One-to-One Functions and the Horizontal Line Test, precalculus lesson

Give every student a locker. If no two students share a locker, then a locker number tells you exactly whose locker it is. Functions face the same question. When two different inputs can produce the same output, the output alone cannot tell you where it came from, and the process cannot be run backward. One-to-one functions are the ones where every output leads back to a single input.

Key terms

One-to-one function

Definition A function $f$ with domain $D$ is one-to-one if no two different inputs in $D$ have the same output. Equivalently, each number in the range of $f$ is the output of exactly one input in $D$ . Whether a function is one-to-one depends on its domain as well as on its rule.

In plain words No output is used twice. If you know an output, you know the input it came from.

Example The function $f(x) = 2x + 1$ on $\mathbb{R}$ is one-to-one, because different numbers $x$ have different doubles and so different values $2x + 1$ . The function $g(x) = x^2$ on $\mathbb{R}$ is not, since $g(-3) = g(3) = 9$ . With the domain $[0, \infty)$ the same rule is one-to-one. If $0 \le a < b$ , then

b^2 - a^2 = (b - a)(b + a) > 0,

so two different nonnegative numbers always have different squares.

Injective

Definition Injective is the standard name for one-to-one in the language of sets. A function $f\colon A \to B$ is injective, or an injection, if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1$ and $x_2$ in $A$ . Being injective puts no condition on how much of the codomain $B$ is reached. An injective function need not reach every element of $B$ , and its range can be a proper subset of $B$ . A one-to-one function should not be confused with a one-to-one correspondence, which is a function that is injective and whose range is all of $B$ .

In plain words Injective and one-to-one mean the same thing. The word is about outputs never repeating, not about covering every possible output.

Example The function $s\colon [0, \infty) \to \mathbb{R}$ given by $s(x) = \sqrt{x}$ is injective. If $\sqrt{x_1} = \sqrt{x_2}$ , squaring both sides gives $x_1 = x_2$ . Its range is $[0, \infty)$ , so it never reaches $-1$ , and it is not a one-to-one correspondence between $[0, \infty)$ and $\mathbb{R}$ .

Distinct inputs give distinct outputs

Definition The defining condition can be written in two equivalent forms. For all $x_1$ and $x_2$ in the domain of $f$ ,

\begin{gathered} x_1 \ne x_2 \implies f(x_1) \ne f(x_2), \\[0.4em] \text{or equivalently} \\[0.4em] f(x_1) = f(x_2) \implies x_1 = x_2. \end{gathered}

The second form is the contrapositive of the first, so the two say the same thing, and it is the form used in proofs. Assume two outputs are equal and show by algebra that the inputs must be equal.

In plain words Different in means different out. To prove it, start from equal outputs and show there was only one input all along.

Example Let $f(x) = \frac{2x + 1}{x - 3}$ for $x \ne 3$ , and suppose $f(a) = f(b)$ with $a \ne 3$ and $b \ne 3$ . Multiplying both sides by the nonzero number $(a - 3)(b - 3)$ gives

(2a + 1)(b - 3) = (2b + 1)(a - 3).

Expanding gives

\begin{aligned} &2ab - 6a + b - 3 \\[0.4em] &\quad = 2ab - 6b + a - 3, \end{aligned}

and collecting terms leaves $-7a = -7b$ . So $a = b$ , and $f$ is one-to-one.

Showing a function is not one-to-one

Definition A function $f$ fails to be one-to-one exactly when there are inputs $x_1 \ne x_2$ in its domain with $f(x_1) = f(x_2)$ . A single such pair is a complete proof. The reverse does not work. Checking many inputs and finding no repeated output is evidence, not a proof that $f$ is one-to-one.

In plain words One repeated output is enough to say no. Saying yes needs an argument that covers every input.

Example For $f(x) = x^2 - 4x$ the values $f(1) = -3$ and $f(3) = -3$ already settle the question. To see every such pair, set $f(a) = f(b)$ and factor,

\begin{aligned} &a^2 - b^2 - 4(a - b) \\[0.4em] &\quad = (a - b)(a + b - 4) = 0. \end{aligned}

So $f(a) = f(b)$ with $a \ne b$ happens exactly when $a + b = 4$ , that is, for inputs placed symmetrically about $2$ .

