The problem given is:

Given $f(x) = \frac{1 - 5x}{x + 2}$, with $x \neq -2$, find the value of the inverse function $f^{-1}(-3)$.

Solution:

To find $f^{-1}(-3)$, we need to determine the inverse of $f(x)$ and then substitute $-3$ as the input of the inverse function.

1. Set up the function $f(x) = \frac{1 - 5x}{x + 2}$ and replace $f(x)$ with $y$: $y = \frac{1 - 5x}{x + 2}$

2. Solve for $x$ in terms of $y$: Multiply both sides by $x + 2$ to eliminate the denominator: $y(x + 2) = 1 - 5x$ Expand and rearrange to isolate $x$: $yx + 2y = 1 - 5x$ $yx + 5x = 1 - 2y$ Factor out $x$ on the left side: $x(y + 5) = 1 - 2y$ $x = \frac{1 - 2y}{y + 5}$

   Thus, the inverse function is: $f^{-1}(y) = \frac{1 - 2y}{y + 5}$

3. Evaluate $f^{-1}(-3)$ by substituting $y = -3$ into the inverse function: $f^{-1}(-3) = \frac{1 - 2(-3)}{-3 + 5}$ Simplify the expression: $f^{-1}(-3) = \frac{1 + 6}{2} = \frac{7}{2}$

Answer:

The value of $f^{-1}(-3)$ is: $\boxed{\frac{7}{2}}$

The correct answer is (e) $\frac{7}{2}$.

Would you like further clarification or details on any step?

Here are some related questions to expand understanding:

1. How do we find the inverse of a function in general?
2. What conditions must be met for a function to have an inverse?
3. How do we verify if our inverse function is correct?
4. What are some practical applications of inverse functions?
5. Why do we need to restrict $x \neq -2$ in the original function $f(x)$?

Tip: Always verify your inverse function by substituting it back into the original function to see if you get back the identity $f(f^{-1}(x)) = x$.