Chapter 4, Problem 3RE

a.

To determine

To find: To find the increasing intervals for the function ${\text{y = x}}^{\text{2}}{\text{e}}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{${\text{x}}^{\text{2}}$}\right.}$ .

Answer to Problem 3RE

The function is increasing at $\text{[-1, 0) and [1, }\infty \text{)}$ .

Explanation of Solution

Given information: The given function is ${\text{y = x}}^{\text{2}}{\text{e}}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{${\text{x}}^{\text{2}}$}\right.}$

Formula used: Product rule: $\frac{\text{d}}{\text{dx}}{\text{ = u}}^{\text{'}}{\text{v + uv}}^{\text{'}}$

Calculation:

The above equation is rewritten as,

  $\begin{array}{c}{\text{y = x}}^{\text{2}}{\text{e}}^{\frac{\text{1}}{{\text{x}}^{\text{2}}}}\\ {\text{= x}}^{\text{2}}{\text{e}}^{{\text{x}}^{\text{-2}}}\end{array}$

Using the product rule finding the derivative for the above equation,

  $\frac{\text{d}}{\text{dx}}{\text{ = u}}^{\text{'}}{\text{v + uv}}^{\text{'}}$

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

Simplifying to get,

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

Since ${\text{2e}}^{{\text{x}}^{\text{-2}}}\ne \text{ 0}$ , there is no critical value for this factor.

  $\begin{array}{c}\text{x - }\frac{\text{1}}{\text{x}}\text{ = 0}\\ {\text{x}}^{\text{2}}\text{ - 1 = 0}\\ \text{x = ±1}\end{array}$

Therefore, the critical values are $\text{x = -1, 0 and 1}$ .

The possible intervals for the critical values are $\text{(-}\infty \text{, -1], [-1, 0), (0,1] or [1, }\infty \text{)}$

Now testing the above intervals to determine the increasing functions,

Take $\text{x = -2}$ for $\text{(-}\infty \text{, -1]}$

  $\begin{array}{c}{\text{y(-2) = 2e}}^{{\text{(-2)}}^{\text{-2}}}\left(\text{-2 - }\frac{\text{1}}{\text{-2}}\right)\\ \approx \text{ -2}\text{.568 < 0}\end{array}$

Take $\text{x = -0}\text{.5}$ for $\text{[-1, 0)}$

  $\begin{array}{c}\text{y(-0}{\text{.5) = 2e}}^{{\text{(-0}\text{.5)}}^{\text{-2}}}\left(\text{-0}\text{.5 - }\frac{\text{1}}{\text{-0}\text{.5}}\right)\\ \approx \text{ 382}\text{.187 > 0}\end{array}$

Take $\text{x = 0}\text{.5}$ for $\text{(0,1]}$

  $\begin{array}{c}\text{y(0}{\text{.5) = 2e}}^{{\text{(0}\text{.5)}}^{\text{-2}}}\left(\text{0}\text{.5 - }\frac{\text{1}}{\text{0}\text{.5}}\right)\\ \approx \text{ -382}\text{.187 < 0}\end{array}$

Take $\text{x = 2}$ for $\text{[1, }\infty \text{)}$

  $\begin{array}{c}{\text{y(2) = 2e}}^{{\text{(2)}}^{\text{-2}}}\left(\text{2 - }\frac{\text{1}}{\text{2}}\right)\\ \approx \text{2}\text{.568 > 0}\end{array}$

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

b.

To determine

To find: To find the decreasing intervals for the function ${\text{y = x}}^{\text{2}}{\text{e}}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{${\text{x}}^{\text{2}}$}\right.}$ .

Answer to Problem 3RE

The function is decreasing at $\text{(-}\infty \text{, -1] and (0, 1)}$ .

Explanation of Solution

Given information: The given function is ${\text{y = x}}^{\text{2}}{\text{e}}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{${\text{x}}^{\text{2}}$}\right.}$

Calculation:

The given function is,

  ${\text{y = x}}^{\text{2}}{\text{e}}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{${\text{x}}^{\text{2}}$}\right.}$

By using product and chain rule,

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

The factor ${\text{2e}}^{{\text{1/x}}^{\text{2}}}$ is always positive.

