Chapter 5, Problem 3RE

a. $$

To determine

Tofindthe interval on which the function $y=x^2{e}^{\frac{1}{x^2}}$ is increasing.

Answer to Problem 3RE

The function is then increasing for $[-1,0)$ and $[1,\infty )$ .

Explanation of Solution

Given information:

The given function is $y=x^2{e}^{\frac{1}{x^2}}$ .

Formula:

Chain rule:

  $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$

Consider the function $y=x^2{e}^{\frac{1}{x^2}}$ ,

  $y=x^2{e}^{\frac{1}{x^2}}=x^2{e}^{x^{-2}}$

Using product and chain rule: $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$ find the derivative,

Here $u=x^2\Rightarrow u^{\prime }=2x$ and $v=e^{x^{-2}}\Rightarrow v^{\prime }=e^{x^{-2}}\left(-2x^{-3}\right)$ ,

  $\begin{array}{l}{y}^{\prime }=2x{e}^{x^{-2}}+x^2{e}^{x^{-2}}\left(-2x^{-3}\right)\\ y^{\prime }=2x{e}^{x^{-2}}-\frac{2}{x}{e}^{x^{-2}}\end{array}$

Now setting the derivative equal to zero to find the critical values,

  $\begin{array}{l}2x{e}^{x^{-2}}-\frac{2}{x}{e}^{x^{-2}}=0\\ 2e^{x^{-2}}\left(x-\frac{1}{x}\right)=0\end{array}$

  $\Rightarrow 2e^{x^{-2}}=0$ or $\left(x-\frac{1}{x}\right)=0$

  $2e^{x^{-2}}\ne 0$ , for all values of $x$ . Hence there are no critical values for this factor.

  $\begin{array}{l}\left(x-\frac{1}{x}\right)=0\\ \Rightarrow x^2-1=0\\ \Rightarrow x^2=1\\ \Rightarrow x=\pm 1\end{array}$

Therefore, the critical values are $x=1,x=-1,x=0$ .

From the critical values the possible intervals are $(-\infty ,-1],[-1,0),(0,1]$ or $[1,\infty )$ . Here $x=0$ is not in the domain of $y$ henceforth it is not included.

Now to determine which of the intervals is increasing,

Consider $x=-2$ for $(-\infty ,-1]$ :

  $y\left(-2\right)=2e^{{\left(-2\right)}^{-2}}\left(-2-\frac{1}{-2}\right)\approx -2.568<0$

Consider $x=-0.5$ for $[-1,0)$ :

  $y\left(-0.5\right)=2e^{{\left(-0.5\right)}^{-2}}\left(-0.5-\frac{1}{-0.5}\right)\approx 382.187>0$

Consider $x=0.5$ for $(0,1]$ :

  $y\left(0.5\right)=2e^{{\left(0.5\right)}^{-2}}\left(0.5-\frac{1}{0.5}\right)\approx -382.187<0$

Consider $x=2$ for $[1,\infty )$ :

  $y\left(2\right)=2e^{{\left(2\right)}^{-2}}\left(2-\frac{1}{2}\right)\approx 2.568>0$

Therefore, the function is increasing for $[-1,0)$ and $[1,\infty )$ .

b. $$

To determine

To find the interval on which the function $y=x^2{e}^{\frac{1}{x^2}}$ is decreasing.

Answer to Problem 3RE

The function is decreasing for $(-\infty ,-1]$ and $(0,1]$ .

Explanation of Solution

Given information:

The given function is $y=x^2{e}^{\frac{1}{x^2}}$ .

Formula:

Chain rule:

  $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$

Consider the function $y=x^2{e}^{\frac{1}{x^2}}$ ,

  $y=x^2{e}^{\frac{1}{x^2}}=x^2{e}^{x^{-2}}$

Using product and chain rule: $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$ find the derivative,

Here $u=x^2\Rightarrow u^{\prime }=2x$ and $v=e^{x^{-2}}\Rightarrow v^{\prime }=e^{x^{-2}}\left(-2x^{-3}\right)$ ,

  $\begin{array}{l}{y}^{\prime }=2x{e}^{x^{-2}}+x^2{e}^{x^{-2}}\left(-2x^{-3}\right)\\ y^{\prime }=2x{e}^{x^{-2}}-\frac{2}{x}{e}^{x^{-2}}\end{array}$

Now setting the derivative equal to zero to find the critical values,

  $\begin{array}{l}2x{e}^{x^{-2}}-\frac{2}{x}{e}^{x^{-2}}=0\\ 2e^{x^{-2}}\left(x-\frac{1}{x}\right)=0\end{array}$

  $\Rightarrow 2e^{x^{-2}}=0$ or $\left(x-\frac{1}{x}\right)=0$

  $2e^{x^{-2}}\ne 0$ , for all values of $x$ . Hence there are no critical values for this factor.

