For which real numbers $a$ is the function $f(x)=x^2+ax$ one‑to‑one on $mathbbR$? Justify your answer.

Step-by-step solution

1. 1. Observe $f(x)$ is a quadratic polynomial, which opens upwards since the coefficient of $x^2$ is positive.
2. 2. A quadratic function is symmetric about its vertex at $x=-\frac{a}{2}$.
3. 3. For $x<-\frac{a}{2}$ the function decreases, and for $x>-\frac{a}{2}$ it increases, producing the same $y$–values for two different $x$–values unless the domain is restricted.
4. 4. Thus for any real $a$, there exist $x_1\ne x_2$ such that $f(x_1)=f(x_2)$, e.g., take $x_1=-\frac{a}{2}+\delta$ and $x_2=-\frac{a}{2}-\delta$ for any $\delta \ne 0$.
5. 5. Therefore $f$ cannot be one‑to‑one on the entire real line for any real $a$.

Concept tested: Injectivity of quadratic functions

Published 6 September 2026 · How this solution was prepared and checked