Chapter 4.3, Problem 33E

(a)

To determine

To find : The vertical and horizontal asymptotes.

Explanation of Solution

Given: The function is $f\left(x\right)=\frac{x^2}{x^2-1}$

Consider the function.

  $f\left(x\right)=\frac{x^2}{x^2-1}$

The horizontal asymptote is calculated as,

  $\begin{array}{c}\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\frac{x^2}{x^2-1}\\ =\underset{x\to \infty }{\lim }\frac{x^2}{x^2\left(1-\frac{1}{x^2}\right)}\\ =\underset{x\to \infty }{\lim }\frac{1}{1-\frac{1}{x^2}}\\ =\frac{1}{1-0}\\ =1\end{array}$

So, the horizontal asymptote is $f\left(x\right)=1$ .

The vertical asymptote is calculated as,

  $\begin{array}{c}\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }\frac{x^2}{x^2-1}\\ =\underset{x\to 1^{+}}{\lim }\frac{x^2}{x^2\left(1-\frac{1}{x^2}\right)}\\ =\infty \end{array}$

And,

  $\begin{array}{c}\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{-}}{\lim }\frac{x^2}{x^2-1}\\ =\underset{x\to 1^{-}}{\lim }\frac{x^2}{x^2\left(1-\frac{1}{x^2}\right)}\\ =-\infty \end{array}$

So, the vertical asymptotes are $x=\pm 1$ .

(b)

To determine

To find : The interval of increase or decrease.

Explanation of Solution

Given: The function is $f\left(x\right)=\frac{x^2}{x^2-1}$

Consider the function.

  $f\left(x\right)=\frac{x^2}{x^2-1}$

Differentiate the above expression with respect to $x$ .

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{\left(x^2-1\right)\left(2x\right)-x^2\left(2x\right)}{{\left(x^2-1\right)}^2}\\ =-\frac{2x}{{\left(x^2-1\right)}^2}\end{array}$

Equate the above expression to zero.

  $-\frac{2x}{{\left(x^2-1\right)}^2}=0$

So,

  $x=0,\pm 1$

The function is increasing in the interval $\left(-\infty ,-1\right)$ .

The function is increasing in the interval $\left(-1,0\right)$

The function is decreasing in the interval $\left(0,1\right)$

The function is decreasing in the interval $\left(1,\infty \right)$ .

(c)

To determine

To find : The local maximum and minimum values..

Explanation of Solution

Given: The function is $f\left(x\right)=\frac{x^2}{x^2-1}$

The function is increasing in the interval $\left(-\infty ,-1\right)$ .

The function is increasing in the interval $\left(-1,0\right)$

The function is decreasing in the interval $\left(0,1\right)$

The function is decreasing in the interval $\left(1,\infty \right)$ .

So, the function is maximum at $x=0$ and the local maxima is $\left(0,f\left(0\right)\right)=\left(0,0\right)$ .

There is no local minima.

(d)

To determine

To find : The intervals of concavity and the inflection points.

Explanation of Solution

Given: The function is $f\left(x\right)=\frac{x^2}{x^2-1}$

Differentiate $f^{\prime }\left(x\right)$ with respect to $x$ .

  $\begin{array}{c}{f}^{″}\left(x\right)=-2\left[\frac{\left(x^2-1\right)\left(x^2-1-4x^2\right)}{{\left(x^2-1\right)}^4}\right]\\ =\frac{2\left(3x^2+1\right)}{{\left(x^2-1\right)}^3}\end{array}$

The value of $f^{″}\left(x\right)$ is .positive in for $x<-1$ So the concavity is upward in the interval $\left(-\infty ,-1\right)$ .

The value of $f^{″}\left(c\right)$ is negative for $-1<x<1$ . So the concavity is downward in the interval $\left(-1,1\right)$ .

The value of $f^{″}\left(x\right)$ is positive for $x>1$ . So the concavity is upward in the interval $\left(1,\infty \right)$ .

The function has no inflection point.

(e)

To determine

To sketch : The graph of the function.

Explanation of Solution

Given: The function is $f\left(x\right)=\frac{x^2}{x^2-1}$

Consider the function.

  $f\left(x\right)=\frac{x^2}{x^2-1}$

The graph of the above function is shown in figure below.

  

Figure (1)

Therefore, the graph of the function is shown in Figure (1).