  $\begin{array}{l}\left(x-\frac{1}{x}\right)=0\\ \Rightarrow x^2-1=0\\ \Rightarrow x^2=1\\ \Rightarrow x=\pm 1\end{array}$

Therefore, the critical values are $x=1,x=-1,x=0$ .

From the critical values the possible intervals are $(-\infty ,-1],[-1,0),(0,1]$ or $[1,\infty )$ . Here $x=0$ is not in the domain of $y$ henceforth it is not included.

Now to determine which of the intervals is decreasing,

Consider $x=-2$ for $(-\infty ,-1]$ :

  $y\left(-2\right)=2e^{{\left(-2\right)}^{-2}}\left(-2-\frac{1}{-2}\right)\approx -2.568<0$

Consider $x=-0.5$ for $[-1,0)$ :

  $y\left(-0.5\right)=2e^{{\left(-0.5\right)}^{-2}}\left(-0.5-\frac{1}{-0.5}\right)\approx 382.187>0$

Consider $x=0.5$ for $(0,1]$ :

  $y\left(0.5\right)=2e^{{\left(0.5\right)}^{-2}}\left(0.5-\frac{1}{0.5}\right)\approx -382.187<0$

Consider $x=2$ for $[1,\infty )$ :

  $y\left(2\right)=2e^{{\left(2\right)}^{-2}}\left(2-\frac{1}{2}\right)\approx 2.568>0$

Therefore, the function is decreasing for $(-\infty ,-1]$ and $(0,1]$ .

c. $$

To determine

To find the interval on which the function $y=x^2{e}^{\frac{1}{x^2}}$ is concave up.

Answer to Problem 3RE

The function is concave up for $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ .

Explanation of Solution

Given information:

The given function is $y=x^2{e}^{\frac{1}{x^2}}$ .

Formula:

Chain rule:

  $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$

Consider the function $y=x^2{e}^{\frac{1}{x^2}}$ ,

  $y=x^2{e}^{\frac{1}{x^2}}=x^2{e}^{x^{-2}}$

Using product and chain rule: $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$ find the derivative,

Here $u=x^2\Rightarrow u^{\prime }=2x$ and $v=e^{x^{-2}}\Rightarrow v^{\prime }=e^{x^{-2}}\left(-2x^{-3}\right)$ ,

  $\begin{array}{l}{y}^{\prime }=2x{e}^{x^{-2}}+x^2{e}^{x^{-2}}\left(-2x^{-3}\right)\\ y^{\prime }=2x{e}^{x^{-2}}-\frac{2}{x}{e}^{x^{-2}}\end{array}$

Factor out $2e^{x^{-2}}$ to find second derivative,

  $\begin{array}{l}{y}^{″}=2e^{x^{-2}}\left(-2x^{-3}\right)\left(x-x^{-1}\right)+2e^{x^{-2}}\left(1+x^{-2}\right)\\ y^{″}=2e^{x^{-2}}\left(-2x^{-3}\left(x-x^{-1}\right)+1+x^{-2}\right)\\ y^{″}=2e^{x^{-2}}\left(-2x^{-2}+2x^{-4}+1+x^{-2}\right)\\ y^{″}=2e^{x^{-2}}\left(2x^{-4}+1-x^{-2}\right)\\ y^{″}=2e^{x^{-2}}\left(\frac{2}{x^4}+1-\frac{1}{x^2}\right)\\ y^{″}=2e^{x^{-2}}\left(\frac{x^4-x^2+2}{x^4}\right)\end{array}$

Now to find the possible points of inflection,

  $2e^{x^{-2}}\ne 0$ , for all values of $x$ and is undefined when $x=0$ .

