Chapter 4, Problem 4RE

a.

To determine

To find: To find the increasing intervals for the function $\text{y = x}\sqrt{{\text{4 - x}}^{\text{2}}}$ .

Answer to Problem 4RE

The function is increasing at $\left[\text{-}\sqrt{\text{2}}\text{, }\sqrt{\text{2}}\right]$ .

Explanation of Solution

Given information: The given function is $\text{y = x}\sqrt{{\text{4 - x}}^{\text{2}}}$

Formula used: Product rule: $\frac{\text{d}}{\text{dx}}{\text{ = u}}^{\text{'}}{\text{v + uv}}^{\text{'}}$

Calculation:

The above equation is,

  $\text{y = x}\sqrt{{\text{4 - x}}^{\text{2}}}$

Differentiating the function to get,

  $\begin{array}{c}{\text{f(x)}}^{\text{'}}\text{ = }{\left(\text{x}\sqrt{{\text{4 - x}}^{\text{2}}}\right)}^{\text{'}}\\ {\text{= (x)}}^{\text{'}}\sqrt{{\text{4 - x}}^{\text{2}}}\text{ + x}{\left(\sqrt{{\text{4 - x}}^{\text{2}}}\right)}^{\text{'}}\\ =\sqrt{{\text{4 - x}}^{\text{2}}}\text{ + x}\frac{\text{-2x}\left(\frac{\text{1}}{\text{2}}\right)}{\sqrt{{\text{4 - x}}^{\text{2}}}}\\ =\sqrt{{\text{4 - x}}^{\text{2}}}\text{ - }\frac{{\text{x}}^{\text{2}}}{\sqrt{{\text{4 - x}}^{\text{2}}}}\\ =\frac{{\text{4 - 2x}}^{\text{2}}}{\sqrt{{\text{4 - x}}^{\text{2}}}}\end{array}$

When ${\text{4 - 2x}}^{\text{2}}=0$ ,

  $\begin{array}{c}{\text{f(x)}}^{\text{'}}\text{ = 0}\\ \text{x = }\pm \sqrt{2}\end{array}$

When $\sqrt{{\text{4 - x}}^{\text{2}}}=0$ , ${\text{f(x)}}^{\text{'}}\text{ = 0}$ is undefined

  $\text{x = }\pm 2$

Therefore, the function increases at $\left[\text{-}\sqrt{\text{2}}\text{, }\sqrt{\text{2}}\right]$ .

b.

To determine

To find: To find the decreasing intervals for the function $\text{y = x}\sqrt{{\text{4 - x}}^{\text{2}}}$ .

Answer to Problem 4RE

The function is decreasing at $\left[\text{-2, -}\sqrt{\text{2}}\right]\left[\sqrt{\text{2}}\text{, 2}\right]$ .

Explanation of Solution

Given information: The given function is $\text{y = x}\sqrt{{\text{4 - x}}^{\text{2}}}$

Formula used: Product rule: $\frac{\text{d}}{\text{dx}}{\text{ = u}}^{\text{'}}{\text{v + uv}}^{\text{'}}$

Calculation:

The above equation is,

  $\text{y = x}\sqrt{{\text{4 - x}}^{\text{2}}}$

Differentiating the function to get,

  $\begin{array}{c}{\text{f(x)}}^{\text{'}}\text{ = }{\left(\text{x}\sqrt{{\text{4 - x}}^{\text{2}}}\right)}^{\text{'}}\\ {\text{= (x)}}^{\text{'}}\sqrt{{\text{4 - x}}^{\text{2}}}\text{ + x}{\left(\sqrt{{\text{4 - x}}^{\text{2}}}\right)}^{\text{'}}\\ =\sqrt{{\text{4 - x}}^{\text{2}}}\text{ + x}\frac{\text{-2x}\left(\frac{\text{1}}{\text{2}}\right)}{\sqrt{{\text{4 - x}}^{\text{2}}}}\\ =\sqrt{{\text{4 - x}}^{\text{2}}}\text{ - }\frac{{\text{x}}^{\text{2}}}{\sqrt{{\text{4 - x}}^{\text{2}}}}\\ =\frac{{\text{4 - 2x}}^{\text{2}}}{\sqrt{{\text{4 - x}}^{\text{2}}}}\end{array}$

When ${\text{4 - 2x}}^{\text{2}}=0$ ,

  $\begin{array}{c}{\text{f(x)}}^{\text{'}}\text{ = 0}\\ \text{x = }\pm \sqrt{2}\end{array}$

When $\sqrt{{\text{4 - x}}^{\text{2}}}=0$ , ${\text{f(x)}}^{\text{'}}\text{ = 0}$ is undefined

  $\text{x = }\pm 2$

Therefore, the function decreases at $\left[\text{-2, -}\sqrt{\text{2}}\right]\left[\sqrt{\text{2}}\text{, 2}\right]$ .

c.

