Let's analyze the given function $f(x) = e^{-2/x}$:

(a) Vertical and Horizontal Asymptotes

1. Vertical Asymptotes:

   • The function $f(x) = e^{-2/x}$ is undefined at $x = 0$, so there is a vertical asymptote at $x = 0$.

   Answer: $x = 0$

2. Horizontal Asymptotes:

   • As $x \to \infty$, $\frac{-2}{x} \to 0$, and $e^0 = 1$. Hence, $y = 1$ is a horizontal asymptote.
   • Similarly, as $x \to -\infty$, $\frac{-2}{x} \to 0$, and $e^0 = 1$. Hence, $y = 1$ is also the horizontal asymptote.

   Answer: $y = 1$

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(b) Intervals of Increase and Decrease

To find the intervals of increase and decrease:

1. Compute the derivative of $f(x)$: $f'(x) = e^{-2/x} \cdot \frac{2}{x^2}$

2. Analyze $f'(x)$:

   • For $x > 0$, $f'(x) > 0$ (positive), so $f(x)$ is increasing on $(0, \infty)$.
   • For $x < 0$, $f'(x) > 0$ (positive), so $f(x)$ is also increasing on $(-\infty, 0)$.

   Intervals of Increase: $(-\infty, 0) \cup (0, \infty)$
   Intervals of Decrease: $\text{DNE (none)}$

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(c) Local Maximum and Minimum Values

• Since $f(x)$ is increasing everywhere in its domain, there are no local maximum or minimum values.

  Local Maximum Value: DNE
  Local Minimum Value: DNE

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(d) Concavity and Inflection Points

To find concavity and inflection points:

1. Compute the second derivative of $f(x)$: $f''(x) = e^{-2/x} \cdot \left(-\frac{4}{x^3}\right) + e^{-2/x} \cdot \frac{4}{x^4}$ Factorize: $f''(x) = e^{-2/x} \cdot \frac{4}{x^4} \cdot (-x + 1)$

2. Analyze $f''(x)$:

   • $f''(x) > 0$ (concave up) when $x > 1$ or $x < 0$.
   • $f''(x) < 0$ (concave down) when $0 < x < 1$.

   Intervals of Concave Up: $(-\infty, 0) \cup (1, \infty)$
   Intervals of Concave Down: $(0, 1)$

3. Inflection Point:

   • The concavity changes at $x = 1$. Substitute $x = 1$ into $f(x)$: $f(1) = e^{-2/1} = e^{-2}$
   • Inflection point: $(1, e^{-2})$.

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Final Answers:

(a)

• Vertical Asymptote(s): $x = 0$
• Horizontal Asymptote(s): $y = 1$

(b)

• Interval(s) of Increase: $(-\infty, 0) \cup (0, \infty)$
• Interval(s) of Decrease: DNE

(c)

• Local Maximum Value: DNE
• Local Minimum Value: DNE

(d)

• Interval(s) of Concave Up: $(-\infty, 0) \cup (1, \infty)$
• Interval(s) of Concave Down: $(0, 1)$
• Inflection Point: $(1, e^{-2})$

Let me know if you have questions or need further clarifications!

Here are some follow-up questions to explore:

1. What happens to $f(x)$ as $x$ approaches the vertical asymptote?
2. How would shifting the function horizontally or vertically change the asymptotes?
3. Can we generalize this analysis to $f(x) = e^{-a/x}$ for any constant $a$?
4. How does the concavity affect the overall graph shape?
5. What happens to the intervals of increase if the sign of $e^{-2/x}$ is flipped?

Tip: Always check for undefined points in the domain of a function to identify potential vertical asymptotes.