1. **Problem 9:** For each function, find the greatest domain, range, and whether it is one-to-one or many-to-one. **a) Function:** $f(x) = 4x - 5$ - Domain: Since $f$ is a linear function, it is defined for all real numbers. $$\text{Domain} = \mathbb{R}$$ - Range: Linear functions with nonzero slope have range all real numbers. $$\text{Range} = \mathbb{R}$$ - One-to-one or many-to-one: Since the slope is 4 (nonzero), $f$ is one-to-one. **b) Function:** $g(x) = \sqrt{x}$ - Domain: The square root function is defined for $x \geq 0$. $$\text{Domain} = [0, \infty)$$ - Range: The output of $\sqrt{x}$ is also $\geq 0$. $$\text{Range} = [0, \infty)$$ - One-to-one or many-to-one: $g$ is one-to-one on its domain because it is strictly increasing. **c) Function:** $h(x) = \frac{1}{x^2}$ - Domain: $x^2$ is zero at $x=0$, so $h$ is undefined there. $$\text{Domain} = \mathbb{R} \setminus \{0\}$$ - Range: Since $x^2 > 0$ for $x \neq 0$, $1/x^2 > 0$. $$\text{Range} = (0, \infty)$$ - One-to-one or many-to-one: $h$ is many-to-one because $h(x) = h(-x)$. 2. **Problem 10:** Given $g(x) = x^2 + 10x + p$, express in vertex form and find range. **a) Express in vertex form:** - Complete the square: $$g(x) = x^2 + 10x + p = (x^2 + 10x + 25) + p - 25 = (x + 5)^2 + (p - 25)$$ **b) Range:** - Since $(x + 5)^2 \geq 0$, minimum value of $g(x)$ is at $x = -5$: $$g(-5) = (0)^2 + (p - 25) = p - 25$$ - Therefore, $$\text{Range} = [p - 25, \infty)$$