Math Problem Statement

Find all horizontal asymptotes of the following function. \[ f(x) = \begin{cases} \frac{3x - 1}{\sqrt{3x^2 + 2}}, & \text{if } x \leq 2 \\ \frac{7e^{6x} - 2e^{3x} + 2}{2e^{6x} + 3e^{3x} - 1}, & \text{if } x > 2 \end{cases} \]

Solution

We are given the piecewise function f(x)f(x) and are asked to find its horizontal asymptotes. The function is defined as:

\begin{cases} \frac{3x - 1}{\sqrt{3x^2 + 2}}, & \text{if } x \leq 2 \\ \frac{7e^{6x} - 2e^{3x} + 2}{2e^{6x} + 3e^{3x} - 1}, & \text{if } x > 2 \end{cases}$$ ### Step 1: Horizontal Asymptotes for $$x \leq 2$$ For the first piece, $$f(x) = \frac{3x - 1}{\sqrt{3x^2 + 2}}$$, we need to check the behavior of the function as $$x \to -\infty$$. - As $$x \to -\infty$$, the dominant term in both the numerator and denominator is the term involving $$x^2$$. - The numerator behaves like $$3x$$, and the denominator behaves like $$\sqrt{3x^2} = \sqrt{3} |x|$$. Thus, as $$x \to -\infty$$: $$f(x) \approx \frac{3x}{\sqrt{3} |x|} = \frac{3}{\sqrt{3}} = \sqrt{3}$$ Therefore, there is a horizontal asymptote at $$y = \sqrt{3}$$ as $$x \to -\infty$$. ### Step 2: Horizontal Asymptotes for $$x > 2$$ For the second piece, $$f(x) = \frac{7e^{6x} - 2e^{3x} + 2}{2e^{6x} + 3e^{3x} - 1}$$, we need to check the behavior as $$x \to \infty$$. - As $$x \to \infty$$, the exponential terms $$e^{6x}$$ dominate both the numerator and denominator. - In the numerator, the dominant term is $$7e^{6x}$$. - In the denominator, the dominant term is $$2e^{6x}$$. Thus, as $$x \to \infty$$: $$f(x) \approx \frac{7e^{6x}}{2e^{6x}} = \frac{7}{2}$$ Therefore, there is a horizontal asymptote at $$y = \frac{7}{2}$$ as $$x \to \infty$$. ### Final Answer The horizontal asymptotes of the function are: - $$y = \sqrt{3}$$ as $$x \to -\infty$$, - $$y = \frac{7}{2}$$ as $$x \to \infty$$. Would you like a more detailed explanation or have any other questions? Here are 5 related questions to explore this topic further: 1. What are vertical asymptotes, and how do they differ from horizontal asymptotes? 2. How can the dominant term approach be used in more complex rational functions? 3. What happens to the function behavior around $$x = 2$$, where the two pieces of the function connect? 4. How do you find oblique asymptotes, and when do they occur? 5. Can we find the limits of both function pieces as $$x \to 2$$ to check for continuity? **Tip:** When determining horizontal asymptotes, focus on the highest degree terms in the numerator and denominator, as they dominate the behavior at extreme values of $$x$$.

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Math Problem Analysis

Mathematical Concepts

Calculus
Horizontal Asymptotes
Piecewise Functions

Formulas

Horizontal Asymptote Limit: \(\lim_{x \to \pm\infty} f(x)\)

Theorems

Dominant Term Analysis

Suitable Grade Level

Undergraduate Calculus