Chapter 5, Problem 8RE

a. $$

To determine

Tofindthe interval on which the function $y=\frac{x}{x^3-1}$ is increasing.

Answer to Problem 8RE

The function is then increasing for $x<\frac{-1}{\sqrt[3]{2}}$ .

Explanation of Solution

Given information:

The given function is $y=\frac{x}{x^3-1}$ .

Formula:

Chain rule:

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

Consider the function $y=\frac{x}{x^3-1}$ ,

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

  $y^{\prime }=\frac{-2x^3-1}{{\left(x^3-1\right)}^2}$

  $y^{″}=-\frac{6x^2\left(-x^3-2\right)}{{\left(x^3-1\right)}^3}$

Set the derivative equal to zero,

  $\begin{array}{l}{y}^{\prime }=\frac{-2x^3-1}{{\left(x^3-1\right)}^2}=0\\ \Rightarrow x=-\sqrt[3]{\frac{1}{2}},x=\frac{2^{\frac{2}{3}}}{4}-i\frac{2^{\frac{2}{3}}\sqrt{3}}{4},x=\frac{2^{\frac{2}{3}}}{4}+i\frac{2^{\frac{2}{3}}\sqrt{3}}{4}\end{array}$

Here we consider only the real value $x=-\sqrt[3]{\frac{1}{2}}=-0.794$ .

Consider $x>-\sqrt[3]{\frac{1}{2}}$ :

  $y^{\prime }\left(1\right)=\frac{-2{\left(1\right)}^3-1}{{\left({\left(1\right)}^3-1\right)}^2}$

Therefore, numerator is always positive for $x>-\sqrt[3]{\frac{1}{2}}$ .

Therefore, the function is increasing for $x<\frac{-1}{\sqrt[3]{2}}$ .

b. $$

To determine

To find the interval on which the function $y=\frac{x}{x^3-1}$ is decreasing.

Answer to Problem 8RE

The function is then decreasing for $\left(-2^{-\frac{1}{3}},1\right)$ and $\left(1,\infty \right)$ .

Explanation of Solution

Given information:

The given function is $y=\frac{x}{x^3-1}$ .

Formula:

Chain rule:

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

Consider the function $y=\frac{x}{x^3-1}$ ,

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

  $y^{\prime }=\frac{-2x^3-1}{{\left(x^3-1\right)}^2}$

  $y^{″}=-\frac{6x^2\left(-x^3-2\right)}{{\left(x^3-1\right)}^3}$

Set the derivative equal to zero,

  $\begin{array}{l}{y}^{\prime }=\frac{-2x^3-1}{{\left(x^3-1\right)}^2}=0\\ \Rightarrow x=-\sqrt[3]{\frac{1}{2}},x=\frac{2^{\frac{2}{3}}}{4}-i\frac{2^{\frac{2}{3}}\sqrt{3}}{4},x=\frac{2^{\frac{2}{3}}}{4}+i\frac{2^{\frac{2}{3}}\sqrt{3}}{4}\end{array}$

Here we consider only the real value $x=-\sqrt[3]{\frac{1}{2}}=-0.794$ .

Consider $x<-\sqrt[3]{\frac{1}{2}},x=-1$ :

  $y^{\prime }\left(1\right)=\frac{-2{\left(-1\right)}^3-1}{{\left({\left(-1\right)}^3-1\right)}^2}=\frac{1}{4}>0$

Consider $x=0$ :

  $y^{\prime }\left(0\right)=\frac{-2{\left(0\right)}^3-1}{{\left({\left(0\right)}^3-1\right)}^2}=-1<0$

Consider $x>-\sqrt[3]{\frac{1}{2}},x=2$ :

  $y^{\prime }\left(2\right)=\frac{-2{\left(2\right)}^3-1}{{\left({\left(2\right)}^3-1\right)}^2}=-\frac{9}{49}<0$

Therefore, the function is decreasing for $\left(-2^{-\frac{1}{3}},1\right)$ and $\left(1,\infty \right)$ .

c. $$

To determine

To find the interval on which the function $y=\frac{x}{x^3-1}$ is concave up.

