Chapter 1: Q. 24 (page 149)

Find the roots, discontinuous, and horizontal and vertical asymptotes of the function in Exercises 23-24. Support your answers by explicitly computing any relevant limits.
$f (x) = \frac{2 x^{2} - 1}{x^{2} - 2 x + 1}$

Short Answer

Expert verified

The root of $f (x)$ is localid="1649411404296" $\pm \frac{1}{\sqrt{2}}$ $f (x)$ discontinuous at $x = 1 .$ The horizontal asymptotes are at $y = 2 .$ The vertical asymptotes are at $x = 1 .$

Step by step solution

Step 1. Given information

The given function is $f (x) = \frac{2 x^{2} - 1}{x^{2} - 2 x + 1} .$ We need to determine the roots, discontinuous, horizontal, and vertical asymptotes of the function.

Step 2. Solution

$f (x) = \frac{2 x^{2} - 1}{x^{2} - 2 x + 1} .$

$= \frac{2 x^{2} - 1}{{(x - 1)}^{2}} .$

Put numerator equals zero

$2 x^{2} - 1 = 0 \Rightarrow x = \pm \frac{1}{\sqrt{2}}$

The root of $f (x)$ is localid="1649411436092" $\pm \frac{1}{\sqrt{2}}$ , $f (x)$ discontinuous at $x = 1 .$

By comparing the degree of the numerator and denominator the horizontal asymptotes are $y = 2 .$ The value of $x$ that cause the denominator of $f (x)$ to zero are $1 .$ So, the vertical asymptotes are at $x = 1 .$

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Find the roots, discontinuous, and horizontal and vertical asymptotes of the function in Exercises 23-24. Support your answers by explicitly computing any relevant limits.
$f (x) = \frac{2 x^{2} - 1}{x^{2} - 2 x + 1}$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Solution

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Find the roots, discontinuous, and horizontal and vertical asymptotes of the function in Exercises 23-24. Support your answers by explicitly computing any relevant limits.fx=2x2-1x2-2x+1

Short Answer

Step by step solution

Step 1. Given information

Step 2. Solution

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

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Study anywhere. Anytime. Across all devices.

Find the roots, discontinuous, and horizontal and vertical asymptotes of the function in Exercises 23-24. Support your answers by explicitly computing any relevant limits.
$f (x) = \frac{2 x^{2} - 1}{x^{2} - 2 x + 1}$