Chapter 5, Problem 13RE

a.

To determine

To find: the intervals on which the function is increasing.

Answer to Problem 13RE

  $(0,\frac{2\sqrt{3}}{3})$ .

Explanation of Solution

Given information: Given function is $y=e^{-x},x\le 0\text{and}y=4x-x^3,x>0.$

Calculation:

  $\begin{array}{c}y=e^{-x},x\le 0\text{and}y=4x-x^3,x>0.\\ y\text{'}=-e^{-x}x\le 0\text{and}y\text{'}=4-3x^2,x>0\\ -e^{-x}\ne \text{0}\text{for}\text{all}\text{values}\text{of }x.\\ 4-3x^2\to 0\to x^2=\frac{4}{3}\to x=\pm \sqrt{\frac{4}{3}}=\pm \frac{2}{\sqrt{3}}=\pm \frac{2\sqrt{3}}{3}\end{array}$

  $\begin{array}{l}\text{For}(-\infty ,0),\text{test}x=-1\\ y\text{'}(-1)=-e^{-1}=-e<0\\ \text{For}(0,\frac{2\sqrt{3}}{3}),\text{test}x=1\\ $ y\text{'}(1)=4-3{(1)}^2=1>0$\text{For}(0,\frac{2\sqrt{3}}{3}),\text{test}x=2\\ $ y\text{'}(1)=4-3{(2)}^2=-8<0$\end{array}$

Therefore ,y is increasing if y’ > 0 which only occurs on $(0,\frac{2\sqrt{3}}{3})$ .

b.

To determine

To find: the intervals on which the function is decreasing.

Answer to Problem 13RE

  $(-\infty ,0)\text{and}(\frac{2\sqrt{3}}{3},\infty )$ .

Explanation of Solution

Given information: Given function is $y=e^{-x},x\le 0\text{and}y=4x-x^3,x>0.$

Calculation:

  $\begin{array}{c}y=e^{-x},x\le 0\text{and}y=4x-x^3,x>0.\\ y\text{'}=-e^{-x}x\le 0\text{and}y\text{'}=4-3x^2,x>0\\ -e^{-x}\ne \text{0}\text{for}\text{all}\text{values}\text{of }x.\\ 4-3x^2\to 0\to x^2=\frac{4}{3}\to x=\pm \sqrt{\frac{4}{3}}=\pm \frac{2}{\sqrt{3}}=\pm \frac{2\sqrt{3}}{3}\end{array}$

  $\begin{array}{l}\text{For}(-\infty ,0),\text{test}x=-1\\ y\text{'}(-1)=-e^{-1}=-e<0\\ \text{For}(0,\frac{2\sqrt{3}}{3}),\text{test}x=1\\ $ y\text{'}(1)=4-3{(1)}^2=1>0$\text{For}(0,\frac{2\sqrt{3}}{3}),\text{test}x=2\\ $ y\text{'}(1)=4-3{(2)}^2=-8<0$\end{array}$

Therefore, y is decreasing if y’ < 0 which occurs for $(-\infty ,0)\text{and}(\frac{2\sqrt{3}}{3},\infty ).$

c.

To determine

To find: the intervals on which the function is concave up.

Answer to Problem 13RE

  $(-\infty ,0)$ .

Explanation of Solution

Given information: Given function is $y=e^{-x},x\le 0\text{and}y=4x-x^3,x>0.$

Calculation:

  $\begin{array}{c}y=e^{-x},x\le 0\text{and}y=4x-x^3,x>0.\\ y\text{'}=-e^{-x},x\le 0\text{and}y\text{'}=4-3x^2,x>0\\ y\text{'}\text{'}=e^{-x},,x\le 0\text{and}y\text{'}\text{'}=-6x,x>0\end{array}$

  $\begin{array}{l}\text{For}(-\infty ,0),\text{test}x=-1\\ y\text{'}\text{'}(-1)=e^{-1(-1)}=e>0\\ \text{For}(0,\infty ),\text{test}x=1\\ y\text{'}\text{'}(1)=-6(1)=-6<0\end{array}$

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

d.

