construction: there is at most one two-sided undoer, so the symbol $f^{-1}$ names something definite whenever it names anything at all.

What the chain cannot do is produce that rule. Nothing in it says an undoer exists, and a function with no undoer on either side satisfies the statement with nothing to check. Existence is a separate question, settled by whether the outputs of $f$ come from single inputs and whether $f$ reaches every value it has been declared to reach.

Let $a$ and $b$ be allowed inputs with $f(a) = f(b)$. Neither $a - 3$ nor $b - 3$ is zero, so multiplying both sides by both denominators is legal:

$(4a + 1)(b - 3) = (4b + 1)(a - 3)$

Expand each side:

$4ab - 12a + b - 3 = 4ab - 12b + a - 3$

Subtract $4ab - 3$ from both sides, leaving $-12a + b = -12b + a$. Gathering the letters gives $13b = 13a$, so $a = b$.

Equal outputs force equal inputs, which is exactly the definition of one-to-one.

Solve $f(x) = y$ for $x$ and watch which value of $y$ breaks the step:

$y(x - 3) = 4x + 1 \;\Longrightarrow\; x(y - 4) = 1 + 3y$

When $y \ne 4$ this gives $x = \dfrac{3y + 1}{y - 4}$, an allowed input. When $y = 4$ the equation reads $0 = 13$, which no $x$ satisfies, so $4$ is never an output of $f$.

Now suppose a rule $k$ from $\mathbb{R}$ back into the domain satisfied $f \circ k = \operatorname{id}$ on all of $\mathbb{R}$. At $y = 4$ that demands $f(k(4)) = 4$. But $f(k(4))$ is by construction an output of $f$, and $4$ is not one, so the demand cannot be met and no such $k$ exists. Being one-to-one, which part A settled, does not rescue this: the failure sits on the other composition of the definition.

Each composition in the definition asks for a different thing. One-to-one is what lets a rule send every output back to a single input, and onto is what gives every declared output something to be sent back from. Demanding both at once is demanding a bijection, and that is what a two-sided inverse needs when a codomain has been declared in advance.

Taking the codomain to be the set of values actually produced is not a different theorem; it is the declaration that makes onto true by construction, which leaves one-to-one as the only condition left to check. That is why the clean statement, invertible exactly when one-to-one, always travels with that hypothesis attached. For this $f$, part B showed the produced values are

$\mathbb{R} \setminus \{4\}$

Declared into that set, $f$ is one-to-one and onto, so it is invertible. The inputs of the inverse are exactly the outputs of $f$, so the domain of $f^{-1}$ is that same set, every real number except $4$.

A point $(a, b)$ lies on the graph of $f$ when $f(a) = b$, and it lies on the graph of $f^{-1}$ when $f^{-1}(a) = b$, which says the same thing as $f(b) = a$. So a shared point must satisfy both conditions at once:

$f(a) = b \quad \text{and} \quad f(b) = a$

That is the test: apply the rule to $a$, then apply it again to the result, and see whether $a$ comes back. Run all five inputs. From $0$: $f(0) = 3$ and $f(3) = 8$, not $0$. From $2$: $f(2) = 7$ and $f(7) = 2$, which is the input. From $3$: $f(3) = 8$ and $f(8) = 0$, not $3$. From $7$: $f(7) = 2$ and $f(2) = 7$, which is the input. From $8$: $f(8) = 0$ and $f(0) = 3$, not $8$.

Only $a = 2$ and $a = 7$ survive, giving the shared points $(2, 7)$ and $(7, 2)$. Neither has equal coordinates, so neither one sits on $y = x$.

Suppose $(a, b)$ lies on both graphs. Lying on the graph of $f$ gives $f(a) = b$, and lying on the graph of $f^{-1}$ gives $f(b) = a$. Now test the swapped point $(b, a)$ against the same two requirements. It lies on the graph of $f$ if $f(b) = a$, and on the graph of $f^{-1}$ if $f(a) = b$:

$f(b) = a \quad \text{and} \quad f(a) = b$

Those are the two facts already in hand, written in the other order, so $(b, a)$ lies on both graphs as well.