The value of ${\text{y}}^{\text{'}}$ depends on the expression $\frac{\text{(x - 1)(x + 1)}}{\text{x}}$

By entering any value from the observed period into the expression and noting the sign, the sign of over four intervals are

$\text{x}$	$\text{(-}\infty \text{, -1)}$	$\text{(-1, 0)}$	$\text{(0, 1)}$	$\text{(1, }\infty \text{)}$
${\text{y}}^{\text{'}}$	$-$	$\text{+}$	$-$	$\text{+}$

Therefore, the function is then decreasing for $\text{(-}\infty \text{, -1) and (0, 1)}$ .

c.

To determine

To find: To find the concave up for the function ${\text{y = x}}^{\text{2}}{\text{e}}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{${\text{x}}^{\text{2}}$}\right.}$ .

Answer to Problem 3RE

The function is concave up for $\text{(-}\infty \text{, 0) }\cup \text{ (0, }\infty \text{)}$ .

Explanation of Solution

Given information: The given function is ${\text{y = x}}^{\text{2}}{\text{e}}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{${\text{x}}^{\text{2}}$}\right.}$

Calculation:

The above equation is rewritten as,

  $\begin{array}{c}{\text{y = x}}^{\text{2}}{\text{e}}^{\frac{\text{1}}{{\text{x}}^{\text{2}}}}\\ {\text{= x}}^{\text{2}}{\text{e}}^{{\text{x}}^{\text{-2}}}\end{array}$

Using the product rule finding the derivative for the above equation,

  $\frac{\text{d}}{\text{dx}}{\text{ = u}}^{\text{'}}{\text{v + uv}}^{\text{'}}$

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

Simplifying to get,

  $\begin{array}{l}{\text{y}}^{\text{'}}{\text{ = 2xe}}^{{\text{x}}^{\text{-2}}}{\text{ - 2x}}^{\text{-1}}{\text{e}}^{{\text{x}}^{\text{-2}}}\\ {\text{y}}^{\text{'}}{\text{ = 2e}}^{{\text{x}}^{\text{-2}}}\left({\text{x - x}}^{\text{-1}}\right)\end{array}$

To find the second derivative,

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

To find the value of $\text{x}$ , put $\text{y'' = 0}$

  $\begin{array}{c}\frac{{\text{x}}^{\text{4}}{\text{ - x}}^{\text{2}}\text{ + 2}}{{\text{x}}^{\text{4}}}\text{ = 0}\\ {\text{x}}^{\text{4}}{\text{ - x}}^{\text{2}}\text{ + 2 = 0}\\ {\text{x}}^{\text{4}}{\text{ - x}}^{\text{2}}\text{ = -2}\\ {\text{x}}^{\text{4}}{\text{ - x}}^{\text{2}}\text{ + }\frac{\text{1}}{\text{4}}\text{ = -2 + }\frac{\text{1}}{\text{4}}\\ {\left({\text{x}}^{\text{2}}\text{ - }\frac{\text{1}}{\text{2}}\right)}^{\text{2}}\text{ = -}\frac{\text{7}}{\text{4}}\\ {\text{x}}^{\text{2}}\text{ - }\frac{\text{1}}{\text{2}}\text{ = ±}\sqrt{\text{-}\frac{\text{7}}{\text{4}}}\end{array}$

As a result, there are no x values because this has no real solutions.

Substitute the value less than $\text{0}$ and greater than $\text{0}$ to see the intervals where it is concave up

  $\begin{array}{c}{\text{y}}^{\text{'}\text{'}}{\text{(-1) = 2e}}^{{\text{(-1)}}^{\text{-2}}}\left(\frac{{\text{(-1)}}^{\text{4}}{\text{ - (-1)}}^{\text{2}}\text{ + 2}}{{\text{(-1)}}^{\text{4}}}\right)\\ \approx \text{ 10}\text{.873 > 0}\\ {\text{y}}^{\text{'}\text{'}}{\text{(1) = 2e}}^{{\text{(1)}}^{\text{-2}}}\left(\frac{{\text{(1)}}^{\text{4}}{\text{ - (1)}}^{\text{2}}\text{ + 2}}{{\text{(1)}}^{\text{4}}}\right)\\ \approx \text{ 10}\text{.873 > 0}\end{array}$

The function ${\text{y}}^{\text{'}\text{'}}\text{ > 0}$ for $\text{x > 0 and x < 0}$

Therefore, the function is concave up for $\text{(-}\infty \text{, 0) }\cup \text{ (0, }\infty \text{)}$ .

d.

To determine

To find: To find the concave down for the function ${\text{y = x}}^{\text{2}}{\text{e}}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{${\text{x}}^{\text{2}}$}\right.}$ .

Answer to Problem 3RE

The function does not have concave down intervals.

Explanation of Solution

Given information: The given function is ${\text{y = x}}^{\text{2}}{\text{e}}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{${\text{x}}^{\text{2}}$}\right.}$

Calculation:

The above equation is rewritten as,

  $\begin{array}{c}{\text{y = x}}^{\text{2}}{\text{e}}^{\frac{\text{1}}{{\text{x}}^{\text{2}}}}\\ {\text{= x}}^{\text{2}}{\text{e}}^{{\text{x}}^{\text{-2}}}\end{array}$

Using the product rule finding the derivative for the above equation,

  $\frac{\text{d}}{\text{dx}}{\text{ = u}}^{\text{'}}{\text{v + uv}}^{\text{'}}$

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

Simplifying to get,

  $\begin{array}{l}{\text{y}}^{\text{'}}{\text{ = 2xe}}^{{\text{x}}^{\text{-2}}}{\text{ - 2x}}^{\text{-1}}{\text{e}}^{{\text{x}}^{\text{-2}}}\\ {\text{y}}^{\text{'}}{\text{ = 2e}}^{{\text{x}}^{\text{-2}}}\left({\text{x - x}}^{\text{-1}}\right)\end{array}$

To find the second derivative,

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

To find the value of $\text{x}$ , put $\text{y'' = 0}$

  $\begin{array}{c}\frac{{\text{x}}^{\text{4}}{\text{ - x}}^{\text{2}}\text{ + 2}}{{\text{x}}^{\text{4}}}\text{ = 0}\\ {\text{x}}^{\text{4}}{\text{ - x}}^{\text{2}}\text{ + 2 = 0}\\ {\text{x}}^{\text{4}}{\text{ - x}}^{\text{2}}\text{ = -2}\\ {\text{x}}^{\text{4}}{\text{ - x}}^{\text{2}}\text{ + }\frac{\text{1}}{\text{4}}\text{ = -2 + }\frac{\text{1}}{\text{4}}\\ {\left({\text{x}}^{\text{2}}\text{ - }\frac{\text{1}}{\text{2}}\right)}^{\text{2}}\text{ = -}\frac{\text{7}}{\text{4}}\\ {\text{x}}^{\text{2}}\text{ - }\frac{\text{1}}{\text{2}}\text{ = ±}\sqrt{\text{-}\frac{\text{7}}{\text{4}}}\end{array}$

As a result, there are no x values because this has no real solutions.

Substitute the value less than $\text{0}$ and greater than $\text{0}$ to see the intervals where it is concave up

  $\begin{array}{c}{\text{y}}^{\text{'}\text{'}}{\text{(-1) = 2e}}^{{\text{(-1)}}^{\text{-2}}}\left(\frac{{\text{(-1)}}^{\text{4}}{\text{ - (-1)}}^{\text{2}}\text{ + 2}}{{\text{(-1)}}^{\text{4}}}\right)\\ \approx \text{ 10}\text{.873 > 0}\\ {\text{y}}^{\text{'}\text{'}}{\text{(1) = 2e}}^{{\text{(1)}}^{\text{-2}}}\left(\frac{{\text{(1)}}^{\text{4}}{\text{ - (1)}}^{\text{2}}\text{ + 2}}{{\text{(1)}}^{\text{4}}}\right)\\ \approx \text{ 10}\text{.873 > 0}\end{array}$

The function ${\text{y}}^{\text{'}\text{'}}\text{ > 0}$ for $\text{x > 0 and x < 0}$

Therefore, the function does not have intervals at concave down.

e.

