Chapter 4.3, Problem 42E

(a)

To determine

The right end behavior model.

Answer to Problem 42E

The right behavior model shows that as x approaches $\infty$ , y approaches 0.

Explanation of Solution

Given Information: The function is $y={\cot }^{-1}\left(x\right)$ .

Calculation:

The graph of the function looks like this,

  

The right-hand behaviour model is,

  $\begin{array}{c}y=\underset{x\to \infty }{\lim }{\cot }^{-1}x\\ =\underset{\frac{1}{m}\to \infty }{\lim }{\cot }^{-1}\frac{1}{m}\\ =\underset{m\to 0}{\lim }{\cot }^{-1}\frac{1}{m}\\ ={\cot }^{-1}\frac{1}{0}\\ \cot y=\frac{1}{0}\\ \frac{\cos y}{\sin y}=\frac{1}{0}\\ \sin y=0\text{,}\cos y=1\\ y=0\end{array}$

Conclusion:

The right behavior model shows that as x approaches $\infty$ , y approaches 0.

(b)

To determine

To calculate: The left end behavior model.

Answer to Problem 42E

The right behavior model shows that as x approaches $-\infty$ , y approaches $\pi$ .

Explanation of Solution

Given Information: The function is $y={\cot }^{-1}\left(x\right)$ .

Calculation:

The graph of the function looks like this,

  

The left-hand behaviour model is,

  $\begin{array}{c}y=\underset{x\to -\infty }{\lim }{\cot }^{-1}x\\ =\underset{-\frac{1}{m}\to -\infty }{\lim }{\cot }^{-1}\left(-\frac{1}{m}\right)\\ =\underset{m\to 0}{\lim }{\cot }^{-1}\left(-\frac{1}{m}\right)\\ ={\tan }^{-1}\frac{-1}{0}\\ \cot y=\frac{-1}{0}\\ \frac{\cos y}{\sin y}=\frac{-1}{0}\\ \sin y=0\text{,}\cos y=-1\\ y=\pi \end{array}$

Conclusion:

The right behavior model shows that as x approaches $-\infty$ , y approaches $\pi$ .

(c)

To determine

To find: Any horizontal tangents.

Answer to Problem 42E

The function has no horizontal tangent.

Explanation of Solution

Given Information: The function is $y={\cot }^{-1}\left(x\right)$ .

Calculation:

The graph of the function looks like this,

  

The horizontal tangent is obtained by equating derivative of the function to 0.

The derivative of the function is,

  $\begin{array}{c}y={\cot }^{-1}\left(x\right)\\ \frac{dy}{dx}=-\frac{1}{1+x^2}\end{array}$

Equate the derivative to 0,

  $\begin{array}{c}\frac{dy}{dx}=0\\ -\frac{1}{1+x^2}=0\\ -1=0\end{array}$

The above condition is not possible for any value of x, hence the function has no horizontal tangent.

Conclusion:

The function has no horizontal tangent.