One-to-One Functions

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Idea

A one-to-one, or injective, function never sends two distinct inputs to the same output. The horizontal-line test is its graphical version: each output height occurs at most once.

Visual guide

VISUAL GUIDENo two inputs share an output

One-to-one

Not one-to-one

The left rule is injective; each output has at most one incoming arrow. The right rule is not injective because inputs −1 and 1 both map to 1.

Method

f(a)=f(b)\Longrightarrow a=b.

Strictly increasing or strictly decreasing functions on an interval are injective there. Restricting the domain can make a noninjective rule injective.

Worked example

For $f(x)=5x+1$ , equality $5a+1=5b+1$ forces $a=b$ . For $g(x)=x^2$ on $\mathbb R$ , $g(-2)=g(2)$ , so it is not injective. On $[0,\infty)$ , $a^2=b^2$ and nonnegativity force $a=b$ , making the restriction injective.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

Is f(x)=x² one-to-one on ℝ?

Yes: squares are nonnegative.

No: f(−1)=f(1).

Yes: each input has one output.

Hint 1 · Find a starting point

One-to-one asks whether different inputs can share an output.

Hint 2 · Take the next step

Compare inputs −1 and 1.

Show the reasoning

Answer: No: f(−1)=f(1).

Two distinct inputs produce 1. Being a function does not imply being one-to-one.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Common mistake

A graph passing the vertical-line test can still fail the horizontal-line test. It is a function, but may not have an inverse function on the full domain.

Check your understanding

Is $|x|$ one-to-one on $(-\infty,0]$ ?

Show answer

Yes. On that domain it equals $-x$ , a strictly decreasing rule.

MAKE IT YOURS

Pause before the next idea.

Can you explain how the visualization connects to this lesson’s goal? If a step still feels uncertain, put this lesson on your review list and try the check again another day.

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