Chapter 4.3, Problem 41E

(a)

To determine

To calculate: The right end behavior model.

Answer to Problem 41E

The right behavior model shows that as x approaches $\infty$, y approaches $\frac{\pi }{2}$.

Explanation of Solution

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

Calculation:

The graph of the function looks like this,

  

The right-hand behaviour model is,

  $\begin{array}{c}y=\underset{x\to \infty }{\lim }{\tan }^{-1}x\\ =\underset{\frac{1}{m}\to \infty }{\lim }{\tan }^{-1}\frac{1}{m}\\ =\underset{m\to 0}{\lim }{\tan }^{-1}\frac{1}{m}\\ ={\tan }^{-1}\frac{1}{0}\\ \tan y=\frac{1}{0}\\ \frac{\sin y}{\cos y}=\frac{1}{0}\\ \sin y=1\text{,}\cos y=0\\ y=\frac{\pi }{2}\end{array}$

Conclusion:

The right behavior model shows that as x approaches $\infty$, y approaches $\frac{\pi }{2}$.

(b)

To determine

To calculate: The left end behavior model.

Answer to Problem 41E

The right behavior model shows that as x approaches $-\infty$, y approaches $-\frac{\pi }{2}$.

Explanation of Solution

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

Calculation:

The graph of the function looks like this,

  

Theleft-hand behaviour model is,

  $\begin{array}{c}y=\underset{x\to -\infty }{\lim }{\tan }^{-1}x\\ =\underset{-\frac{1}{m}\to -\infty }{\lim }{\tan }^{-1}\left(-\frac{1}{m}\right)\\ =\underset{m\to 0}{\lim }{\tan }^{-1}\left(-\frac{1}{m}\right)\\ ={\tan }^{-1}\frac{-1}{0}\\ \tan y=\frac{-1}{0}\\ \frac{\sin y}{\cos y}=\frac{-1}{0}\\ \sin y=-1\text{,}\cos y=0\\ y=-\frac{\pi }{2}\end{array}$

Conclusion:

The right behavior model shows that as x approaches $-\infty$, y approaches $-\frac{\pi }{2}$.

(c)

To determine

To find: If there are any horizontal tangents.

Answer to Problem 41E

The function has no horizontal tangent.

Explanation of Solution

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

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,

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

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.