To solve the problem, we need to find the formula for the composite function $\frac{f}{g}(x)$, where $f(x) = x + 1$ and $g(x) = x^2 - 1$. Here are the steps:

Step 1: Understand the function

The division of two functions $\frac{f}{g}(x)$ is defined as: $\frac{f}{g}(x) = \frac{f(x)}{g(x)}$

Step 2: Substitute $f(x)$ and $g(x)$

Given: $f(x) = x + 1, \quad g(x) = x^2 - 1$ Substitute these into the formula: $\frac{f}{g}(x) = \frac{x + 1}{x^2 - 1}$

Step 3: Simplify the denominator

Note that $x^2 - 1$ can be factored as a difference of squares: $x^2 - 1 = (x - 1)(x + 1)$ Thus: $\frac{f}{g}(x) = \frac{x + 1}{(x - 1)(x + 1)}$

Step 4: Simplify the fraction

Cancel $x + 1$ in the numerator and denominator, provided $x \neq -1$ (to avoid division by zero): $\frac{f}{g}(x) = \frac{1}{x - 1}, \quad x \neq -1$

Final Answer:

The formula for $\frac{f}{g}(x)$ is: $\frac{f}{g}(x) = \frac{1}{x - 1}, \quad x \neq -1$

Would you like a detailed breakdown of this explanation or have any questions?

Here are 5 related questions:

1. How do you handle undefined points in a function, such as $x = -1$ in this case?
2. What is the domain of $\frac{f}{g}(x)$ given this solution?
3. How would this change if $f(x) = x^2 + 2x + 1$?
4. What happens if $g(x)$ has a repeated factor in its denominator?
5. How can this concept be extended to division of more complex functions?

Tip: Always check for restrictions in the domain, especially where the denominator equals zero.