To analyze the given function $f(x) = \frac{1}{x} - \frac{1}{x-2}$, let's determine the following:

1. Domain of the function:

The domain of $f(x)$ is all $x$ values except those that make the denominators $x$ or $x - 2$ equal to zero.

• For $\frac{1}{x}$, $x = 0$ is excluded.
• For $\frac{1}{x-2}$, $x = 2$ is excluded.

Domain: $x \in (-\infty, 0) \cup (0, 2) \cup (2, \infty)$.

2. Transition Points:

Transition points occur where the function is undefined or undergoes a major structural change. Here, the transition points are:

• $x = 0$: A discontinuity (undefined at $x = 0$).
• $x = 2$: Another discontinuity (undefined at $x = 2$).

Transition Points: $x = 0$ and $x = 2$.

3. Asymptotes:

Vertical Asymptotes:

Vertical asymptotes occur at the values of $x$ that make the denominator zero, leading to infinite behavior in the function:

• $x = 0$: The term $\frac{1}{x}$ dominates.
• $x = 2$: The term $\frac{1}{x-2}$ dominates.

Horizontal Asymptotes:

To find the horizontal asymptote, analyze the behavior of $f(x)$ as $x \to \pm\infty$: $f(x) = \frac{1}{x} - \frac{1}{x-2}.$ For large $x$, both terms approach zero. Therefore, the horizontal asymptote is: $y = 0.$

Asymptotes:

• Vertical asymptotes: $x = 0$, $x = 2$.
• Horizontal asymptote: $y = 0$.

Would you like further clarification or a graph of this function?

Related Questions:

1. How does the behavior of $f(x)$ change near the vertical asymptotes?
2. What happens to $f(x)$ as $x \to 0^-$ and $x \to 0^+$?
3. How can $f(x)$ be rewritten to simplify analysis?
4. Can $f(x)$ have oblique asymptotes instead of horizontal ones?
5. What is the symmetry of this function?

Tip:

To verify asymptotes and transitions visually, graph the function and examine it near critical points.