Given the functions:

• $f(x) = \sqrt{x}$
• $g(x) = -3x - 2$

We need to find both:

1. $(f \circ g)(x) = f(g(x))$
2. $(g \circ f)(x) = g(f(x))$

1. Finding $(f \circ g)(x) = f(g(x))$

Substitute $g(x)$ into $f(x)$:

$(f \circ g)(x) = f(g(x)) = f(-3x - 2)$ Now, apply $f(x) = \sqrt{x}$ to the expression:

$(f \circ g)(x) = \sqrt{-3x - 2}$

So, $(f \circ g)(x) = \sqrt{-3x - 2}$

2. Finding $(g \circ f)(x) = g(f(x))$

Substitute $f(x)$ into $g(x)$:

$(g \circ f)(x) = g(f(x)) = g(\sqrt{x})$ Now, apply $g(x) = -3x - 2$ to $\sqrt{x}$:

$(g \circ f)(x) = -3\sqrt{x} - 2$

Thus, $(g \circ f)(x) = -3\sqrt{x} - 2$

Final Answer:

• $(f \circ g)(x) = \sqrt{-3x - 2}$
• $(g \circ f)(x) = -3\sqrt{x} - 2$

Would you like more details or have any questions about these functions?

Here are five related questions to deepen understanding:

1. How do you compute the domain of $(f \circ g)(x)$?
2. What is the domain of $(g \circ f)(x)$?
3. Can the composition $(f \circ g)(x)$ ever be negative?
4. What is the range of $(f \circ g)(x)$ for specific values of $x$?
5. How does the behavior of these compositions change as $x \to \infty$?

Tip: Always verify the domain of composite functions to ensure that the operations are valid at each stage.