1. **Problem Statement:** We have a function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x + 1$. We need to check if $f$ is injective, surjective, or bijective. 2. **Definitions:** - A function is **injective** (one-to-one) if $f(a) = f(b)$ implies $a = b$. - A function is **surjective** (onto) if for every $y$ in the codomain, there exists an $x$ in the domain such that $f(x) = y$. - A function is **bijective** if it is both injective and surjective. 3. **Check Injectivity:** Assume $f(a) = f(b)$. $$a + 1 = b + 1$$ Subtract 1 from both sides: $$a = b$$ Since $a = b$ follows from $f(a) = f(b)$, $f$ is injective. 4. **Check Surjectivity:** For any $y \in \mathbb{R}$, solve $f(x) = y$: $$x + 1 = y$$ Subtract 1: $$x = y - 1$$ Since $x = y - 1$ is always a real number, for every $y$ there exists an $x$ such that $f(x) = y$. Thus, $f$ is surjective. 5. **Conclusion:** Since $f$ is both injective and surjective, it is bijective. **Final answer:** $f$ is injective, surjective, and therefore bijective.