Math Problem Statement

Given that 𝑓(π‘₯) = 1/π‘₯Β²βˆ’1 , sketch the graph of 𝑦 = 𝑓(π‘₯).

Solution

To sketch the graph of the function y=f(x)=1x2βˆ’1y = f(x) = \frac{1}{x^2 - 1}, let's analyze its key features step by step:

1. Domain

The function is undefined where the denominator is zero, so let's find those points:

x2βˆ’1=0β€…β€ŠβŸΉβ€…β€Šx=Β±1x^2 - 1 = 0 \implies x = \pm 1

Thus, the domain of the function is all real numbers except x=Β±1x = \pm 1.

2. Asymptotes

  • Vertical Asymptotes: Occur where the function is undefined:

    • x=1x = 1
    • x=βˆ’1x = -1
  • Horizontal Asymptote: As xβ†’βˆžx \to \infty or xβ†’βˆ’βˆžx \to -\infty, the function approaches zero because the numerator is constant and the denominator grows without bound:

    • Horizontal asymptote: y=0y = 0.

3. Intercepts

  • x-intercepts: The function has no x-intercepts because the numerator is always 1, never zero.

  • y-intercept:

    • Set x=0x = 0:

    y=f(0)=102βˆ’1=βˆ’1y = f(0) = \frac{1}{0^2 - 1} = -1

    So, the y-intercept is (0,βˆ’1)(0, -1).

4. Behavior Near Asymptotes

  • As xβ†’1+x \to 1^+ or xβ†’1βˆ’x \to 1^-, f(x)β†’βˆžf(x) \to \infty.
  • As xβ†’βˆ’1+x \to -1^+ or xβ†’βˆ’1βˆ’x \to -1^-, f(x)β†’βˆžf(x) \to \infty.

5. Sign of the Function

  • For x>1x > 1, x2βˆ’1>0x^2 - 1 > 0, so f(x)>0f(x) > 0.
  • For βˆ’1<x<1-1 < x < 1, x2βˆ’1<0x^2 - 1 < 0, so f(x)<0f(x) < 0.
  • For x<βˆ’1x < -1, x2βˆ’1>0x^2 - 1 > 0, so f(x)>0f(x) > 0.

6. Sketch of the Graph

  • The graph has two vertical asymptotes at x=1x = 1 and x=βˆ’1x = -1.
  • There is a horizontal asymptote at y=0y = 0.
  • The graph passes through the point (0,βˆ’1)(0, -1).

The graph will have:

  • Two branches in the regions x<βˆ’1x < -1 and x>1x > 1 approaching the asymptotes.
  • A section between βˆ’1<x<1-1 < x < 1 where the function is negative and approaches negative infinity near the vertical asymptotes.

Summary of the Graph

  • Vertical asymptotes at x=1x = 1 and x=βˆ’1x = -1.
  • Horizontal asymptote at y=0y = 0.
  • Decreasing and approaching infinity near vertical asymptotes.
  • The function is negative between βˆ’1-1 and 11.

Would you like a graphical representation of this description, or do you have any specific questions?


Related Questions:

  1. How do we determine asymptotes for more complex rational functions?
  2. What changes in the graph if f(x)=1x2+1f(x) = \frac{1}{x^2 + 1}?
  3. How can we find points of inflection for rational functions?
  4. What effect does a shift f(x)=1x2βˆ’1+2f(x) = \frac{1}{x^2 - 1} + 2 have on the graph?
  5. How does multiplying the function by a constant affect its graph?

Tip: When dealing with rational functions, always identify the vertical and horizontal asymptotes first as they help in sketching the general shape of the graph.

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Math Problem Analysis

Mathematical Concepts

Rational functions
Asymptotes
Function behavior

Formulas

Formula for rational functions

Theorems

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Suitable Grade Level

High School