Equating $y^{″}=0$ to find the value of $x$ ,

  $\begin{array}{l}\frac{x^4-x^2+2}{x^4}=0\\ \Rightarrow x^4-x^2+2=0\end{array}$

This polynomial does not appear to be factorable so solve for $x$ by completing the square.

  $\begin{array}{l}{x}^4-x^2=-2\\ x^4-x^2+\frac{1}{4}=-2+\frac{1}{4}\\ {\left(x^2-\frac{1}{2}\right)}^2=-\frac{7}{4}\\ x^2-\frac{1}{2}=\pm \sqrt{-\frac{7}{4}}\end{array}$

Therefore, this has no real solution so there are no values of $x$ that makes $y^{″}=0$ .

Hence, $x=0$ is the only value found in the previous step so substitute a value less than $0$ and a value greater than $0$ into $y^{″}$ .

  $\begin{array}{l}{y}^{″}\left(-1\right)=2e^{{\left(-1\right)}^{-2}}\left(\frac{{\left(-1\right)}^4-{\left(-1\right)}^2+2}{{\left(-1\right)}^4}\right)\approx 10.873>0\\ y^{″}\left(1\right)=2e^{{\left(1\right)}^{-2}}\left(\frac{{\left(1\right)}^4-{\left(1\right)}^2+2}{{\left(1\right)}^4}\right)\approx 10.873>0\end{array}$

Since $y^{″}>0$ for $x<0$ and $x>0$ ,the function is concave up for $\left(-\infty ,0\right)\cup \left(0,\infty \right)$

Therefore, the function is concave up for $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ .

d. $$

To determine

To find the interval on which the function $y=x^2{e}^{\frac{1}{x^2}}$ is concave down.

Answer to Problem 3RE

The function is concave down for no values of $x$ .

Explanation of Solution

Given information:

The given function is $y=x^2{e}^{\frac{1}{x^2}}$ .

Formula:

Chain rule:

  $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$

Consider the function $y=x^2{e}^{\frac{1}{x^2}}$ ,

  $y=x^2{e}^{\frac{1}{x^2}}=x^2{e}^{x^{-2}}$

Using product and chain rule: $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$ find the derivative,

Here $u=x^2\Rightarrow u^{\prime }=2x$ and $v=e^{x^{-2}}\Rightarrow v^{\prime }=e^{x^{-2}}\left(-2x^{-3}\right)$ ,

  $\begin{array}{l}{y}^{\prime }=2x{e}^{x^{-2}}+x^2{e}^{x^{-2}}\left(-2x^{-3}\right)\\ y^{\prime }=2x{e}^{x^{-2}}-\frac{2}{x}{e}^{x^{-2}}\end{array}$

Factor out $2e^{x^{-2}}$ to find second derivative,

  $\begin{array}{l}{y}^{″}=2e^{x^{-2}}\left(-2x^{-3}\right)\left(x-x^{-1}\right)+2e^{x^{-2}}\left(1+x^{-2}\right)\\ y^{″}=2e^{x^{-2}}\left(-2x^{-3}\left(x-x^{-1}\right)+1+x^{-2}\right)\\ y^{″}=2e^{x^{-2}}\left(-2x^{-2}+2x^{-4}+1+x^{-2}\right)\\ y^{″}=2e^{x^{-2}}\left(2x^{-4}+1-x^{-2}\right)\\ y^{″}=2e^{x^{-2}}\left(\frac{2}{x^4}+1-\frac{1}{x^2}\right)\\ y^{″}=2e^{x^{-2}}\left(\frac{x^4-x^2+2}{x^4}\right)\end{array}$

Now to find the possible points of inflection,

  $2e^{x^{-2}}\ne 0$ , for all values of $x$ and is undefined when $x=0$ .

Equating $y^{″}=0$ to find the value of $x$ ,

  $\begin{array}{l}\frac{x^4-x^2+2}{x^4}=0\\ \Rightarrow x^4-x^2+2=0\end{array}$

This polynomial does not appear to be factorable so solve for $x$ by completing the square.

  $\begin{array}{l}{x}^4-x^2=-2\\ x^4-x^2+\frac{1}{4}=-2+\frac{1}{4}\\ {\left(x^2-\frac{1}{2}\right)}^2=-\frac{7}{4}\\ x^2-\frac{1}{2}=\pm \sqrt{-\frac{7}{4}}\end{array}$

Therefore, this has no real solution so there are no values of $x$ that makes $y^{″}=0$ .

