Let's go through these questions step by step.

Question 1:
If $f(x) = \frac{1}{3x - 5}$, what is the inverse of $f$?

To find the inverse of a function, we follow these steps:

1. Replace $f(x)$ with $y$:
   $y = \frac{1}{3x - 5}$
2. Swap $x$ and $y$:
   $x = \frac{1}{3y - 5}$
3. Solve for $y$:
   $x(3y - 5) = 1$
   $3y - 5 = \frac{1}{x}$
   $3y = \frac{1}{x} + 5$
   $y = \frac{1}{3} \left( \frac{1}{x} + 5 \right)$

Thus, the inverse function is $f^{-1}(x) = \frac{1}{3} \left( \frac{1}{x} + 5 \right)$, which corresponds to option d.

Question 2:
For $f(x) = \sqrt{x + 5} + 2$, what is the domain of $f^{-1}$?

1. The original function has a domain where $x + 5 \geq 0$, i.e., $x \geq -5$.
2. The range of $f(x)$ will be $[2, \infty)$, because the square root term $\sqrt{x + 5}$ is non-negative and $+2$ shifts it up by 2.

Since the range of $f(x)$ becomes the domain of $f^{-1}(x)$, the domain of $f^{-1}(x)$ is [2, ∞), which corresponds to option d.

Question 3:
If $f(x) = 2x + 3$, and $g(x) = x^2 + x$, what is $f \circ g$?

1. $f \circ g$ means $f(g(x))$, or applying $f$ to $g(x)$.
2. $g(x) = x^2 + x$, so we substitute this into $f$:
   $f(g(x)) = f(x^2 + x) = 2(x^2 + x) + 3 = 2x^2 + 2x + 3$

Thus, $f \circ g(x) = 2x^2 + 2x + 3$, which corresponds to option d.

Question 4:
Let $f(x) = x^2 - 4$, where $x < 0$. What is $f^{-1}$?

1. To find the inverse, we start with $y = x^2 - 4$.
2. Swap $x$ and $y$:
   $x = y^2 - 4$
3. Solve for $y$:
   $y^2 = x + 4$
   $y = -\sqrt{x + 4}$, because $x < 0$, we choose the negative root.

Thus, the inverse is $f^{-1}(x) = -\sqrt{x + 4}$, which corresponds to option c.

Question 5:
If $f(x) = 2x + 3$, and $g(x) = x^2 + x$, what is $g \circ f$?

1. $g \circ f$ means $g(f(x))$, or applying $g$ to $f(x)$.
2. $f(x) = 2x + 3$, so we substitute this into $g$:
   $g(f(x)) = g(2x + 3) = (2x + 3)^2 + (2x + 3)$
3. Expanding the terms:
   $(2x + 3)^2 = 4x^2 + 12x + 9$
   $g(2x + 3) = 4x^2 + 12x + 9 + 2x + 3 = 4x^2 + 14x + 12$

Thus, $g \circ f(x) = 4x^2 + 14x + 12$, which corresponds to option a.

Do you want further details or have any questions?

5 Related Questions:

1. How do you find the inverse of a general linear function?
2. What is the difference between the domain and range of a function?
3. How do composition of functions work in different contexts?
4. Can the inverse of a function always be found, and what determines this?
5. What is the role of the range when determining the domain of an inverse?

Tip: Always check if the function is one-to-one before finding the inverse. Functions must be injective to have an inverse.