1. **Problem Statement:** Determine which of the given functions are one-to-one functions. 2. **Recall:** A function is one-to-one (injective) if each output corresponds to exactly one input. Graphically, it passes the Horizontal Line Test. 3. **Functions:** - $f(x) = \sqrt{x - 3}$ - $f(x) = \sqrt{-8 + 2x}$ - $f(x) = \sqrt{20 - x}$ - $f(x) = \sqrt{37 - 3x} - 4$ 4. **Analyze each function:** **(a) $f(x) = \sqrt{x - 3}$** - Domain: $x \geq 3$ - The square root function is increasing on its domain. - Since it is strictly increasing, it is one-to-one. **(b) $f(x) = \sqrt{-8 + 2x}$** - Domain: $-8 + 2x \geq 0 \Rightarrow x \geq 4$ - Inside the square root is a linear function with positive slope 2. - The square root of an increasing linear function is increasing. - Hence, $f$ is one-to-one. **(c) $f(x) = \sqrt{20 - x}$** - Domain: $20 - x \geq 0 \Rightarrow x \leq 20$ - Inside the root is a decreasing linear function. - The square root of a decreasing function is decreasing. - Since it is strictly decreasing, it is one-to-one. **(d) $f(x) = \sqrt{37 - 3x} - 4$** - Domain: $37 - 3x \geq 0 \Rightarrow x \leq \frac{37}{3}$ - Inside the root is a decreasing linear function. - Square root of decreasing function is decreasing. - Subtracting 4 shifts the graph down but does not affect one-to-one property. - So, $f$ is one-to-one. 5. **Summary:** All four functions are one-to-one because they are either strictly increasing or strictly decreasing on their domains. **Final answer:** All given functions are one-to-one.