To find the domain of the function $f(x) = \frac{x^2 + x}{x - 1}$, we need to determine the values of $x$ for which the function is defined.

Step-by-step solution:

1. Identify restrictions on the denominator:

   • The function is undefined wherever the denominator equals zero, since division by zero is not allowed.
   • Here, the denominator is $x - 1$.

2. Set the denominator equal to zero: $x - 1 = 0 \implies x = 1$

3. Determine the domain:

   • The domain includes all real numbers except $x = 1$.
   • Therefore, the domain is: $\text{Domain of } f(x): \, (-\infty, 1) \cup (1, \infty)$

The function is defined for all values of $x$ except where $x = 1$.

Would you like further details or have any other questions?

Related Questions:

1. What is the range of the function $f(x) = \frac{x^2 + x}{x - 1}$?
2. How do you find the vertical asymptote of the function $f(x) = \frac{x^2 + x}{x - 1}$?
3. What happens to $f(x)$ as $x$ approaches 1?
4. How can we simplify the function $f(x) = \frac{x^2 + x}{x - 1}$ if possible?
5. What is the behavior of $f(x)$ as $x \to \infty$?

Tip:

Always check for points where the denominator of a rational function becomes zero, as these points are excluded from the domain.