To determine

To find: To find the local extreme values for the function ${\text{y = x}}^{\text{2}}{\text{e}}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{${\text{x}}^{\text{2}}$}\right.}$ .

Answer to Problem 3RE

The local extreme values have local minimums at $\text{(-1, e) and (1, e)}$ .

Explanation of Solution

Given information: The given function is ${\text{y = x}}^{\text{2}}{\text{e}}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{${\text{x}}^{\text{2}}$}\right.}$

Calculation:

The above equation is rewritten as,

  $\begin{array}{c}{\text{y = x}}^{\text{2}}{\text{e}}^{\frac{\text{1}}{{\text{x}}^{\text{2}}}}\\ {\text{= x}}^{\text{2}}{\text{e}}^{{\text{x}}^{\text{-2}}}\end{array}$

Using the product rule finding the derivative for the above equation,

  $\frac{\text{d}}{\text{dx}}{\text{ = u}}^{\text{'}}{\text{v + uv}}^{\text{'}}$

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

Simplifying to get,

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

Since ${\text{2e}}^{{\text{x}}^{\text{-2}}}\ne \text{ 0}$ , there is no critical value for this factor.

  $\begin{array}{c}\text{x - }\frac{\text{1}}{\text{x}}\text{ = 0}\\ {\text{x}}^{\text{2}}\text{ - 1 = 0}\\ \text{x = ±1}\end{array}$

Therefore, the critical values are $\text{x = -1, 0 and 1}$ .

The possible intervals for the critical values are $\text{(-}\infty \text{, -1], [-1, 0), (0,1] or [1, }\infty \text{)}$

Now testing the above intervals to determine the increasing functions,

Take $\text{x = -2}$ for $\text{(-}\infty \text{, -1]}$

  $\begin{array}{c}{\text{y}}^{\text{'}}{\text{(-2) = 2e}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${(-2)}^2$}\right.}\left(\text{x - }\frac{\text{1}}{\text{-2}}\right)\\ \approx 3.9\text{ < 0}\end{array}$

Take $\text{x = -0}\text{.5}$ for $\text{[-1, 0)}$

  $\begin{array}{c}{\text{y}}^{\text{'}}\text{(-0}{\text{.5) = 2e}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${(-0.5)}^2$}\right.}\left(\text{x - }\frac{\text{1}}{\text{-0}\text{.5}}\right)\\ \approx 163.8\text{ > 0}\end{array}$

Take $\text{x = 0}\text{.5}$ for $\text{(0,1]}$

  $\begin{array}{c}{\text{y}}^{\text{'}}\text{(0}{\text{.5) = 2e}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${(0.5)}^2$}\right.}\left(\text{x - }\frac{\text{1}}{\text{0}\text{.5}}\right)\\ \approx 163.8\text{ < 0}\end{array}$

Take $\text{x = 2}$ for $\text{[1, }\infty \text{)}$

  $\begin{array}{c}{\text{y}}^{\text{'}}{\text{(2) = 2e}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${(2)}^2$}\right.}\left(\text{x - }\frac{\text{1}}{\text{2}}\right)\\ \approx \text{3}\text{.9 > 0}\end{array}$

For all the values of $\text{x}$ , there is a local extrema for $\text{y}$ .

The value of ${\text{y}}^{\text{'}}$ changes from negative and positive values at $\text{x = -1 and x = 1 }$

So, there will be local minimum values.

