Chapter 5, Problem 4RE

a. $$

To determine

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

Answer to Problem 4RE

The function is then increasing for $x\in \left[-\sqrt{2},\sqrt{2}\right]$ .

Explanation of Solution

Given information:

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

Formula:

Chain rule:

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

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

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

Here $u=x\Rightarrow u^{\prime }=1$ and $v=\sqrt{4-x^2}\Rightarrow v^{\prime }=-\frac{x}{\sqrt{4-x^2}}$ ,

  $\begin{array}{l}{y}^{\prime }=1\cdot \sqrt{4-x^2}+\left(-\frac{x}{\sqrt{4-x^2}}\right)x\\ y^{\prime }=\sqrt{4-x^2}-\frac{x^2}{\sqrt{4-x^2}}\\ y^{\prime }=\frac{4-x^2-x^2}{\sqrt{4-x^2}}=\frac{4-2x^2}{\sqrt{4-x^2}}\end{array}$

Now setting the derivative equal to zero to find the critical values,

  $\begin{array}{l}\frac{4-2x^2}{\sqrt{4-x^2}}=0\\ 4-2x^2=0\\ 2x^2=4\\ x^2=2\\ x=\pm 2\end{array}$

  $y^{\prime }$ is undefined when

  $\begin{array}{l}\sqrt{4-x^2}=0\\ 4-x^2=0\\ 4=x^2\Rightarrow x=\pm 2\end{array}$

Therefore, the critical values are $x=2,x=-2,x=\sqrt{2},x=-\sqrt{2}$ .

From the critical values the possible intervals are $\left[-2,-\sqrt{2}\right],\left[-\sqrt{2},\sqrt{2}\right],\left[\sqrt{2},2\right]$ .

Now to determine which of the intervals is increasing,

Consider $x=-2$ for $\left[-2,-\sqrt{2}\right]$ :

  $y^{\prime }=\frac{4-2{\left(-2\right)}^2}{\sqrt{4-{\left(-2\right)}^2}}=undefined$

Consider $x=\sqrt{2}$ for $\left[-\sqrt{2},\sqrt{2}\right]$ :

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

Consider $x=2$ for $\left[\sqrt{2},2\right]$ :

  $y^{\prime }=\frac{4-2{\left(-2\right)}^2}{\sqrt{4-{\left(-2\right)}^2}}=undefined$

Therefore, the function is increasing for $x\in \left[-\sqrt{2},\sqrt{2}\right]$ .

b. $$

To determine

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

Answer to Problem 4RE

The function is then decreasing for $(-\infty ,-2]$ and $[2,\infty )$ .

Explanation of Solution

Given information:

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

Formula:

Chain rule:

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

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

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

Here $u=x\Rightarrow u^{\prime }=1$ and $v=\sqrt{4-x^2}\Rightarrow v^{\prime }=-\frac{x}{\sqrt{4-x^2}}$ ,

  $\begin{array}{l}{y}^{\prime }=1\cdot \sqrt{4-x^2}+\left(-\frac{x}{\sqrt{4-x^2}}\right)x\\ y^{\prime }=\sqrt{4-x^2}-\frac{x^2}{\sqrt{4-x^2}}\\ y^{\prime }=\frac{4-x^2-x^2}{\sqrt{4-x^2}}=\frac{4-2x^2}{\sqrt{4-x^2}}\end{array}$

Now setting the derivative equal to zero to find the critical values,

  $\begin{array}{l}\frac{4-2x^2}{\sqrt{4-x^2}}=0\\ 4-2x^2=0\\ 2x^2=4\\ x^2=2\\ x=\pm 2\end{array}$

  $y^{\prime }$ is undefined when

  $\begin{array}{l}\sqrt{4-x^2}=0\\ 4-x^2=0\\ 4=x^2\Rightarrow x=\pm 2\end{array}$

Therefore, the critical values are $x=2,x=-2,x=\sqrt{2},x=-\sqrt{2}$ .

From the critical values the possible intervals are $\left[-2,-\sqrt{2}\right],\left[-\sqrt{2},\sqrt{2}\right],\left[\sqrt{2},2\right]$ .

Now to determine which of the intervals is decreasing,

Consider $x=-2$ for $\left[-2,-\sqrt{2}\right]$ :

  $y^{\prime }=\frac{4-2{\left(-2\right)}^2}{\sqrt{4-{\left(-2\right)}^2}}=undefined$

Consider $x=\sqrt{2}$ for $\left[-\sqrt{2},\sqrt{2}\right]$ :

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

Consider $x=2$ for $\left[\sqrt{2},2\right]$ :

  $y^{\prime }=\frac{4-2{\left(-2\right)}^2}{\sqrt{4-{\left(-2\right)}^2}}=undefined$

Therefore, the function is decreasing for $\left[-2,-\sqrt{2}\right]$ and $\left[\sqrt{2},2\right]$ .

c. $$

To determine

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

Answer to Problem 4RE

The function is concave up for $\left(-2,0\right)$ .