Horizontal line test

Definition A function $f$ is one-to-one if and only if every horizontal line meets the graph of $f$ in at most one point. The reason is that the line $y = c$ meets the graph at the points $(x, c)$ with $f(x) = c$ , one point for each input whose output is $c$ . The test applies to the graph over the whole domain. A sketch or a calculator picture can suggest the answer but does not prove it.

In plain words Slide a horizontal line up and down. If it ever touches the graph twice, two inputs share an output. The vertical line test asks whether a graph belongs to a function at all, and the horizontal line test asks whether that function is one-to-one.

Example The line $y = 4$ meets the graph of $y = x^2$ at $(-2, 4)$ and $(2, 4)$ , so squaring on $\mathbb{R}$ fails the test. A picture can mislead. On the window $-2 \le x \le 2$ the graph of $f(x) = x^3 - 0.01x$ looks just like the graph of $x^3$ , which passes the test. But

f(x) = x(x - 0.1)(x + 0.1),

so the line $y = 0$ meets the graph at $x = -0.1$ , $x = 0$ and $x = 0.1$ , and $f$ is not one-to-one.

Strict monotonicity as a sufficient condition

Definition A function $f$ is strictly increasing on a set $S$ in its domain if $f(x_1) < f(x_2)$ whenever $x_1 < x_2$ in $S$ , and strictly decreasing on $S$ if $f(x_1) > f(x_2)$ whenever $x_1 < x_2$ in $S$ . These are the conditions that the lesson on reading graphs calls increasing and decreasing. If $f$ is strictly increasing on its whole domain, or strictly decreasing on its whole domain, then $f$ is one-to-one. Indeed, two different inputs can be named so that $x_1 < x_2$ , and then $f(x_1)$ and $f(x_2)$ are strictly ordered, so they differ. The inequalities must be strict. Allowing $f(x_1) \le f(x_2)$ would permit a flat piece, and a flat piece repeats outputs. The converse is false in general. For a function whose domain is an interval and whose graph has no breaks it does hold, which follows from the intermediate value theorem.

In plain words A graph that always climbs, or always falls, can never come back to a height it already reached. So it passes the horizontal line test automatically.

Example Let $f(x) = x^3 + x$ . Solving $y = x^3 + x$ for $x$ is hard, but if $x_1 < x_2$ then $x_1^3 < x_2^3$ , and adding $x_1 < x_2$ gives $f(x_1) < f(x_2)$ . So $f$ is strictly increasing on $\mathbb{R}$ and therefore one-to-one. The converse fails for $g(x) = \frac{1}{x}$ , whose domain has a gap at $0$ . It is one-to-one, since $\frac{1}{x_1} = \frac{1}{x_2}$ forces $x_1 = x_2$ . Yet it is neither increasing nor decreasing on its domain, because $g(-1) = -1 < 1 = g(1)$ while $g(1) = 1 > \frac{1}{2} = g(2)$ .

Check yourself

Is $f(x) = (x + 1)^2$ one-to-one on $\mathbb{R}$ ? Is it one-to-one on $[-1, \infty)$ ?

Show the answer

Not on $\mathbb{R}$ , since $f(-3) = 4 = f(1)$ . On $[-1, \infty)$ it is. If $-1 \le a < b$ then $0 \le a + 1 < b + 1$ , so $(a + 1)^2 < (b + 1)^2$ . The function is strictly increasing there and therefore one-to-one.

Prove that $g(x) = \frac{3x - 1}{x + 2}$ , with $x \ne -2$ , is one-to-one.

Show the answer

Suppose $g(a) = g(b)$ with $a, b \ne -2$ . Multiplying by the nonzero number $(a + 2)(b + 2)$ gives

(3a - 1)(b + 2) = (3b - 1)(a + 2),

that is

\begin{aligned} &3ab + 6a - b - 2 \\[0.4em] &\quad = 3ab + 6b - a - 2. \end{aligned}

This simplifies to $7a = 7b$ , so $a = b$ .

Does $h(x) = x^3 - 3x$ pass the horizontal line test?

Show the answer

No. The values $h(-1) = 2$ and $h(2) = 2$ are equal, so the line $y = 2$ meets the graph at $(-1, 2)$ and $(2, 2)$ . One repeated output is enough, and $h$ is not one-to-one.

One-to-One Functions and the Horizontal Line Test

Key terms

One-to-one function

Injective

Distinct inputs give distinct outputs

Showing a function is not one-to-one

Horizontal line test

Strict monotonicity as a sufficient condition

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus