Chapter 4.5, Problem 56E

(a)

To determine

To graph: The function of $f(x)=x{e}^{1/x}$ .

Explanation of Solution

Given information:

The given function is $f(x)=x{e}^{1/x}$ .

Graph:

Consider the function $f(x)=x{e}^{1/x}$ .

(a)

Sketch the graph of $f(x)=x{e}^{1/x}$

is as follows:

  

Figure (1)

Interpretation: Graph for the functions $f(x)=x{e}^{1/x}$ shows in figure (1).

(b)

To determine

To explain: The behavior as $x\to 0.$is $x=0$ .

Answer to Problem 56E

The behavior as $x\to 0.$

Explanation of Solution

Given information:

The given function is $f(x)=x{e}^{1/x}$ .

Calculation:

Note that $y=x{e}^{\frac{1}{2}}$ is not defined at $x=0$ .

From the graph it is clear that as $x$ approaches 0 from right side, $f(x)$ approaches to 0.

Use I ‘Hospital’s Rule to explain this behavior.

Note that as $x\to 0^{}$ , we have $e^{\frac{1}{x}}\to \infty$

Since the limit is an indeterminate form of type $0(\infty )$ , I ‘Hospital’s Rule is applicable $\underset{x\to 0^{\text{'}}}{\lim }f(x)=0.$

Rewrite the limit as,

  $\begin{array}{l}\underset{x\to 0}{\lim }(x{e}^{\frac{1}{x}})=\underset{x\to 0}{\lim }(\frac{e^{\frac{1}{x}}}{1/x})\\ =\underset{x\to 0}{\lim }\left[\frac{\frac{d}{dx}(e^{\frac{1}{x}})}{\frac{d}{dx}(1/x)}\right]\\ =\underset{x\to 0}{\lim }\left(\frac{e^{\frac{1}{x}}(-\frac{1}{x^2})}{-\frac{1}{x^2}}\right)\\ =\underset{x\to 0}{\lim }{e}^{\begin{array}{l}\frac{1}{x}\\ \end{array}}\\ =0\end{array}$

(c)

To determine

To find: The minimum value and intervals.

Answer to Problem 56E

There is no inflection points of $f^{"}(x)=x{e}^{\frac{1}{x}}$ .

Explanation of Solution

Given information:

The given function is $f(x)=x{e}^{1/x}$ .

Calculation:

From the graph, the minimum value occurs $x=1$ and is estimated as $f(1)=e$ .

The inflection point no occurs

Use derivative concept to explain these features.

Differentiate the given function with respect to x,

  $\begin{array}{c}f(x)=\frac{d}{dx}\left(x{e}^{\frac{1}{x}}\right)\\ f^{\text{'}}(x)=x.e^{\frac{1}{x}}\left(-\frac{1}{x^2}\right)+e^{\frac{1}{x}}.1\\ =e^{\frac{1}{x}}\left(-\frac{1}{x}+1\right)\end{array}$

For maxima or minima,

  $\begin{array}{c}{f}^{\text{'}}(x)=0\\ e^{\frac{1}{x}}\left(-\frac{1}{x}+1\right)=0\end{array}$

  $y=x{e}^{\frac{1}{x}}$ is not defined at $x=0$ , take $-\frac{1}{x}+1=0$ and solve for x.

  $\begin{array}{c}\frac{1}{x}=1\\ x=1\end{array}$

Therefore, the local minimum occurs $x=1$ and is the minimum value is $f(1)=e$ .

Differentiate $f^{\text{'}}(x)=e^{\frac{1}{x}}\left(-\frac{1}{x}+1\right)$ with respect to x,

  $\begin{array}{c}{f}^{\text{'}}(x)=e^{\frac{1}{x}}\left(\frac{1}{x^2}\right)+\left(-\frac{1}{x+}+1\right)e^{\frac{1}{x}}\left(-\frac{1}{x^2}\right)\\ =e^{\frac{1}{x}\left(\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^2}\right)}\\ =e^{\frac{1}{x}}.\frac{1}{x^3}\end{array}$

To find the inflection points,

  $f^{"}(x)=0$

Observe that not possible values for x if $f^{"}(x)=0$ .

Therefore,

The no inflection points of $f^{"}(x)=x{e}^{\frac{1}{x}}$ .