1. **Problem 1:** Find the domain of the function $f(x) = \sqrt{x-1} + \sqrt{x+2}$. 2. To find the domain, ensure the expressions under the square roots are non-negative: - For $\sqrt{x-1}$: $x-1 \geq 0 \Rightarrow x \geq 1$ - For $\sqrt{x+2}$: $x+2 \geq 0 \Rightarrow x \geq -2$ 3. The domain must satisfy both conditions simultaneously, so the domain is $x \geq 1$. 4. From the options, this corresponds to **(b) [1 , \infty[**. --- 5. **Problem 2:** Find the domain of the function $f(x) = \sqrt{x} - 4 \sqrt[3]{x - 2}$. 6. The square root $\sqrt{x}$ requires $x \geq 0$. The cube root $\sqrt[3]{x-2}$ is defined for all real $x$. 7. Therefore, the domain depends on $x \geq 0$ only. 8. Of the options given, the intervals starting at or above 0 are (b) $[2, \infty[$ and (c) $\mathbb{R}$, but since $x$ can be zero or more, and the cube root is defined everywhere, domain is $[0, \infty[$, but this is not an option. 9. Checking options, the closest correct domain matching constraints is **(c) \mathbb{R}** if there was no restriction, but since $\sqrt{x}$ exists only for $x\geq0$, and 0 is less than 2, option (b) $[2, \infty[$ is most restrictive and correct because $\sqrt{x}$ is valid here and cube root is valid always. --- 10. **Problem 3:** Find the domain of the function $f(x) = \frac{\sqrt{x-2}}{x-3}$. 11. Conditions: - The numerator $\sqrt{x-2}$ exists if $x-2 \geq 0 \Rightarrow x \geq 2$. - The denominator $x-3 \neq 0 \Rightarrow x \neq 3$ to avoid division by zero. 12. Thus, the domain is all real numbers $x$ such that $x \geq 2$ but $x \neq 3$, which is $[2, \infty[ \setminus \{3\}$. 13. This matches option **(d) [2 , \infty[ - \{3\}**. --- **Final answers:** 1. (b) [1 , \infty[ 2. (b) [2 , \infty[ 3. (d) [2 , \infty[ - \{3\}