To determine

To find: To find the concave up for the function $\text{y = x}\sqrt{{\text{4 - x}}^{\text{2}}}$ .

Answer to Problem 4RE

The interval on which the function is concave up is $\text{(-2, 0)}$ .

Explanation of Solution

Given information: The given function is $\text{y = x}\sqrt{{\text{4 - x}}^{\text{2}}}$

Formula used: Product rule: $\frac{\text{d}}{\text{dx}}{\text{ = u}}^{\text{'}}{\text{v + uv}}^{\text{'}}$

Calculation:

The above equation is,

  $\text{y = x}\sqrt{{\text{4 - x}}^{\text{2}}}$

Differentiating the function to get,

  $\begin{array}{c}{\text{f(x)}}^{\text{'}}\text{ = }{\left(\text{x}\sqrt{{\text{4 - x}}^{\text{2}}}\right)}^{\text{'}}\\ {\text{= (x)}}^{\text{'}}\sqrt{{\text{4 - x}}^{\text{2}}}\text{ + x}{\left(\sqrt{{\text{4 - x}}^{\text{2}}}\right)}^{\text{'}}\\ =\sqrt{{\text{4 - x}}^{\text{2}}}\text{ + x}\frac{\text{-2x}\left(\frac{\text{1}}{\text{2}}\right)}{\sqrt{{\text{4 - x}}^{\text{2}}}}\\ =\sqrt{{\text{4 - x}}^{\text{2}}}\text{ - }\frac{{\text{x}}^{\text{2}}}{\sqrt{{\text{4 - x}}^{\text{2}}}}\\ =\frac{{\text{4 - 2x}}^{\text{2}}}{\sqrt{{\text{4 - x}}^{\text{2}}}}\end{array}$

Using quotient and chain rule, find $\text{y''}$

  $\begin{array}{c}{\text{y}}^{\text{''}}\text{ = }\frac{{\left({\text{4 - 2x}}^{\text{2}}\right)}^{\text{'}}\sqrt{{\text{4 - x}}^{\text{2}}}\text{ - }\left({\text{4 - 2x}}^{\text{2}}\right)\text{×}{\left(\sqrt{{\text{4 - x}}^{\text{2}}}\right)}^{\text{'}}}{{\left(\sqrt{{\text{4 - x}}^{\text{2}}}\right)}^{\text{2}}}\\ \text{=}\frac{\text{-4x}\sqrt{{\text{4 - x}}^{\text{2}}}\text{ - }\left({\text{4 - 2x}}^{\text{2}}\right)\text{×}\frac{\text{1}}{\text{2}\sqrt{{\text{4 - x}}^{\text{2}}}}\text{(-2x)}}{{\left(\sqrt{{\text{4 - x}}^{\text{2}}}\right)}^{\text{2}}}\\ \text{=}\frac{\text{-4x}\sqrt{{\text{4 - x}}^{\text{2}}}\text{+}\frac{\text{x}\left({\text{4 - 2x}}^{\text{2}}\right)}{\sqrt{{\text{4 - x}}^{\text{2}}}}}{{\text{4 - x}}^{\text{2}}}\\ \text{=}\frac{\text{-4x}\left({\text{4 - x}}^{\text{2}}\right)\text{+ x}\left({\text{4 - 2x}}^{\text{2}}\right)}{\left({\text{4 - x}}^{\text{2}}\right)\sqrt{{\text{4 - x}}^{\text{2}}}}\\ \text{=}\frac{{\text{-16x + 4x}}^{\text{3}}{\text{ + 4x - 2x}}^{\text{3}}}{\left({\text{4 - x}}^{\text{2}}\right)\sqrt{{\text{4 - x}}^{\text{2}}}}\\ \text{=}\frac{{\text{2x}}^{\text{3}}\text{ - 12x}}{\left({\text{4 - x}}^{\text{2}}\right)\sqrt{{\text{4 - x}}^{\text{2}}}}\\ \text{=}\frac{\text{2x}\left({\text{x}}^{\text{2}}\text{ - 6}\right)}{\left({\text{4 - x}}^{\text{2}}\right)\sqrt{{\text{4 - x}}^{\text{2}}}}\end{array}$

The value of $\text{y'}$ depends on the expression ${\text{x(x}}^{\text{2}}\text{ - 6)}$ in the numerator at $\text{x = 0}$ .

