Q. 24 - Calculus

Question

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. 

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

Step-by-Step Solution

Answer

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

Step 1. Given information

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

Step 2. Solution

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

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

Put numerator equals zero 

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

The root of $f\left(x\right)$ is $\pm \frac{1}{\sqrt{2}}$ ,$f\left(x\right)$ 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 $f\left(x\right)$ to zero are $1.$ So, the vertical asymptotes are at $x=1.$