Answer to Problem 8RE

The function is concave up for $\left(-\infty ,-1.26\right)$ and $\left(1,\infty \right)$ .

Explanation of Solution

Given information:

The given function is $y=\frac{x}{x^3-1}$ .

Formula:

Chain rule:

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

Consider the function $y=\frac{x}{x^3-1}$ ,

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

  $y^{\prime }=\frac{-2x^3-1}{{\left(x^3-1\right)}^2}$

  $y^{″}=-\frac{6x^2\left(-x^3-2\right)}{{\left(x^3-1\right)}^3}$

Set the second derivative equal to zero,

  $y^{″}=-\frac{6x^2\left(-x^3-2\right)}{{\left(x^3-1\right)}^3}=0$

  $\begin{array}{l}{y}^{″}=-6x^2\left(x^3+2\right)=0\\ 6x^2\left(x^3+2\right)=0\end{array}$

  $x=0$ or $x=\sqrt[3]{-2}\approx -1.26$

Consider $\left(-\infty ,-1.26\right)$ :

  $y^{″}\left(-2\right)=\frac{16}{81}>0$

Consider $\left(-1.26,0\right)$ :

  $y^{″}\left(-1\right)=-\frac{3}{4}<0$

Consider $\left(0,1\right)$ :

  $y^{″}\left(0.5\right)\approx -4.8<0$

Consider $\left(1,\infty \right)$ :

  $y^{″}\left(02\right)\approx -\frac{240}{343}>0$

Therefore, the function isconcave up for $\left(-\infty ,-1.26\right)$ and $\left(1,\infty \right)$ .

d. $$

To determine

To find the interval on which the function $y=\frac{x}{x^3-1}$ is concave down.

Answer to Problem 8RE

The function is concave down for $\left(-1.26,0\right)$ and $\left(0,1\right)$ .

Explanation of Solution

Given information:

The given function is $y=\frac{x}{x^3-1}$ .

Formula:

Chain rule:

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

Consider the function $y=\frac{x}{x^3-1}$ ,

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

  $y^{\prime }=\frac{-2x^3-1}{{\left(x^3-1\right)}^2}$

  $y^{″}=-\frac{6x^2\left(-x^3-2\right)}{{\left(x^3-1\right)}^3}$

Set the second derivative equal to zero,

  $y^{″}=-\frac{6x^2\left(-x^3-2\right)}{{\left(x^3-1\right)}^3}=0$

  $\begin{array}{l}{y}^{″}=-6x^2\left(x^3+2\right)=0\\ 6x^2\left(x^3+2\right)=0\end{array}$

  $x=0$ or $x=\sqrt[3]{-2}\approx -1.26$

Consider $\left(-\infty ,-1.26\right)$ :

  $y^{″}\left(-2\right)=\frac{16}{81}>0$

Consider $\left(-1.26,0\right)$ :

  $y^{″}\left(-1\right)=-\frac{3}{4}<0$

Consider $\left(0,1\right)$ :

  $y^{″}\left(0.5\right)\approx -4.8<0$

Consider $\left(1,\infty \right)$ :

  $y^{″}\left(02\right)\approx -\frac{240}{343}>0$

Therefore, the function is concave down for $\left(-1.26,0\right)$ and $\left(0,1\right)$ .

e. $$

To determine

To find the interval on which the function $y=\frac{x}{x^3-1}$ has local extreme values.

Answer to Problem 8RE

The function haslocal maximum at $\left(-2^{-\frac{1}{3}},\frac{2}{3}\cdot 2^{-\frac{1}{3}}\right)$ .

Explanation of Solution

Given information:

The given function is $y=\frac{x}{x^3-1}$ .