To find the value of $\text{y-}$ coordinate evaluate $\text{y}$ at $\text{x = -1 and x = 1 }$

  $\begin{array}{c}{\text{y(-1) = (-1)}}^2{\text{e}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${(-1)}^2$}\right.}\\ \text{= e}\\ {\text{y(1) = (1)}}^2{\text{e}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${(1)}^2$}\right.}\\ \text{= e}\end{array}$

Therefore, the local extreme values have local minimums at $\text{(-1, e) and (1, e)}$ .

f.

To determine

To find: To find the inflection points for the function ${\text{y = x}}^{\text{2}}{\text{e}}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{${\text{x}}^{\text{2}}$}\right.}$ .

Answer to Problem 3RE

The function does not have inflection points.

Explanation of Solution

Given information: The given function is ${\text{y = x}}^{\text{2}}{\text{e}}^{\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{${\text{x}}^{\text{2}}$}\right.}$

Calculation:

The above equation is rewritten as,

  $\begin{array}{c}{\text{y = x}}^{\text{2}}{\text{e}}^{\frac{\text{1}}{{\text{x}}^{\text{2}}}}\\ {\text{= x}}^{\text{2}}{\text{e}}^{{\text{x}}^{\text{-2}}}\end{array}$

Using the product rule finding the derivative for the above equation,

  $\frac{\text{d}}{\text{dx}}{\text{ = u}}^{\text{'}}{\text{v + uv}}^{\text{'}}$

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

Simplifying to get,

  $\begin{array}{l}{\text{y}}^{\text{'}}{\text{ = 2xe}}^{{\text{x}}^{\text{-2}}}{\text{ - 2x}}^{\text{-1}}{\text{e}}^{{\text{x}}^{\text{-2}}}\\ {\text{y}}^{\text{'}}{\text{ = 2e}}^{{\text{x}}^{\text{-2}}}\left({\text{x - x}}^{\text{-1}}\right)\end{array}$

To find the second derivative,

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

To find the value of $\text{x}$ , put $\text{y'' = 0}$

  $\begin{array}{c}\frac{{\text{x}}^{\text{4}}{\text{ - x}}^{\text{2}}\text{ + 2}}{{\text{x}}^{\text{4}}}\text{ = 0}\\ {\text{x}}^{\text{4}}{\text{ - x}}^{\text{2}}\text{ + 2 = 0}\\ {\text{x}}^{\text{4}}{\text{ - x}}^{\text{2}}\text{ = -2}\\ {\text{x}}^{\text{4}}{\text{ - x}}^{\text{2}}\text{ + }\frac{\text{1}}{\text{4}}\text{ = -2 + }\frac{\text{1}}{\text{4}}\\ {\left({\text{x}}^{\text{2}}\text{ - }\frac{\text{1}}{\text{2}}\right)}^{\text{2}}\text{ = -}\frac{\text{7}}{\text{4}}\\ {\text{x}}^{\text{2}}\text{ - }\frac{\text{1}}{\text{2}}\text{ = ±}\sqrt{\text{-}\frac{\text{7}}{\text{4}}}\end{array}$

As a result, there are no x values because this has no real solutions.

Substitute the value less than $\text{0}$ and greater than $\text{0}$ to see the intervals where it is concave up

  $\begin{array}{c}{\text{y}}^{\text{'}\text{'}}{\text{(-1) = 2e}}^{{\text{(-1)}}^{\text{-2}}}\left(\frac{{\text{(-1)}}^{\text{4}}{\text{ - (-1)}}^{\text{2}}\text{ + 2}}{{\text{(-1)}}^{\text{4}}}\right)\\ \approx \text{ 10}\text{.873 > 0}\\ {\text{y}}^{\text{'}\text{'}}{\text{(1) = 2e}}^{{\text{(1)}}^{\text{-2}}}\left(\frac{{\text{(1)}}^{\text{4}}{\text{ - (1)}}^{\text{2}}\text{ + 2}}{{\text{(1)}}^{\text{4}}}\right)\\ \approx \text{ 10}\text{.873 > 0}\end{array}$

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

Therefore, the function does not have any point of inflection.