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 different inputs map to different outputs. - A function is **surjective** (onto) if every element in the codomain has a preimage in the domain. - A function is **bijective** if it is both injective and surjective. 3. **Check injectivity:** Assume $f(a) = f(b)$ for some $a,b \in \mathbb{R}$. $$a + 1 = b + 1$$ Subtract 1 from both sides: $$a = b$$ Since $a = b$ whenever $f(a) = f(b)$, the function is injective. 4. **Check surjectivity:** For any $y \in \mathbb{R}$, we want to find $x \in \mathbb{R}$ such that: $$f(x) = y$$ Substitute $f(x) = x + 1$: $$x + 1 = y$$ Solve for $x$: $$x = y - 1$$ Since $x = y - 1$ is always a real number for any real $y$, every $y$ in the codomain has a preimage. Thus, $f$ is surjective. 5. **Conclusion:** Since $f$ is both injective and surjective, it is bijective. **Final answer:** The function $f(x) = x + 1$ is injective, surjective, and therefore bijective.