Math Problem Statement

Tìm tiệm cận của các đường cong sau: a) y = (1 + x^3)^(1/3) b) y = ln(1 + e^(-x)) c) y = (x^3 * arccot(x)) / (1 + x^2)

Solution

The task in the image is to find the asymptotes of the following functions:

a) y=1+x33y = \sqrt[3]{1 + x^3}

b) y=ln(1+ex)y = \ln(1 + e^{-x})

c) y=x3\arccot(x)1+x2y = \frac{x^3 \arccot(x)}{1 + x^2}

Let's analyze each one to find the asymptotes.


a) y=1+x33y = \sqrt[3]{1 + x^3}

  • As xx \to \infty: yx33=x.y \sim \sqrt[3]{x^3} = x. So, the function behaves like y=xy = x, implying there is a slant asymptote with equation y=xy = x as xx \to \infty.

  • As xx \to -\infty: yx33=x.y \sim \sqrt[3]{x^3} = x. Similarly, yxy \sim x, so the asymptote for xx \to -\infty is also y=xy = x.

Thus, the asymptote for large xx values is y=xy = x.


b) y=ln(1+ex)y = \ln(1 + e^{-x})

  • As xx \to \infty: ex0yln(1)=0.e^{-x} \to 0 \quad \Rightarrow \quad y \sim \ln(1) = 0. Therefore, there is a horizontal asymptote at y=0y = 0 as xx \to \infty.

  • As xx \to -\infty: exyln(ex)=x.e^{-x} \to \infty \quad \Rightarrow \quad y \sim \ln(e^{-x}) = -x. So the asymptote as xx \to -\infty is a slant asymptote with equation y=xy = -x.


c) y=x3\arccot(x)1+x2y = \frac{x^3 \arccot(x)}{1 + x^2}

  • As xx \to \infty: \arccot(x)1xyx31x1+x2=x2x2+11.\arccot(x) \sim \frac{1}{x} \quad \Rightarrow \quad y \sim \frac{x^3 \cdot \frac{1}{x}}{1 + x^2} = \frac{x^2}{x^2 + 1} \sim 1. Therefore, there is a horizontal asymptote at y=1y = 1 as xx \to \infty.

  • As xx \to -\infty: Since \arccot(x)π1x\arccot(x) \sim \pi - \frac{1}{x}, we get: yx3(π1x)1+x2πx3x2=πx.y \sim \frac{x^3 (\pi - \frac{1}{x})}{1 + x^2} \sim \frac{\pi x^3}{x^2} = \pi x. So as xx \to -\infty, there is a slant asymptote at y=πxy = \pi x.


Let me know if you'd like further details on any of the steps. Here are some follow-up questions:

  1. What are the general conditions for the existence of horizontal asymptotes?
  2. How would you approach finding vertical asymptotes for a rational function?
  3. Can a function have more than one horizontal asymptote?
  4. What is the significance of the behavior of \arccot(x)\arccot(x) as x±x \to \pm\infty?
  5. How do we derive slant asymptotes when the degree of the numerator exceeds that of the denominator?

Tip: Slant asymptotes occur when the degree of the numerator is exactly one more than the denominator in rational functions.

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Math Problem Analysis

Mathematical Concepts

Asymptotes
Limits
Logarithmic Functions
Inverse Trigonometric Functions

Formulas

y ~ (1 + x^3)^(1/3) for large x
y = ln(1 + e^(-x))
y ~ (x^3 * arccot(x)) / (1 + x^2)

Theorems

Asymptotes at infinity
Behavior of logarithmic and inverse trigonometric functions

Suitable Grade Level

Grade 12 and higher (Calculus level)