We are asked to find $f \circ g (x)$ and $g \circ f (x)$ for the given functions:

1. $f(x) = \frac{x + 7}{2}$
2. $g(x) = \sqrt{3 - x}$

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

The notation $f \circ g(x)$ means $f(g(x))$, which indicates we need to substitute $g(x)$ into $f(x)$.

$f(g(x)) = f\left( \sqrt{3 - x} \right)$ Substitute $\sqrt{3 - x}$ into $f(x) = \frac{x + 7}{2}$:

$f\left( \sqrt{3 - x} \right) = \frac{\sqrt{3 - x} + 7}{2}$

Thus: $f \circ g (x) = \frac{\sqrt{3 - x} + 7}{2}$

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

The notation $g \circ f(x)$ means $g(f(x))$, so we need to substitute $f(x)$ into $g(x)$.

$g(f(x)) = g\left( \frac{x + 7}{2} \right)$ Substitute $\frac{x + 7}{2}$ into $g(x) = \sqrt{3 - x}$:

$g\left( \frac{x + 7}{2} \right) = \sqrt{3 - \frac{x + 7}{2}}$

Simplify inside the square root:

$g\left( \frac{x + 7}{2} \right) = \sqrt{\frac{6 - (x + 7)}{2}} = \sqrt{\frac{-x - 1}{2}}$

Since this expression involves a negative value under the square root for certain values of $x$, this composition is not defined for all $x$. It is only defined when $-x - 1 \geq 0$, meaning $x \leq -1$.

Thus: $g \circ f (x) = \sqrt{\frac{-x - 1}{2}}, \quad \text{for } x \leq -1$

Summary:

• $f \circ g (x) = \frac{\sqrt{3 - x} + 7}{2}$
• $g \circ f (x) = \sqrt{\frac{-x - 1}{2}}, \text{ for } x \leq -1$

Do you have any questions, or would you like more details?

Here are 5 related questions for further exploration:

1. What is the domain of $f \circ g (x)$?
2. What is the domain of $g \circ f (x)$?
3. Can the function $f \circ g (x)$ have any discontinuities?
4. How would you graph $f \circ g (x)$?
5. Can you find the inverse of $f(x)$ or $g(x)$?

Tip: When combining functions, always check the domain of the resulting composition to avoid undefined expressions.