Math Problem Statement

Реши задачу с картинки и построй единый график

Solution

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

  1. y=x3x29y = \frac{x^3}{x^2 - 9}
  2. y=e1xy = e^{\frac{1}{x}}
  3. y=e2cosxy = 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=x3x29y = \frac{x^3}{x^2 - 9}

  • Domain: The function is undefined when the denominator is zero, i.e., x29=0x^2 - 9 = 0. Solving this, we get x=±3x = \pm 3. So, the domain is xR{3,3}x \in \mathbb{R} \setminus \{-3, 3\}.
  • Asymptotes:
    • Vertical asymptotes: At x=±3x = \pm 3.
    • Horizontal asymptote: As x±x \to \pm \infty, yxy \approx x, indicating no horizontal asymptote. Instead, there is an oblique asymptote.
    • Oblique asymptote: Divide x3x^3 by x29x^2 - 9. Using polynomial division: the oblique asymptote is y=xy = x.
  • Derivative (Critical points): y=(3x2)(x29)x3(2x)(x29)2=x2(3x2272x2)(x29)2=x2(x227)(x29)2y' = \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=0x = 0 and x=±27±5.2x = \pm \sqrt{27} \approx \pm 5.2.

2. Function Analysis: y=e1xy = e^{\frac{1}{x}}

  • Domain: The function is defined for all x0x \neq 0, so the domain is xR{0}x \in \mathbb{R} \setminus \{0\}.
  • Asymptotes:
    • Vertical asymptote: At x=0x = 0.
    • Horizontal asymptote: None. As xx \to \infty, y1y \to 1, and as xx \to -\infty, y1y \to 1 as well.
  • Derivative (Critical points): y=e1xx2y' = -\frac{e^{\frac{1}{x}}}{x^2}.
    • For x>0x > 0, y<0y' < 0 (decreasing function).
    • For x<0x < 0, y>0y' > 0 (increasing function).

3. Function Analysis: y=e2cosxy = e^{\sqrt{2} \cos x}

  • Domain: The function is defined for all xx, Domain=R\text{Domain} = \mathbb{R}.
  • Range: Since exe^x is always positive, y>0y > 0. The range of y=e2cosxy = e^{\sqrt{2} \cos x} depends on the range of cosx\cos x, which is [1,1][-1, 1]. Hence, yy varies from e2e^{-\sqrt{2}} to e2e^{\sqrt{2}}.
  • Periodicity: This function is periodic with period 2π2\pi, same as cosx\cos x.
  • Derivative (Critical points): y=2sinxe2cosxy' = -\sqrt{2} \sin x \cdot e^{\sqrt{2} \cos x}.
    • y=0y' = 0 when sinx=0\sin x = 0, i.e., x=nπx = n\pi (where nn 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.

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Math Problem Analysis

Mathematical Concepts

Functions
Exponential Functions
Trigonometric Functions
Graphing Functions
Asymptotes
Derivatives

Formulas

Exponential function formula: y = e^x
Derivative formulas for rational and exponential functions

Theorems

Properties of exponential functions
Behavior of trigonometric functions

Suitable Grade Level

Advanced High School