Chapter 5, Problem 7RE

a. $$

To determine

Tofindthe interval on which the function $y=\frac{1}{\sqrt[4]{1-x^2}}$ is increasing.

Answer to Problem 7RE

The function is then increasing for $\left(0,1\right)$ .

Explanation of Solution

Given information:

The given function is $y=\frac{1}{\sqrt[4]{1-x^2}}$ .

Formula:

Chain rule:

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

Consider the function $y=\frac{1}{\sqrt[4]{1-x^2}}$ ,

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

  $y^{\prime }=\frac{x}{2{\left(-x^2+1\right)}^{\frac{5}{4}}}$

  $y^{″}=\frac{3x^2+2}{4{\left(-x^2+1\right)}^{\frac{9}{4}}}$

Set the derivative equal to zero,

  $y^{\prime }=\frac{x}{2{\left(-x^2+1\right)}^{\frac{5}{4}}}=0$

  $y^{\prime }=0$ only when $x=0$ . Hence the domain $y$ is $-1<x<1$

Consider $x=0.5$ :

  $y^{\prime }\left(0.5\right)=y^{\prime }=\frac{x}{2{\left(-x^2+1\right)}^{\frac{5}{4}}}\approx -0.358<0$

Consider $x=0.5$ :

  $y^{\prime }\left(-0.5\right)=y^{\prime }=\frac{x}{2{\left(-x^2+1\right)}^{\frac{5}{4}}}\approx 0.358>0$

Therefore, the function is increasing for $\left(0,1\right)$ .

b. $$

To determine

To find the interval on which the function $y=\frac{1}{\sqrt[4]{1-x^2}}$ is decreasing.

Answer to Problem 7RE

The function is then decreasing for $\left(-1,0\right)$ .

Explanation of Solution

Given information:

The given function is $y=\frac{1}{\sqrt[4]{1-x^2}}$ .

Formula:

Chain rule:

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

Consider the function $y=\frac{1}{\sqrt[4]{1-x^2}}$ ,

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

  $y^{\prime }=\frac{x}{2{\left(-x^2+1\right)}^{\frac{5}{4}}}$

  $y^{″}=\frac{3x^2+2}{4{\left(-x^2+1\right)}^{\frac{9}{4}}}$

Set the derivative equal to zero,

  $y^{\prime }=\frac{x}{2{\left(-x^2+1\right)}^{\frac{5}{4}}}=0$

  $y^{\prime }=0$ only when $x=0$ . Hence the domain $y$ is $-1<x<1$

Consider $x=0.5$ :

  $y^{\prime }\left(0.5\right)=y^{\prime }=\frac{x}{2{\left(-x^2+1\right)}^{\frac{5}{4}}}\approx -0.358<0$

Consider $x=0.5$ :

  $y^{\prime }\left(-0.5\right)=y^{\prime }=\frac{x}{2{\left(-x^2+1\right)}^{\frac{5}{4}}}\approx 0.358>0$

Therefore, the function is decreasing for $\left(-1,0\right)$ .

c. $$

To determine

To find the interval on which the function $y=\frac{1}{\sqrt[4]{1-x^2}}$ is concave up.

Answer to Problem 7RE

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

Explanation of Solution

Given information:

The given function is $y=\frac{1}{\sqrt[4]{1-x^2}}$ .

Formula:

Chain rule:

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

Consider the function $y=\frac{1}{\sqrt[4]{1-x^2}}$ ,

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

  $y^{\prime }=\frac{x}{2{\left(-x^2+1\right)}^{\frac{5}{4}}}$

  $y^{″}=\frac{3x^2+2}{4{\left(-x^2+1\right)}^{\frac{9}{4}}}$

Set the second derivative equal to zero,

  $\begin{array}{l}{y}^{″}=\frac{3x^2+2}{4{\left(-x^2+1\right)}^{\frac{9}{4}}}=0\\ \Rightarrow x=0\end{array}$

  $y^{″}>0$ throughout the entire domain of $y$ , so $y$ is always concave up for $\left(-1,1\right)$

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

d. $$

To determine

To find the interval on which the function $y=\frac{1}{\sqrt[4]{1-x^2}}$ is concave down.

Answer to Problem 7RE

The function is neverconcave down.

Explanation of Solution

Given information:

The given function is $y=\frac{1}{\sqrt[4]{1-x^2}}$ .

Formula:

Chain rule:

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

Consider the function $y=\frac{1}{\sqrt[4]{1-x^2}}$ ,

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

  $y^{\prime }=\frac{x}{2{\left(-x^2+1\right)}^{\frac{5}{4}}}$

  $y^{″}=\frac{3x^2+2}{4{\left(-x^2+1\right)}^{\frac{9}{4}}}$

Set the second derivative equal to zero,

  $\begin{array}{l}{y}^{″}=\frac{3x^2+2}{4{\left(-x^2+1\right)}^{\frac{9}{4}}}=0\\ \Rightarrow x=0\end{array}$

  $y^{″}>0$ throughout the entire domain of $y$ , so $y$ is never concave down.

Therefore, the function is never concave down.

e. $$

To determine

To find the interval on which the function $y=\frac{1}{\sqrt[4]{1-x^2}}$ has local extreme values.

Answer to Problem 7RE

The function haslocal minimum at $\left(0,1\right)$ .

Explanation of Solution

Given information:

The given function is $y=\frac{1}{\sqrt[4]{1-x^2}}$ .

Formula:

Chain rule:

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

Consider the function $y=\frac{1}{\sqrt[4]{1-x^2}}$ ,

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

  $y^{\prime }=\frac{x}{2{\left(-x^2+1\right)}^{\frac{5}{4}}}$

  $y^{″}=\frac{3x^2+2}{4{\left(-x^2+1\right)}^{\frac{9}{4}}}$

Set the derivative equal to zero,

  $y^{\prime }=\frac{x}{2{\left(-x^2+1\right)}^{\frac{5}{4}}}=0$ when $x=0$ .

  $y^{″}>0$ , throughout the entire domain of $y$ .

Hence at $x=0$ the local minimum occurs

  $y\left(0\right)=1$

Therefore, the function has local minimum at $\left(0,1\right)$ .

f. $$

To determine

To find the interval on which the function $y=\frac{1}{\sqrt[4]{1-x^2}}$ has inflection points.

Answer to Problem 7RE

The function has noinflection point.

Explanation of Solution

Given information:

The given function is $y=\frac{1}{\sqrt[4]{1-x^2}}$ .

Formula:

Chain rule:

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

Consider the function $y=\frac{1}{\sqrt[4]{1-x^2}}$ ,

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

  $y^{\prime }=\frac{x}{2{\left(-x^2+1\right)}^{\frac{5}{4}}}$

  $y^{″}=\frac{3x^2+2}{4{\left(-x^2+1\right)}^{\frac{9}{4}}}$

Set the second derivative equal to zero,

  $y^{″}=\frac{3x^2+2}{4{\left(-x^2+1\right)}^{\frac{9}{4}}}=0$

Here, the second derivative is negative for all values of $x$ .

Hence, the function is defined for $-1<x<1$ and the first derivative has the same sign as $x$ .

  $\begin{array}{l}{y}^{″}=e^{x-1}=\text{no solution}\\ \Rightarrow x=1\end{array}$

Since $y^{″}>0$ for any $x$ , the function has no inflection points.

Therefore, the function has no inflection point.