1. Problem 14: Find functions $f$ and $g$ such that $h(x) = (f \circ g)(x)$ where $h(x) = \sqrt{x + 4}$. Step 1: Recall that $(f \circ g)(x) = f(g(x))$. We want to express $\sqrt{x + 4}$ as $f(g(x))$. Step 2: Choose $g(x)$ to be the inside function and $f$ to be the outside function. Here, the inside function is $x + 4$, so let $g(x) = x + 4$. Step 3: The outside function applies the square root, so let $f(x) = \sqrt{x}$. Step 4: Verify: $(f \circ g)(x) = f(g(x)) = f(x + 4) = \sqrt{x + 4} = h(x)$. 2. Problem 15: Find functions $f$ and $g$ such that $h(x) = (g \circ f)(x)$ where $h(x) = \frac{1}{x + 2}$. Step 1: Recall $(g \circ f)(x) = g(f(x))$. We want $g(f(x)) = \frac{1}{x + 2}$. Step 2: Let $f(x)$ be the inside function. Choose $f(x) = x + 2$. Step 3: Then $g(x)$ must be the outside function that takes input $f(x)$ and outputs $\frac{1}{f(x)}$. So $g(x) = \frac{1}{x}$. Step 4: Verify: $(g \circ f)(x) = g(f(x)) = g(x + 2) = \frac{1}{x + 2} = h(x)$. 3. Problem 16: Find functions $f$ and $g$ such that $h(x) = (f \circ g)(x)$ where $h(x) = -3|x + 1| - 4$. Step 1: $(f \circ g)(x) = f(g(x))$. We want to write $-3|x + 1| - 4$ as $f(g(x))$. Step 2: Let $g(x) = |x + 1|$ (inside function). Step 3: Then $f(x) = -3x - 4$ (outside function). Step 4: Verify: $(f \circ g)(x) = f(g(x)) = f(|x + 1|) = -3|x + 1| - 4 = h(x)$. 4. Problem 17: Find functions $f$ and $g$ such that $h(x) = (g \circ f)(x)$ where $h(x) = \frac{1}{(x - 5)^2}$. Step 1: $(g \circ f)(x) = g(f(x))$. We want $g(f(x)) = \frac{1}{(x - 5)^2}$. Step 2: Let $f(x) = x - 5$ (inside function). Step 3: Then $g(x) = \frac{1}{x^2}$ (outside function). Step 4: Verify: $(g \circ f)(x) = g(f(x)) = g(x - 5) = \frac{1}{(x - 5)^2} = h(x)$.