Chapter 4, Problem 7RE

$\text{(a)}$

To determine

To find: The interval on that the function is increasing.

Answer to Problem 7RE

The answer: Increasing $(0,1)$

Explanation of Solution

Given information:

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

Calculation:

  $\begin{array}{c}y={\left(1-x^2\right)}^{-1/4}\\ y^{\prime }=-\frac{1}{4}{\left(1-x^2\right)}^{-5/4}(-2x)\\ =\frac{1}{2}x{\left(1-x^2\right)}^{-5/4}\\ y^{″}=\frac{1}{2}\left({\left(1-x^2\right)}^{-5/4}+x\left(-\frac{5}{4}{\left(1-x^2\right)}^{-9/4}(-2x)\right)\right)\\ =\frac{1}{2}\left(\frac{1}{{\left(1-x^2\right)}^{5/4}}+\frac{5x^2}{2{\left(1-x^2\right)}^{9/4}}\right)\\ =\frac{1}{2}\left(\frac{2-2x^2+5x^2}{2{\left(1-x^2\right)}^{9/4}}\right)\\ =\frac{3x^2+2}{4{\left(1-x^2\right)}^{9/4}}\end{array}$

Interval

  $(-1,0)$ ,

Then,

  $y^{\prime }(-0.5)\approx -0.35819<0$

Decreasing.

interval $(0,1)$ ,

So,

  $y^{\prime }(0.5)\approx 0.35819>0$

Increasing.

Therefore, the required interval on that the function is increasing $(0,1)$ .

$\text{(b)}$

To determine

To find: The interval on that the function is decreasing

Answer to Problem 7RE

The answer: Decreasing $(-1,0)$

Explanation of Solution

Given information:

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

Calculation:

  $\begin{array}{c}y={\left(1-x^2\right)}^{-1/4}\\ y^{\prime }=-\frac{1}{4}{\left(1-x^2\right)}^{-5/4}(-2x)\\ =\frac{1}{2}x{\left(1-x^2\right)}^{-5/4}\\ y^{″}=\frac{1}{2}\left({\left(1-x^2\right)}^{-5/4}+x\left(-\frac{5}{4}{\left(1-x^2\right)}^{-9/4}(-2x)\right)\right)\\ =\frac{1}{2}\left(\frac{1}{{\left(1-x^2\right)}^{5/4}}+\frac{5x^2}{2{\left(1-x^2\right)}^{9/4}}\right)\\ =\frac{1}{2}\left(\frac{2-2x^2+5x^2}{2{\left(1-x^2\right)}^{9/4}}\right)\\ =\frac{3x^2+2}{4{\left(1-x^2\right)}^{9/4}}\end{array}$

Interval

  $(-1,0)$ ,

Then,

  $y^{\prime }(-0.5)\approx -0.35819<0$

Decreasing.

interval $(0,1)$ ,

So,

  $y^{\prime }(0.5)\approx 0.35819>0$

Increasing.

Therefore, the required interval on that the function is decreasing $(-1,0)$ .

$(c)$

To determine

To find: The interval on that the function is concave up.

Answer to Problem 7RE

The answer: Concave up in $(-1,1)$

Explanation of Solution

Given information:

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

Calculation:

  $\begin{array}{c}y={\left(1-x^2\right)}^{-1/4}\\ y^{\prime }=-\frac{1}{4}{\left(1-x^2\right)}^{-5/4}(-2x)\\ =\frac{1}{2}x{\left(1-x^2\right)}^{-5/4}\\ y^{″}=\frac{1}{2}\left({\left(1-x^2\right)}^{-5/4}+x\left(-\frac{5}{4}{\left(1-x^2\right)}^{-9/4}(-2x)\right)\right)\\ =\frac{1}{2}\left(\frac{1}{{\left(1-x^2\right)}^{5/4}}+\frac{5x^2}{2{\left(1-x^2\right)}^{9/4}}\right)\\ =\frac{1}{2}\left(\frac{2-2x^2+5x^2}{2{\left(1-x^2\right)}^{9/4}}\right)\\ =\frac{3x^2+2}{4{\left(1-x^2\right)}^{9/4}}\end{array}$

Then,

  $y^{″}$ is $>0$ throughout the entire domain of $y$ , so $y$ is always concave up in $(-1,1)$ .

