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The problem in the image asks for a complete investigation and plotting of the following three functions:

1. \( y = \frac{x^3}{x^2 - 9} \)
2. \( y = e^{\frac{1}{x}} \)
3. \( y = e^{\sqrt{2} \cos x} \)

To solve this problem, we need to analyze each function separately and then plot them on a single graph. Let's start by examining each function.

### **1. Function Analysis: \( y = \frac{x^3}{x^2 - 9} \)**

- **Domain:** The function is undefined when the denominator is zero, i.e., \( x^2 - 9 = 0 \). Solving this, we get \( x = \pm 3 \). So, the domain is \( x \in \mathbb{R} \setminus \{-3, 3\} \).
- **Asymptotes:**
  - **Vertical asymptotes**: At \( x = \pm 3 \).
  - **Horizontal asymptote**: As \( x \to \pm \infty \), \( y \approx x \), indicating no horizontal asymptote. Instead, there is an oblique asymptote.
  - **Oblique asymptote**: Divide \( x^3 \) by \( x^2 - 9 \). Using polynomial division: the oblique asymptote is \( y = x \).
- **Derivative (Critical points):** \( y' = \frac{(3x^2)(x^2 - 9) - x^3(2x)}{(x^2 - 9)^2} = \frac{x^2(3x^2 - 27 - 2x^2)}{(x^2 - 9)^2} = \frac{x^2(x^2 - 27)}{(x^2 - 9)^2} \).
  - Critical points: \( x = 0 \) and \( x = \pm \sqrt{27} \approx \pm 5.2 \).

### **2. Function Analysis: \( y = e^{\frac{1}{x}} \)**

- **Domain:** The function is defined for all \( x \neq 0 \), so the domain is \( x \in \mathbb{R} \setminus \{0\} \).
- **Asymptotes:**
  - **Vertical asymptote**: At \( x = 0 \).
  - **Horizontal asymptote**: None. As \( x \to \infty \), \( y \to 1 \), and as \( x \to -\infty \), \( y \to 1 \) as well.
- **Derivative (Critical points):** \( y' = -\frac{e^{\frac{1}{x}}}{x^2} \).
  - For \( x > 0 \), \( y' < 0 \) (decreasing function).
  - For \( x < 0 \), \( y' > 0 \) (increasing function).

### **3. Function Analysis: \( y = e^{\sqrt{2} \cos x} \)**

- **Domain:** The function is defined for all \( x \), \( \text{Domain} = \mathbb{R} \).
- **Range:** Since \( e^x \) is always positive, \( y > 0 \). The range of \( y = e^{\sqrt{2} \cos x} \) depends on the range of \( \cos x \), which is \([-1, 1]\). Hence, \( y \) varies from \( e^{-\sqrt{2}} \) to \( e^{\sqrt{2}} \).
- **Periodicity:** This function is periodic with period \( 2\pi \), same as \( \cos x \).
- **Derivative (Critical points):** \( y' = -\sqrt{2} \sin x \cdot e^{\sqrt{2} \cos x} \).
  - \( y' = 0 \) when \( \sin x = 0 \), i.e., \( x = n\pi \) (where \( n \) is an integer).

### **Graphing All Functions Together**

- Plot each function, marking the critical points, asymptotes, and behavior at infinity.
- Ensure the graph captures the periodic nature, increasing and decreasing intervals, and vertical asymptotes where applicable.

### Would you like more details on a specific part of the analysis or see the graphs for these functions?

### Here are some related questions you might find useful:

1. What are the steps to find the oblique asymptote for a rational function?
2. How does the derivative determine the increasing and decreasing behavior of a function?
3. What methods are used to find critical points of a function?
4. How do periodic functions behave differently from non-periodic functions?
5. What are the key characteristics to look for when sketching graphs of exponential functions?

### Tip: For periodic functions, always check for the range of values over one complete period to understand the repeating behavior.

Explore a detailed analysis and plotting of three complex functions: y = x^3 / (x^2 - 9), y = e^(1/x), and y = e^(sqrt(2) * cos(x)). Learn about their domains, asymptotes, derivatives, and graphing characteristics. Suitable for advanced high school students and beyond.

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