To determine

To find: the intervals on which the function is concave down.

Answer to Problem 13RE

  $(0,\infty )$ .

Explanation of Solution

Given information: Given function is $y=e^{-x},x\le 0\text{and}y=4x-x^3,x>0.$

Calculation:

  $\begin{array}{c}y=e^{-x},x\le 0\text{and}y=4x-x^3,x>0.\\ y\text{'}=-e^{-x},x\le 0\text{and}y\text{'}=4-3x^2,x>0\\ y\text{'}\text{'}=e^{-x},x\le 0\text{and}y\text{'}\text{'}=-6x,x>0\end{array}$

  $\begin{array}{l}\text{For}(-\infty ,0),\text{test}x=-1\\ y\text{'}\text{'}(-1)=e^{-1(-1)}=e>0\\ \text{For}(0,\infty ),\text{test}x=1\\ y\text{'}\text{'}(1)=-6(1)=-6<0\end{array}$

Therefore, the function is concave down for $(0,\infty )$ .

e.

To determine

To find: the intervals on which the function is local extreme values.

Answer to Problem 13RE

local maximum at $(\frac{2\sqrt{3}}{3},\frac{16\sqrt{3}}{3})$ .

Explanation of Solution

Given information: Given function is $y=e^{-x},x\le 0\text{and}y=4x-x^3,x>0.$

Calculation:

  $\begin{array}{c}y=e^{-x},x\le 0\text{and}y=4x-x^3,x>0.\\ y\text{'}=-e^{-x}x\le 0\text{and}y\text{'}=4-3x^2,x>0\\ -e^{-x}\ne \text{0}\text{for}\text{all}\text{values}\text{of }x.\\ 4-3x^2\to 0\to x^2=\frac{4}{3}\to x=\pm \sqrt{\frac{4}{3}}=\pm \frac{2}{\sqrt{3}}=\pm \frac{2\sqrt{3}}{3}\end{array}$

  $\begin{array}{l}\text{For}(-\infty ,0),\text{test}x=-1\\ y\text{'}(-1)=-e^{-1}=-e<0\\ \text{For}(0,\frac{2\sqrt{3}}{3}),\text{test}x=1\\ $ y\text{'}(1)=4-3{(1)}^2=1>0$\text{For}(0,\frac{2\sqrt{3}}{3}),\text{test}x=2\\ $ y\text{'}(1)=4-3{(2)}^2=-8<0$\end{array}$

  $\begin{array}{c}$ y(\frac{2\sqrt{3}}{3})=4(\frac{2\sqrt{3}}{3})-{(\frac{2\sqrt{3}}{3})}^3$=\frac{8\sqrt{3}}{3}-\frac{8\sqrt{3}}{9}\\ =\frac{16\sqrt{3}}{3}\end{array}$

Therefore, local maximum at $(\frac{2\sqrt{3}}{3},\frac{16\sqrt{3}}{3})$ .

e.

To determine

To find: the intervals on which the function is inflection points.

Answer to Problem 13RE

None.

Explanation of Solution

Given information: Given function is $y=e^{-x},x\le 0\text{and}y=4x-x^3,x>0.$

Calculation:

  $\begin{array}{c}y=e^{-x},x\le 0\text{and}y=4x-x^3,x>0.\\ y\text{'}=-e^{-x},x\le 0\text{and}y\text{'}=4-3x^2,x>0\\ y\text{'}\text{'}=e^{-x},,x\le 0\text{and}y\text{'}\text{'}=-6x,x>0\end{array}$

  $\begin{array}{l}\text{For}(-\infty ,0),\text{test}x=-1\\ y\text{'}\text{'}(-1)=e^{-1(-1)}=e>0\\ \text{For}(0,\infty ),\text{test}x=1\\ y\text{'}\text{'}(1)=-6(1)=-6<0\end{array}$

If y’’ switch signs, then x- values is the location of an inflection point. y’’ does switch sign at x =0 but y is discontinuous at this value so it cannot be the location of an inflection point. y therefore has no inflection points.