Chapter 4.3, Problem 4E

To determine

The local extreme values and any absolute extremum of the given function.

Answer to Problem 4E

It has been determined that $x=1$ is a point of local minima for the given function. However, there are no absolute extrema for the given function.

Explanation of Solution

Given:

The function, $y=x{e}^{\frac{1}{x}}$ .

Concept used:

The critical points of a function $y\left(x\right)$ are the points $x=c$ such that $y^{\prime }\left(c\right)=0$ .

If $y^{\prime }\left(x\right)<0$ for $x<c$ and $y^{\prime }\left(x\right)>0$ for $x>c$ , then the function $y\left(x\right)$ has a local minima at $x=c$ and vice versa.

The local minima (or maxima) where the function attains the least (or greatest) value is called the absolute minima (or maxima) of the function.

Calculation:

The given function is $y=x{e}^{\frac{1}{x}}$ .

Differentiating,

  $y^{\prime }\left(x\right)=e^{\frac{1}{x}}+x{e}^{\frac{1}{x}}\left(-\frac{1}{x^2}\right)$

Simplifying,

  $y^{\prime }\left(x\right)=\left(1-\frac{1}{x}\right)e^{\frac{1}{x}}$

Equating the first derivative to zero to obtain the critical points,

  $\left(1-\frac{1}{x}\right)e^{\frac{1}{x}}=0$

Solving, $x=1$ .

So, the only critical point of the given function is $x=1$ .

Let $x<1$ .

It follows that,

  $\begin{array}{c}x<1\\ \frac{1}{x}>1\\ 1-\frac{1}{x}<0\end{array}$

Then, $y^{\prime }\left(x\right)=\left(1-\frac{1}{x}\right)e^{\frac{1}{x}}<0$ .

Similarly, let $x>1$ .

It follows that,

  $\begin{array}{c}x>1\\ \frac{1}{x}<1\\ 1-\frac{1}{x}>0\end{array}$

Then, $y^{\prime }\left(x\right)=\left(1-\frac{1}{x}\right)e^{\frac{1}{x}}>0$ .

So, $y^{\prime }\left(x\right)<0$ for $x<1$ and $y^{\prime }\left(x\right)>0$ for $x>1$ , which implies that $x=1$ is a local minima of the given function.

It should be noted that $y=x{e}^{\frac{1}{x}}\to -\infty$ as $x\to -\infty$ .

This implies that there is no absolute minimum value of the given function.

Conclusion:

It has been determined that $x=1$ is a point of local minima for the given function. However, there are no absolute extrema for the given function.