We are given two functions $f(x) = x + 1$ and $g(x) = x^2 - 1$. The task is to calculate the formula for $\frac{f}{g}(x)$.

Solution:

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

$\frac{f}{g}(x) = \frac{f(x)}{g(x)}$

Substitute $f(x)$ and $g(x)$ into the formula:

$\frac{f}{g}(x) = \frac{x + 1}{x^2 - 1}$

The denominator $x^2 - 1$ can be factored as:

$x^2 - 1 = (x - 1)(x + 1)$

So the expression becomes:

$\frac{f}{g}(x) = \frac{x + 1}{(x - 1)(x + 1)}$

Cancel out $x + 1$ (for $x \neq -1$):

$\frac{f}{g}(x) = \frac{1}{x - 1}, \quad \text{for } x \neq \pm 1$

Final Answer:

$\frac{f}{g}(x) = \frac{1}{x - 1}, \quad \text{with } x \neq \pm 1$

Would you like further clarification or detailed steps?

Related Questions:

1. How can we determine the domain of $\frac{f}{g}(x)$?
2. What happens if $x = -1$? Why must it be excluded?
3. Can you graph $\frac{f}{g}(x) = \frac{1}{x-1}$?
4. How would the result change if $f(x)$ or $g(x)$ were different?
5. What are the implications of a removable discontinuity at $x = -1$?

Tip:

When simplifying rational expressions, always check for restrictions in the domain due to denominators or undefined points.