So,

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

Therefore, the required interval on that the function is concave up $(-1,1)$ .

$(d)$

To determine

To find: The interval on that the function is concave down.

Answer to Problem 7RE

The answer: Never concave down

Explanation of Solution

Given information:

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

Calculation:

  $\begin{array}{c}y={\left(1-x^2\right)}^{-1/4}\\ y^{\prime }=-\frac{1}{4}{\left(1-x^2\right)}^{-5/4}(-2x)\\ =\frac{1}{2}x{\left(1-x^2\right)}^{-5/4}\\ y^{″}=\frac{1}{2}\left({\left(1-x^2\right)}^{-5/4}+x\left(-\frac{5}{4}{\left(1-x^2\right)}^{-9/4}(-2x)\right)\right)\\ =\frac{1}{2}\left(\frac{1}{{\left(1-x^2\right)}^{5/4}}+\frac{5x^2}{2{\left(1-x^2\right)}^{9/4}}\right)\\ =\frac{1}{2}\left(\frac{2-2x^2+5x^2}{2{\left(1-x^2\right)}^{9/4}}\right)\\ =\frac{3x^2+2}{4{\left(1-x^2\right)}^{9/4}}\end{array}$

Then,

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

So,

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

Therefore, the required interval on that the function is never concave down.

$\text{(e)}$

To determine

To find: The any local extreme value.

Answer to Problem 7RE

The answer: Local minimum $(0,1)$

Explanation of Solution

Given information:

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

Calculation:

  $\begin{array}{c}y={\left(1-x^2\right)}^{-1/4}\\ y^{\prime }=-\frac{1}{4}{\left(1-x^2\right)}^{-5/4}(-2x)\\ =\frac{1}{2}x{\left(1-x^2\right)}^{-5/4}\\ y^{″}=\frac{1}{2}\left({\left(1-x^2\right)}^{-5/4}+x\left(-\frac{5}{4}{\left(1-x^2\right)}^{-9/4}(-2x)\right)\right)\\ =\frac{1}{2}\left(\frac{1}{{\left(1-x^2\right)}^{5/4}}+\frac{5x^2}{2{\left(1-x^2\right)}^{9/4}}\right)\\ =\frac{1}{2}\left(\frac{2-2x^2+5x^2}{2{\left(1-x^2\right)}^{9/4}}\right)\\ =\frac{3x^2+2}{4{\left(1-x^2\right)}^{9/4}}\end{array}$

Based on the information from previous steps, $x=0$ is the local minimum.

  $y(0)=1$

So,

  $y^{\prime }=0$

When

  $x=0$

And

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

Therefore, the required value of local minimum $(0,1)$ .

$\text{(f)}$

To determine

To find: The inflection point.

Answer to Problem 7RE

The answer: No inflection point.

Explanation of Solution

Given information:

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

Calculation:

  $\begin{array}{c}{y}^{\prime }=\frac{-x}{2{\left(1-x^2\right)}^{5/4}}\\ y^{″}=\frac{-2{\left(1-x^2\right)}^{5/4}-5x^2{\left(1-x^2\right)}^{1/4}}{4{\left(1-x^2\right)}^{10/4}}\\ =\frac{-2-3x^2}{4{\left(1-x^2\right)}^{9/4}}<0\end{array}$

The function is defined for $\text{r}-1<x<1$ and the 1st derivative has the same sign as $x$ . Its 2nd derivative is negative for all $x$ .

Therefore, the required no inflection point of the given equation.