Formula:

Chain rule:

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

Consider the function $y=\frac{x}{x^3-1}$ ,

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

  $y^{\prime }=\frac{-2x^3-1}{{\left(x^3-1\right)}^2}$

  $y^{″}=-\frac{6x^2\left(-x^3-2\right)}{{\left(x^3-1\right)}^3}$

Set the derivative equal to zero,

  $y^{\prime }=\frac{-2x^3-1}{{\left(x^3-1\right)}^2}=0$

  $\begin{array}{l}\frac{-2x^3-1}{{\left(x^3-1\right)}^2}=0\\ -2x^3-1=0\\ 1=-2x^3\\ x^3=-\frac{1}{2}\\ x=-2^{-\frac{1}{3}}\end{array}$

  $y^{\prime }$ is undefined if it has a division by zero so it will be undefined if

  $\begin{array}{l}\left(x^3-1\right)=0\\ \Rightarrow x=1\end{array}$

The two critical values are $x=-2^{-\frac{1}{3}}$ and $x=1$ which gives the intervals $x<-2^{-\frac{1}{3}},-2^{-\frac{1}{3}}<x<1$ and $x>1$ .

Consider $x=-1$

  $y^{\prime }\left(-1\right)=\frac{-2{\left(-1\right)}^3-1}{{\left({\left(-1\right)}^3-1\right)}^2}=\frac{1}{4}>0$

Consider $x=0$

  $y^{\prime }\left(0\right)=\frac{-2{\left(0\right)}^3-1}{{\left({\left(0\right)}^3-1\right)}^2}=-1<0$

Consider $x=2$

  $y^{\prime }\left(2\right)=\frac{-2{\left(2\right)}^3-1}{{\left({\left(2\right)}^3-1\right)}^2}=-\frac{9}{49}<0$

  $y^{\prime }$ only switches signs at $x=-2^{-\frac{1}{3}}$ . Since it switches from positive to negative, it is local maximum.

  $\begin{array}{l}y=\frac{x}{x^3-1}\\ \text{ =}\frac{-2^{-\frac{1}{3}}}{{\left(-2^{-\frac{1}{3}}\right)}^3-1}\\ =\frac{-2^{-\frac{1}{3}}}{-\frac{1}{2}-1}\\ =\frac{-2^{-\frac{1}{3}}}{-\frac{3}{2}}\\ =\frac{2}{3}\cdot 2^{-\frac{1}{3}}\end{array}$

Therefore, the function has local maximum at $\left(-2^{-\frac{1}{3}},\frac{2}{3}\cdot 2^{-\frac{1}{3}}\right)$ .

f. $$

To determine

To find the interval on which the function $y=\frac{x}{x^3-1}$ has inflection points.

Answer to Problem 8RE

The function hasinflection point $x=\sqrt[3]{-2}\approx -1.26$ .

Explanation of Solution

Given information:

The given function is $y=\frac{x}{x^3-1}$ .

Formula:

Chain rule:

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

Consider the function $y=\frac{x}{x^3-1}$ ,

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

  $y^{\prime }=\frac{-2x^3-1}{{\left(x^3-1\right)}^2}$

  $y^{″}=-\frac{6x^2\left(-x^3-2\right)}{{\left(x^3-1\right)}^3}$

Set the second derivative equal to zero,

  $y^{″}=-\frac{6x^2\left(-x^3-2\right)}{{\left(x^3-1\right)}^3}=0$

  $\begin{array}{l}6x^2\left(-x^3-2\right)=0\\ x=0\text{ and }x=-2^{-\frac{1}{3}}\approx -1.26\end{array}$

Direction of concavity changes only around $x\approx -1.26$ so that is the only inflection point.

Consider $\left(-\infty ,-1.26\right)$

  $y^{″}\left(2\right)=\frac{16}{81}>0$

Consider $\left(-\infty ,-1.26\right)$

  $y^{″}\left(-1\right)=-\frac{3}{4}<0$

Consider $\left(0,1\right)$

  $y^{″}\left(0.5\right)=-4.8<0$

Therefore, the function hasinflection point $x=\sqrt[3]{-2}\approx -1.26$ .