Hence, $x=0$ is the only value found in the previous step so substitute a value less than $0$ and a value greater than $0$ into $y^{″}$ .

  $\begin{array}{l}{y}^{″}\left(-1\right)=2e^{{\left(-1\right)}^{-2}}\left(\frac{{\left(-1\right)}^4-{\left(-1\right)}^2+2}{{\left(-1\right)}^4}\right)\approx 10.873>0\\ y^{″}\left(1\right)=2e^{{\left(1\right)}^{-2}}\left(\frac{{\left(1\right)}^4-{\left(1\right)}^2+2}{{\left(1\right)}^4}\right)\approx 10.873>0\end{array}$

Since $y^{″}>0$ for $x<0$ and $x>0$ ,the function is concave up for $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ so there are no values of $x$ .

Therefore, the function is concave down for no values of $x$ .

e. $$

To determine

To find the interval on which the function $y=x^2{e}^{\frac{1}{x^2}}$ has local extreme values.

Answer to Problem 3RE

The function haslocal extreme values are at $\left(-1,e\right)$ and $\left(1,e\right)$ .

Explanation of Solution

Given information:

The given function is $y=x^2{e}^{\frac{1}{x^2}}$ .

Formula:

Chain rule:

  $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$

Consider the function $y=x^2{e}^{\frac{1}{x^2}}$ ,

  $y=x^2{e}^{\frac{1}{x^2}}=x^2{e}^{x^{-2}}$

Using product and chain rule: $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$ find the derivative,

Here $u=x^2\Rightarrow u^{\prime }=2x$ and $v=e^{x^{-2}}\Rightarrow v^{\prime }=e^{x^{-2}}\left(-2x^{-3}\right)$ ,

  $\begin{array}{l}{y}^{\prime }=2x{e}^{x^{-2}}+x^2{e}^{x^{-2}}\left(-2x^{-3}\right)\\ y^{\prime }=2x{e}^{x^{-2}}-\frac{2}{x}{e}^{x^{-2}}\end{array}$

Here the derivative is undefined if $x=0$ since it will cause division by $0$ .

Now setting the derivative equal to zero to find the critical values,

  $\begin{array}{l}2x{e}^{\frac{1}{x^2}}-\frac{2e^{\frac{1}{x^2}}}{x}=0\\ 2e^{\frac{1}{x^2}}\left(x-\frac{1}{x}\right)=0\end{array}$

  $\Rightarrow 2e^{x^{-2}}=0$ or $\left(x-\frac{1}{x}\right)=0$

  $2e^{x^{-2}}\ne 0$ , for all values of $x$ . Hence there are no critical values for this factor.

  $\begin{array}{l}\left(x-\frac{1}{x}\right)=0\\ \Rightarrow x^2-1=0\\ \Rightarrow x^2=1\\ \Rightarrow x=\pm 1\end{array}$

The critical values include all the values that makes $y^{\prime }=0$ or undefined so the critical values are then $\pm 1$ and $0$ .

Then the possible intervals are given by $(-\infty ,-1],[-1,0),(0,1]$ or $[1,\infty )$ .

Now to determine whether the function is increasing or decreasing on these intervals by substituting a value for $x$ ,

Consider $x=-2$ for $(-\infty ,-1]$ :

  $y^{\prime }\left(-2\right)=2e^{\frac{1}{{\left(-2\right)}^2}}\left(-2-\frac{1}{-2}\right)\approx -3.85<0$

Consider $x=-0.5$ for $[-1,0)$ :

  $y\left(-0.5\right)=2e^{\frac{1}{{\left(-0.5\right)}^2}}\left(-0.5-\frac{1}{-0.5}\right)\approx 163.8>0$

Consider $x=0.5$ for $(0,1]$ :

  $y\left(0.5\right)=2e^{\frac{1}{{\left(0.5\right)}^2}}\left(0.5-\frac{1}{0.5}\right)\approx -163.8<0$

Consider $x=2$ for $[1,\infty )$ :

  $y^{\prime }\left(2\right)=2e^{\frac{1}{{\left(2\right)}^2}}\left(2-\frac{1}{2}\right)\approx 3.85>0$

Therefore, $y$ has local extreme at all values of $x$ in the domain where $y^{\prime }$ switches its signs.

