The image shows a set definition involving $A = \{ x \ | \ \frac{1}{x} \leq 1 \}$.

This set notation implies that we are interested in the values of $x$ such that the inequality $\frac{1}{x} \leq 1$ holds true. Let's solve this inequality step by step.

Solving $\frac{1}{x} \leq 1$:

1. First, subtract 1 from both sides to get: $\frac{1}{x} - 1 \leq 0$

2. Combine the terms on the left-hand side: $\frac{1 - x}{x} \leq 0$

3. Now, solve the inequality $\frac{1 - x}{x} \leq 0$. To determine the intervals where this expression is non-positive (i.e., less than or equal to 0), consider the critical points where the numerator or denominator is zero:

   • The numerator $1 - x = 0$ when $x = 1$.
   • The denominator $x = 0$ gives a vertical asymptote, meaning $x = 0$ is not included in the domain.

4. Analyze the sign of $\frac{1 - x}{x}$ across the intervals determined by these points:

   • For $x > 1$, both $1 - x$ and $x$ are negative, so $\frac{1 - x}{x} > 0$.
   • For $0 < x < 1$, $1 - x$ is positive and $x$ is positive, so $\frac{1 - x}{x} > 0$.
   • For $x < 0$, both $1 - x$ and $x$ have different signs, so $\frac{1 - x}{x} < 0$.

   Therefore, the inequality holds in the intervals: $(-\infty, 0) \cup [1, \infty)$

Conclusion:

The set $A$ includes values of $x$ in $(-\infty, 0) \cup [1, \infty)$. Therefore, $A$ represents these intervals.

Would you like further details on the solution, or do you have any questions?

Here are some related questions to consider:

1. How do we graph the function $\frac{1}{x}$?
2. What happens to the inequality if we change it to $\frac{1}{x} \geq 1$?
3. How can we solve inequalities involving more complex rational functions?
4. What is the significance of vertical asymptotes in rational inequalities?
5. How does set notation help in defining solution intervals for inequalities?

Tip: When solving rational inequalities, always check for discontinuities in the function (like dividing by zero), which affect the domain.