One-One Functions

A function is one-one (injective) when every output comes from exactly one input. $f(x) = 2x + 1$ is one-one; $f(x) = x^2$ is not, because 3 and $-3$ both map to 9. The concept matters for one reason the exam cares about: only one-one functions have inverses.

The horizontal line test

On a graph: $f$ is one-one if no horizontal line crosses it more than once. Increasing-only or decreasing-only functions pass automatically; anything with a turning point fails, unless the domain is cut to one side of it.

When 0606 asks “explain why $f$ has an inverse” (or doesn’t), the expected answer uses the words: ”$f$ is one-one” (or ”$f$ is not one-one, since $f(a) = f(b)$ for $a \ne b$”). A sketch with a horizontal line drawn through twice is acceptable support; the stated term is what the mark scheme wants.

Restricting the domain, the standard question

Any many-one function becomes one-one if you keep only a piece without a turning point. The exam’s favourite phrasing:

$f(x) = x^2 - 8x + 19$ for $x \ge k$. Find the smallest value of $k$ for which $f$ is one-one. Complete the square: $f(x) = (x - 4)^2 + 3$, vertex at $x = 4$. The parabola decreases before $x = 4$ and increases after, so one-one requires the domain to start at (or after) the vertex. Smallest $k = 4$.

The answer is always the turning point’s $x$-coordinate, and completing the square is the route to it. Show the completed square (M), state $k$ (A).

The follow-up: find the inverse on that domain

These parts chain: once $k = 4$, “find $f^{-1}$ and state its domain” runs the inverse routine with the positive root chosen because $x \ge 4$, the sign-reason mark again, and domain of $f^{-1} =$ range of $f = x \ge 3$.

Common mistakes

• “One-one” explained as “passes the vertical line test” (that tests whether it’s a function at all, wrong test)
• $k$ taken from the $y$-coordinate of the vertex instead of the $x$-coordinate
• The strictness fumbled: $x \ge 4$ works; so does $x > 4$, but the smallest $k$ is 4, achieved with $\ge$
• Forgetting that the chosen domain then drives the $\pm$ decision in the inverse

Full topic context: Functions notes · feeds directly into inverse functions and domain & range.