Chapter 3.1, Problem 65E

Solution

a.

To graph: The given function and to analyze it.

  

The domain is $\left(-\infty ,\infty \right)$ .

The range is $\left[-\frac{1}{e},\infty \right)$ .

The function is decreasing on $\left(-\infty ,-1\right]$ and decreasing on $\left[-1,\infty \right)$ .

Bounded below by $y=-\frac{1}{e}$ and not bounded above.

Local minimum at $\left(-1,-\frac{1}{e}\right)$ .

Horizontal asymptote is $y=0$ .

No vertical asymptote.

End behavior:

  $\underset{x\to -\infty }{\lim }f\left(x\right)=0,\underset{x\to \infty }{\lim }f\left(x\right)=\infty$ .

Given:

  $f\left(x\right)=x{e}^x$

Calculation:

The graph is,

  

The given function is well defined for all real numbers, therefore the domain is $\left(-\infty ,\infty \right)$ .

The range is $\left[-\frac{1}{e},\infty \right)$ .

From the graph, it can be seen that the value of $y$ decreases as the value of $x$ increases from $-\infty$ to $-1$ and the value of $y$ increases as the value of $x$ increases from $-1$ to $\infty$ .

The function is decreasing on $\left(-\infty ,-1\right]$ and increasing on $\left[-1,\infty \right)$ .

Bounded below by $y=-\frac{1}{e}$ and not bounded above.

The curve decreases to the left of $x=-1$ and increases to the right of $x=-1$ , therefore the local minimum is at $\left(-1,-\frac{1}{e}\right)$ .

Horizontal asymptote is $y=0$ .

No vertical asymptote.

End behavior:

From the graph, it can be seen that $y\to 0$ as $x\to -\infty$ and $y\to \infty$ as $x\to \infty$ . So,

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

b.

To graph: The given function and to analyze it.

  

The domain is $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ .

The range is $\left(-\infty ,-e\right]\cup \left(0,\infty \right)$ .

The function is decreasing on $\left[-1,0\right)\cup \left(0,\infty \right)$ and increasing on $\left(-\infty ,-1\right]$ .

Not bounded.

The local maximum is at $\left(-1,-e\right)$ .

Horizontal asymptote is $y=0$ , and the vertical asymptote is $x=0$ .

End behavior:

  $\underset{x\to -\infty }{\lim }f\left(x\right)=-\infty ,\underset{x\to \infty }{\lim }f\left(x\right)=0$ .

Given:

  $g\left(x\right)=\frac{e^{-x}}{x}$

Calculation:

The graph is,

  

The given function is well defined for all real numbers except 0, therefore the domain is $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ .

The range is $\left(-\infty ,-e\right]\cup \left(0,\infty \right)$ .

The function is decreasing on $\left[-1,0\right)\cup \left(0,\infty \right)$ and increasing on $\left(-\infty ,-1\right]$ .

Not bounded.

The curve increases to the left of $x=-1$ and decreases to the right of $x=-1$ , therefore the local maximum is at $\left(-1,-e\right)$ .

Horizontal asymptote is $y=0$ , and the vertical asymptote is $x=0$ .

End behavior:

From the graph, it can be seen that $y\to -\infty$ as $x\to -\infty$ and $y\to 0$ as $x\to \infty$ . So,

  $\underset{x\to -\infty }{\lim }f\left(x\right)=-\infty ,\underset{x\to \infty }{\lim }f\left(x\right)=0$ .