Let $f:\mathbb{R}\to \mathbb{R}$ be defined by $f(x)=\{\begin{array}{ll}x+1,& x\le 0\\ 2x-3,& x>0\end{array}.$ Determine whether $f$ is one‑to‑one on $\mathbb{R}$. Justify your answer.

Step-by-step solution

1. Identify the two pieces: for $x\le 0$, $f(x)=x+1$ (strictly increasing); for $x>0$, $f(x)=2x-3$ (also strictly increasing).
2. Check if the ranges of the two pieces overlap.
3. For $x\le 0$, the range is $(-\infty ,1]$ because the largest value occurs at $x=0$, giving $f(0)=1$.
4. For $x>0$, the range is $(-3,\infty )$ because as $x\to 0^{+}$, $f(x)\to -3$ and it increases without bound.
5. The intervals $(-\infty ,1]$ and $(-3,\infty )$ intersect; e.g., $-1\in (-3,1]$.
6. Find two distinct arguments giving the same function value: solve $x+1=2y-3$ with $x\le 0$, $y>0$.
7. Taking $y=1$ gives $x=-2$, and indeed $f(-2)=-1=f(1)$. Hence two different inputs have the same output.
8. Therefore $f$ fails the injectivity test and is not one‑to‑one.

Concept tested: Injectivity of piecewise defined functions

Published 9 September 2026 · How this solution was prepared and checked