1. **Problem statement:** Determine if the function $f(x) = \sqrt{x} + x$ is injective. 2. **Definition:** A function is injective (one-to-one) if for every $x_1, x_2$ in the domain, $f(x_1) = f(x_2)$ implies $x_1 = x_2$. 3. **Domain:** Since $\sqrt{x}$ is defined for $x \geq 0$, the domain is $[0, \infty)$. 4. **Check injectivity:** Assume $f(x_1) = f(x_2)$, that is: $$\sqrt{x_1} + x_1 = \sqrt{x_2} + x_2$$ 5. Rearranging: $$\sqrt{x_1} - \sqrt{x_2} = x_2 - x_1$$ 6. Note that $\sqrt{x}$ is an increasing function and $x$ is also increasing. The sum of two increasing functions is increasing. 7. To confirm, check the derivative: $$f'(x) = \frac{1}{2\sqrt{x}} + 1$$ 8. For $x > 0$, $\frac{1}{2\sqrt{x}} > 0$, so $f'(x) > 0$. At $x=0$, the derivative tends to infinity, so $f$ is strictly increasing on $[0, \infty)$. 9. Since $f$ is strictly increasing, it is injective. **Final answer:** The function $f(x) = \sqrt{x} + x$ is injective on its domain $[0, \infty)$.