Let's solve each part step by step.

Given functions:

• $f(x) = \frac{7}{x + 5}$
• $g(x) = \frac{3}{x}$

Part a: Find $(f \circ g)(x)$

The notation $(f \circ g)(x)$ means the composition of $f$ and $g$, or $f(g(x))$. This means we substitute $g(x)$ into $f(x)$.

$(f \circ g)(x) = f(g(x)) = f\left(\frac{3}{x}\right)$

Now, substitute $\frac{3}{x}$ into the expression for $f(x)$:

$f\left(\frac{3}{x}\right) = \frac{7}{\frac{3}{x} + 5}$

To simplify, we combine the terms in the denominator:

$\frac{3}{x} + 5 = \frac{3 + 5x}{x}$

Now, the expression becomes:

$f\left(\frac{3}{x}\right) = \frac{7}{\frac{3 + 5x}{x}} = \frac{7x}{3 + 5x}$

Thus, the composition $(f \circ g)(x) = \frac{7x}{3 + 5x}$.

Part b: Find the domain of $f \circ g$

The domain of $f \circ g$ is determined by two things:

1. The domain of $g(x)$, which must exclude values that make the denominator zero.
2. The domain of $f(g(x))$, which must also exclude values that make the denominator zero after substituting $g(x)$ into $f(x)$.

Step 1: Domain of $g(x)$

The function $g(x) = \frac{3}{x}$ is undefined when $x = 0$, so $x \neq 0$.

Step 2: Domain of $f(g(x))$

We have $f(g(x)) = \frac{7x}{3 + 5x}$, and this is undefined when the denominator equals zero. So, solve:

$3 + 5x = 0 \quad \Rightarrow \quad x = -\frac{3}{5}$

Thus, $x = -\frac{3}{5}$ must also be excluded from the domain.

Conclusion for the domain:

The domain of $(f \circ g)(x)$ is all real numbers except $x = 0$ and $x = -\frac{3}{5}$. In interval notation, this is:

$\boxed{(-\infty, -\frac{3}{5}) \cup (-\frac{3}{5}, 0) \cup (0, \infty)}$

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Would you like more details on any step? Here are some related questions to deepen understanding:

1. How do you find the composition of two functions in general?
2. What is the domain of $f$ alone, and how does it compare to the domain of $g$?
3. How would the result change if $g(x)$ had a different denominator?
4. Can you explain how to simplify rational expressions in function compositions?
5. What happens if $g(x)$ is a more complex rational function?

Tip: When determining the domain of a composition, always check the domain restrictions of both the inside and outside functions.