Since $y^{\prime }$ switches its value from negative to positive at $x=-1$ and $x=1$ , these will be local minimums.

  $\begin{array}{l}y\left(-1\right)={\left(-1\right)}^2{e}^{\frac{1}{\left(-1^2\right)}}=e\\ y\left(1\right)={\left(1\right)}^2{e}^{\frac{1}{\left(1^2\right)}}=e\end{array}$

The local extreme are then local minimums at $\left(-1,e\right)$ and $\left(1,e\right)$ .

Therefore, the function has local extreme values are at $\left(-1,e\right)$ and $\left(1,e\right)$ .

f. $$

To determine

To find the interval on which the function $y=x^2{e}^{\frac{1}{x^2}}$ has inflection points.

Answer to Problem 3RE

The function has no inflection points.

Explanation of Solution

Given information:

The given function is $y=x^2{e}^{\frac{1}{x^2}}$ .

Formula:

Chain rule:

  $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$

Consider the function $y=x^2{e}^{\frac{1}{x^2}}$ ,

  $y=x^2{e}^{\frac{1}{x^2}}=x^2{e}^{x^{-2}}$

Using product and chain rule: $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$ find the derivative,

Here $u=x^2\Rightarrow u^{\prime }=2x$ and $v=e^{x^{-2}}\Rightarrow v^{\prime }=e^{x^{-2}}\left(-2x^{-3}\right)$ ,

  $\begin{array}{l}{y}^{\prime }=2x{e}^{x^{-2}}+x^2{e}^{x^{-2}}\left(-2x^{-3}\right)\\ y^{\prime }=2x{e}^{x^{-2}}-\frac{2}{x}{e}^{x^{-2}}\end{array}$

Factor out $2e^{x^{-2}}$ to find second derivative,

  $\begin{array}{l}{y}^{″}=2e^{x^{-2}}\left(-2x^{-3}\right)\left(x-x^{-1}\right)+2e^{x^{-2}}\left(1+x^{-2}\right)\\ y^{″}=2e^{x^{-2}}\left(-2x^{-3}\left(x-x^{-1}\right)+1+x^{-2}\right)\\ y^{″}=2e^{x^{-2}}\left(-2x^{-2}+2x^{-4}+1+x^{-2}\right)\\ y^{″}=2e^{x^{-2}}\left(2x^{-4}+1-x^{-2}\right)\\ y^{″}=2e^{x^{-2}}\left(1-\frac{1}{x^2}+\frac{2}{x^4}\right)\\ y^{″}=2e^{x^{-2}}\left(\frac{x^4-x^2+2}{x^4}\right)\end{array}$

Now to find the possible points of inflection,

  $2e^{x^{-2}}\ne 0$ , for all values of $x$ and is undefined when $x=0$ .

Equating $y^{″}=0$ to find the value of $x$ ,

  $\begin{array}{l}\frac{x^4-x^2+2}{x^4}=0\\ \Rightarrow x^4-x^2+2=0\end{array}$

This polynomial does not appear to be factorable so solve for $x$ by completing the square.

  $\begin{array}{l}{x}^4-x^2=-2\\ x^4-x^2+\frac{1}{4}=-2+\frac{1}{4}\\ {\left(x^2-\frac{1}{2}\right)}^2=-\frac{7}{4}\\ x^2-\frac{1}{2}=\pm \sqrt{-\frac{7}{4}}\end{array}$

Therefore, there are no real solutions so there are no values of $x$ that make $y^{″}=0$ .

Hence, $x=0$ is the only value found in the previous step so substitute a value less than $0$ and a value greater than $0$ into $y^{″}$ .

  $\begin{array}{l}{y}^{″}\left(-1\right)=2e^{{\left(-1\right)}^{-2}}\left(\frac{{\left(-1\right)}^4-{\left(-1\right)}^2+2}{{\left(-1\right)}^4}\right)\approx 10.873>0\\ y^{″}\left(1\right)=2e^{{\left(1\right)}^{-2}}\left(\frac{{\left(1\right)}^4-{\left(1\right)}^2+2}{{\left(1\right)}^4}\right)\approx 10.873>0\end{array}$

Since $y^{″}>0$ for $x<0$ and $x>0$ ,the function is concave up for $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ so there are no points of inflection since the concavity doesn’t change signs.

Therefore, the function has no inflection points.