We are given two functions:

1. $f(x) = \frac{5x - 1}{3x + 4}$
2. $g(x) = \frac{5}{2x}$

The problem asks us to find $(f \circ g)(x)$, which is the composition of the functions $f$ and $g$. This means we need to substitute $g(x)$ into $f(x)$.

Step-by-step solution:

To find $(f \circ g)(x)$, we substitute $g(x) = \frac{5}{2x}$ into $f(x)$:

$(f \circ g)(x) = f(g(x)) = f\left( \frac{5}{2x} \right)$

Now, substitute $\frac{5}{2x}$ for $x$ in the expression for $f(x)$:

$f\left( \frac{5}{2x} \right) = \frac{5\left(\frac{5}{2x}\right) - 1}{3\left(\frac{5}{2x}\right) + 4}$

Simplify both the numerator and denominator:

Numerator:

$5\left( \frac{5}{2x} \right) = \frac{25}{2x}$ $\frac{25}{2x} - 1 = \frac{25}{2x} - \frac{2x}{2x} = \frac{25 - 2x}{2x}$

Denominator:

$3\left( \frac{5}{2x} \right) = \frac{15}{2x}$ $\frac{15}{2x} + 4 = \frac{15}{2x} + \frac{8x}{2x} = \frac{15 + 8x}{2x}$

Thus, the composition $(f \circ g)(x)$ becomes:

$(f \circ g)(x) = \frac{\frac{25 - 2x}{2x}}{\frac{15 + 8x}{2x}}$

Since both the numerator and denominator are divided by $2x$, these terms cancel out:

$(f \circ g)(x) = \frac{25 - 2x}{15 + 8x}$

Final Answer:

$(f \circ g)(x) = \frac{25 - 2x}{15 + 8x}$

Would you like further explanation, or have any questions?

Here are 5 related questions to consider:

1. What is the domain of the function $(f \circ g)(x)$?
2. How would you compute $(g \circ f)(x)$ in this case?
3. How can you verify the result of $(f \circ g)(x)$ by plugging in specific values of $x$?
4. What happens to $(f \circ g)(x)$ as $x$ approaches 0?
5. How would you graph the composition $(f \circ g)(x)$?

Tip: When simplifying complex fractions, always look for common factors or terms that can cancel out to simplify the expression!