Chapter 4.3, Problem 34E

(a)

To determine

To find : The vertical and horizontal asymptotes.

Explanation of Solution

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

Consider the function.

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

The horizontal asymptote is calculated as,

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

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

The vertical asymptote is calculated as,

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

But,

  $\underset{x\to 2^{+}}{\lim }{x}^2=4$

And,

  $\underset{x\to 2^{+}}{\lim }{\left(x-2\right)}^2=0^{+}$

Thus,

  $\underset{x\to 2^{+}}{\lim }f\left(x\right)=\infty$

So, the vertical asymptotes is $x=2$ .

(b)

To determine

To find : The interval of increase or decrease.

Explanation of Solution

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

Consider the function.

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

Differentiate the above expression with respect to $x$ .

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

The function is decreasing in the interval $\left(-\infty ,0\right)$ .

The function is increasing in the interval $\left(0,2\right)$

The function is decreasing in the interval $\left(2,\infty \right)$

(c)

To determine

To find : The local maximum and minimum values..

Explanation of Solution

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

The function is decreasing in the interval $\left(-\infty ,0\right)$ .

The function is increasing in the interval $\left(0,2\right)$

The function is decreasing in the interval $\left(2,\infty \right)$

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

There is no local maxima.

(d)

To determine

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

Explanation of Solution

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

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

  $\begin{array}{c}{f}^{″}\left(x\right)=\frac{\left(x-2\right)\left(-4\right)-\left(-4x\right)\left(3\right)}{{\left(x-2\right)}^4}\\ =\frac{-4x+8+12x}{{\left(x-2\right)}^4}\\ =\frac{8\left(1+x\right)}{{\left(x-2\right)}^4}\end{array}$

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

The value of $f^{″}\left(c\right)$ is positive for $-1<x<2$ . So the concavity is upwars in the interval $\left(-1,1\right)$ .

The function has inflection point $\left(-1,\frac{1}{9}\right)$

(e)

To determine

To sketch : The graph of the function.

Explanation of Solution

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

Consider the function.

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

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

  

Figure (1)

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