Explanation of Solution

Given information:

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

Formula:

Chain rule:

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

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

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

  $\begin{array}{l}{y}^{\prime }=1\cdot \sqrt{4-x^2}+\left(-\frac{x}{\sqrt{4-x^2}}\right)x\\ y^{\prime }=\frac{-2x^2+4}{\sqrt{4-x^2}}\end{array}$

Find second derivative,

  $\begin{array}{l}{y}^{″}=\frac{\left(-4x\right)\sqrt{4-x^2}-\left(-\frac{x}{\sqrt{4-x^2}}\right)\left(-2x^2+4\right)}{{\left(\sqrt{4-x^2}\right)}^2}\\ y^{″}=\frac{2x^3-12x}{\left(4-x^2\right)\sqrt{4-x^2}}\end{array}$

The function is undefined whenever the denominator is equal to zero.

  $\begin{array}{l}\left(4-x^2\right)\sqrt{4-x^2}=0\\ \left(4-x^2\right)=0\\ 4=x^2\\ x=\pm 2\end{array}$

Therefore, the inflection points are at $x=\pm 2$ . And the function is zero whenever the numerator is zero.

  $\begin{array}{l}2x^3-12x=0\\ 2x^3=12x\\ 2x^2=12\\ x^2=6\\ x=\pm \sqrt{6}\end{array}$

Inflection point at $x=0$ has no inflection points at $x=\pm \sqrt{6}$ because $\pm \sqrt{6}$

is not in the domain of the function.

Consider $x=-1$ ,

  $\begin{array}{l}{y}^{″}=\frac{2{\left(-1\right)}^3-12\left(-1\right)}{\left(4-{\left(-1\right)}^2\right)\sqrt{4-{\left(-1\right)}^2}}\\ y^{″}=1.92>0\end{array}$

Consider $x=1$ ,

  $\begin{array}{l}{y}^{″}=\frac{2{\left(1\right)}^3-12\left(1\right)}{\left(4-{\left(1\right)}^2\right)\sqrt{4-{\left(1\right)}^2}}\\ y^{″}=-1.92<0\end{array}$

Therefore, the function is concave up from $\left(-2,0\right)$ .

d. $$

To determine

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

Answer to Problem 4RE

The function is concave down for $\left(0,2\right)$ .

Explanation of Solution

Given information:

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

Formula:

Chain rule:

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

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

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

  $\begin{array}{l}{y}^{\prime }=1\cdot \sqrt{4-x^2}+\left(-\frac{x}{\sqrt{4-x^2}}\right)x\\ y^{\prime }=\frac{-2x^2+4}{\sqrt{4-x^2}}\end{array}$

Find second derivative,

  $\begin{array}{l}{y}^{″}=\frac{\left(-4x\right)\sqrt{4-x^2}-\left(-\frac{x}{\sqrt{4-x^2}}\right)\left(-2x^2+4\right)}{{\left(\sqrt{4-x^2}\right)}^2}\\ y^{″}=\frac{2x^3-12x}{\left(4-x^2\right)\sqrt{4-x^2}}\end{array}$

The function is undefined whenever the denominator is equal to zero.

  $\begin{array}{l}\left(4-x^2\right)\sqrt{4-x^2}=0\\ \left(4-x^2\right)=0\\ 4=x^2\\ x=\pm 2\end{array}$

Therefore, the inflection points are at $x=\pm 2$ . And the function is zero whenever the numerator is zero.

  $\begin{array}{l}2x^3-12x=0\\ 2x^3=12x\\ 2x^2=12\\ x^2=6\\ x=\pm \sqrt{6}\end{array}$

Inflection point at $x=0$ has no inflection points at $x=\pm \sqrt{6}$ because $\pm \sqrt{6}$

is not in the domain of the function.

Consider $x=-1$ ,

  $\begin{array}{l}{y}^{″}=\frac{2{\left(-1\right)}^3-12\left(-1\right)}{\left(4-{\left(-1\right)}^2\right)\sqrt{4-{\left(-1\right)}^2}}\\ y^{″}=1.92>0\end{array}$

Consider $x=1$ ,

  $\begin{array}{l}{y}^{″}=\frac{2{\left(1\right)}^3-12\left(1\right)}{\left(4-{\left(1\right)}^2\right)\sqrt{4-{\left(1\right)}^2}}\\ y^{″}=-1.92<0\end{array}$

Therefore, the function is concave down from $\left(0,2\right)$ .

e. $$

To determine

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

Answer to Problem 4RE

The function haslocal maximum values at $\left(-2,0\right)$ and $\left(\sqrt{2},2\right)$ and local minimum values at $\left(-\sqrt{2},-2\right)$ and $\left(2,0\right)$ .