Swapping coordinates is exactly reflection across $y = x$, so the collection of shared points is mapped onto itself by that reflection: the collection is symmetric about the line. Being symmetric about a line is a much weaker condition than lying on it, and that gap is the whole story. The two points found in part A are a pair exchanged by the mirror, not two points resting on it.

Part A produced $(2, 7)$ and $(7, 2)$, each on both graphs and neither on $y = x$. One counterexample refutes a claim about every one-to-one function, and part B explains why these two arrived together rather than singly.

The statement that survives runs in the other direction. If $f(a) = a$, then $(a, a)$ lies on the graph of $f$; applying $f^{-1}$ to $f(a) = a$ gives $f^{-1}(a) = a$, so the same point lies on the graph of $f^{-1}$, and its equal coordinates put it on the mirror line:

$f(a) = a \;\Longrightarrow\; (a, a) \text{ on both graphs}$

Count the inputs of this $f$ satisfying $f(a) = a$: the outputs are $3, 7, 8, 2, 0$ against the inputs $0, 2, 3, 7, 8$, and no input is returned unchanged, so the count is zero. The surviving claim therefore promises nothing at all about this function and is in no danger from part A. The claim in the stem is its converse, and refuting a converse leaves the original standing.

For $x \ge 3$ the quantity $x - 3$ is nonnegative, so the bars come off unchanged:

$f(x) = (x - 3) + 2 = x - 1 \quad (x \ge 3)$

Solving $y = x - 1$ gives $x = y + 1$, so the inverse rule is $f^{-1}(x) = x + 1$. Its inputs are the outputs of the restricted $f$, and as $x$ runs over $x \ge 3$ the value $x - 1$ runs over $y \ge 2$, so the domain of the inverse is $x \ge 2$.

Check one composition on each side: $f^{-1}(f(x)) = (x - 1) + 1 = x$ for $x \ge 3$, and $f(f^{-1}(x)) = (x + 1) - 1 = x$ for $x \ge 2$.

For $x \le 3$ the quantity $x - 3$ is nonpositive, so the bars flip its sign:

$f(x) = (3 - x) + 2 = 5 - x \quad (x \le 3)$

Solving $y = 5 - x$ gives $x = 5 - y$, so this inverse is $f^{-1}(x) = 5 - x$. This restricted rule also produces exactly the outputs $y \ge 2$, so its inverse again has domain $x \ge 2$, and the values it returns run over $x \le 3$, matching the inputs it was built from.

Set the two inverses side by side. Both accept $x \ge 2$, and $x + 1 = 5 - x$ only at $x = 2$, so apart from that one input they disagree: at $x = 6$ one returns $7$ and the other returns $-1$.

Uniqueness is untouched, because the two inverses belong to two different functions. Equality of functions asks for the same domain as well as the same values, and the inputs $x \ge 3$ are not the inputs $x \le 3$, so the two restricted rules are not the same function. Each has exactly one inverse; no function here has two.

Take the two pieces separately, using the rules found in parts A and B.

On $3 \le x \le 5$ the rule is $x - 1$, which climbs from $2$ to $4$, so this piece produces exactly the outputs $2 \le y \le 4$ and repeats none of them. On $x < 1$ the rule is $5 - x$, which exceeds $4$ for every such input, so this piece produces exactly the outputs $y > 4$ and repeats none of them either:

$2 \le y \le 4 \quad \text{and} \quad y > 4$

The two blocks of outputs have no value in common, so no input from one piece can collide with an input from the other, and $f$ is one-to-one across the whole set. The excluded end is doing real work: $f(1) = 4$, and the first piece already produces $4$ at $x = 5$, so admitting $x = 1$ would hand one output two inputs and wreck the arrangement.

Together the blocks make up every $y \ge 2$, so this third restriction produces the same outputs as the other two and its inverse has the same domain, $x \ge 2$. The inverse itself is new: it sends $3$ to $4$ and $6$ to $-1$, where the inverse in part A sends $6$ to $7$ and the one in part B sends $3$ to $2$.

Three workable restrictions, and nothing in the rule prefers one of them. The phrase "the restriction that makes $f$ invertible" names nothing at all. An author chooses a restriction, and which one was chosen has to be reported alongside the inverse, because the inverse changes when the choice does.