Explanation of Solution

Given information:

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

Formula:

Chain rule:

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

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

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

Here $u=x\Rightarrow u^{\prime }=1$ and $v=\sqrt{4-x^2}\Rightarrow v^{\prime }=\frac{1}{2}{\left(4-x^2\right)}^{-\frac{1}{2}}\left(-2x\right)=-\frac{x}{{\left(4-x^2\right)}^{\frac{1}{2}}}$ ,

  $\begin{array}{l}{y}^{\prime }=1\cdot \sqrt{4-x^2}+\left(-\frac{x}{\sqrt{4-x^2}}\right)x\\ y^{\prime }=\frac{-2x^2+4}{\sqrt{4-x^2}}\end{array}$

Now setting the derivative equal to zero to find the critical values,

  $\begin{array}{l}{y}^{\prime }=\frac{-2x^2+4}{\sqrt{4-x^2}}=0\\ -2x^2+4=0\\ 2x^2=4\\ x^2=2\\ x=\pm \sqrt{2}\end{array}$

  $y^{\prime }$ is undefined if the denominator is equal to zero so set the denominator is equal to zero so set the denominator to zero.

  $\begin{array}{l}\sqrt{4-x^2}=0\\ 4-x^2=0\\ x^2=4\\ x=\pm 2\end{array}$

The domain of $\left[-2,2\right]$ and critical values of $x=\pm \sqrt{2}$ then the possible intervals are given by $\left(-2,-\sqrt{2}\right),\left(-\sqrt{2},\sqrt{2}\right),\left(\sqrt{2},2\right)$

Consider $x=-1.5$ for $\left(-2,-\sqrt{2}\right)$ :

  $y^{\prime }\left(-1.5\right)=\frac{-2{\left(-1.5\right)}^2+4}{\sqrt{4-{\left(-1.5\right)}^2}}\approx -0.378<0$

Consider $x=0$ for $\left(-\sqrt{2},\sqrt{2}\right)$ :

  $y^{\prime }\left(0\right)=\frac{-2{\left(0\right)}^2+4}{\sqrt{4-{\left(0\right)}^2}}\approx 2>0$

Consider $x=1.5$ for $\left(\sqrt{2},2\right)$ :

  $y^{\prime }\left(1.5\right)=\frac{-2{\left(1.5\right)}^2+4}{\sqrt{4-{\left(1.5\right)}^2}}\approx -0.378<0$

Since $y^{\prime }<0$ to the right of $x=-2$ then the endpoint at $x=-2$ is a local maximum.

Since $y^{\prime }$ switches from negative to positive at $x=-\sqrt{2}$ then it is a local minimum. Since $y^{\prime }$ switches from positive to negative at $x=\sqrt{2}$ then it is a local maximum.

  $\begin{array}{l}y\left(-2\right)=\left(-2\right)\sqrt{4-{\left(-2\right)}^2}=0\\ y\left(-\sqrt{2}\right)=\left(-\sqrt{2}\right)\sqrt{4-{\left(-\sqrt{2}\right)}^2}=-2\\ y\left(\sqrt{2}\right)=\left(\sqrt{2}\right)\sqrt{4-{\left(\sqrt{2}\right)}^2}=2\\ y\left(2\right)=\left(2\right)\sqrt{4-{\left(2\right)}^2}=0\end{array}$

Therefore, the function has local maximum values at $\left(-2,0\right)$ and $\left(\sqrt{2},2\right)$ and local minimum values at $\left(-\sqrt{2},-2\right)$ and $\left(2,0\right)$ .

f. $$

To determine

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

Answer to Problem 4RE

The function hasinflection point at $x=0$ .

Explanation of Solution

Given information:

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

Formula:

Chain rule:

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

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

Using product and chain rule: 6 $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$ find the derivative,

  $\begin{array}{l}{y}^{\prime }=1\cdot \sqrt{4-x^2}+\left(-\frac{x}{\sqrt{4-x^2}}\right)x\\ y^{\prime }=\frac{-2x^2+4}{\sqrt{4-x^2}}\end{array}$

Find second derivative,

  $\begin{array}{l}{y}^{″}=\frac{\left(-4x\right)\sqrt{4-x^2}-\left(-\frac{x}{\sqrt{4-x^2}}\right)\left(-2x^2+4\right)}{{\left(\sqrt{4-x^2}\right)}^2}\\ y^{″}=\frac{2x^3-12x}{\left(4-x^2\right)\sqrt{4-x^2}}\end{array}$

The sign of $y^{\prime }$ is determined by the sign of $-2x^2+4$ ,

  $\begin{array}{l}-2x^2+4=0\\ x^2=2\\ x=\pm \sqrt{2}\end{array}$

When $x=-2$ ,

  $y^{\prime }=\frac{-2x^2+4}{\sqrt{4-x^2}}=0$ which is minimum.

When $-2<x<2$ ,

  $y^{\prime }>0$ which is increasing.

When $x=2$ ,

  $y^{\prime }=\frac{-2x^2+4}{\sqrt{4-x^2}}=0$ which is maximum.

Now to determine the sign of $2x\left(x^2-6\right)$ in second derivative .

  $-2\le x<0\Rightarrow y^{″}>0$ which is concave up.

  $x=0\Rightarrow y^{″}=0$ which is inflection point.

  $0\le x<2\Rightarrow y^{″}<0$ which is concave down.

Therefore, the function has inflection point at $x=0$ .