So, the value of $\text{y''}$ changes at $\text{x = 0}$

For $\text{-2 < x < 0}$ ,

  $\text{y'' > 0}$

For $\text{0 < x < 2}$ ,

  $\text{y'' < 0}$

Therefore, the interval on which the function is concave up is $\text{(-2, 0)}$ .

d.

To determine

To find: To find the concave down for the function $\text{y = x}\sqrt{{\text{4 - x}}^{\text{2}}}$ .

Answer to Problem 4RE

The interval on which the function is concave down is $\text{(0, 2)}$ .

Explanation of Solution

Given information: The given function is $\text{y = x}\sqrt{{\text{4 - x}}^{\text{2}}}$

Calculation:

The above equation is,

  $\text{y = x}\sqrt{{\text{4 - x}}^{\text{2}}}$

Differentiating the function to get,

  $\begin{array}{c}{\text{f(x)}}^{\text{'}}\text{ = }{\left(\text{x}\sqrt{{\text{4 - x}}^{\text{2}}}\right)}^{\text{'}}\\ {\text{= (x)}}^{\text{'}}\sqrt{{\text{4 - x}}^{\text{2}}}\text{ + x}{\left(\sqrt{{\text{4 - x}}^{\text{2}}}\right)}^{\text{'}}\\ =\sqrt{{\text{4 - x}}^{\text{2}}}\text{ + x}\frac{\text{-2x}\left(\frac{\text{1}}{\text{2}}\right)}{\sqrt{{\text{4 - x}}^{\text{2}}}}\\ =\sqrt{{\text{4 - x}}^{\text{2}}}\text{ - }\frac{{\text{x}}^{\text{2}}}{\sqrt{{\text{4 - x}}^{\text{2}}}}\\ =\frac{{\text{4 - 2x}}^{\text{2}}}{\sqrt{{\text{4 - x}}^{\text{2}}}}\end{array}$

Using quotient and chain rule, find $\text{y''}$

  $\begin{array}{c}{\text{y}}^{\text{''}}\text{ = }\frac{{\left({\text{4 - 2x}}^{\text{2}}\right)}^{\text{'}}\sqrt{{\text{4 - x}}^{\text{2}}}\text{ - }\left({\text{4 - 2x}}^{\text{2}}\right)\text{×}{\left(\sqrt{{\text{4 - x}}^{\text{2}}}\right)}^{\text{'}}}{{\left(\sqrt{{\text{4 - x}}^{\text{2}}}\right)}^{\text{2}}}\\ \text{=}\frac{\text{-4x}\sqrt{{\text{4 - x}}^{\text{2}}}\text{ - }\left({\text{4 - 2x}}^{\text{2}}\right)\text{×}\frac{\text{1}}{\text{2}\sqrt{{\text{4 - x}}^{\text{2}}}}\text{(-2x)}}{{\left(\sqrt{{\text{4 - x}}^{\text{2}}}\right)}^{\text{2}}}\\ \text{=}\frac{\text{-4x}\sqrt{{\text{4 - x}}^{\text{2}}}\text{+}\frac{\text{x}\left({\text{4 - 2x}}^{\text{2}}\right)}{\sqrt{{\text{4 - x}}^{\text{2}}}}}{{\text{4 - x}}^{\text{2}}}\\ \text{=}\frac{\text{-4x}\left({\text{4 - x}}^{\text{2}}\right)\text{+ x}\left({\text{4 - 2x}}^{\text{2}}\right)}{\left({\text{4 - x}}^{\text{2}}\right)\sqrt{{\text{4 - x}}^{\text{2}}}}\\ \text{=}\frac{{\text{-16x + 4x}}^{\text{3}}{\text{ + 4x - 2x}}^{\text{3}}}{\left({\text{4 - x}}^{\text{2}}\right)\sqrt{{\text{4 - x}}^{\text{2}}}}\\ \text{=}\frac{{\text{2x}}^{\text{3}}\text{ - 12x}}{\left({\text{4 - x}}^{\text{2}}\right)\sqrt{{\text{4 - x}}^{\text{2}}}}\\ \text{=}\frac{\text{2x}\left({\text{x}}^{\text{2}}\text{ - 6}\right)}{\left({\text{4 - x}}^{\text{2}}\right)\sqrt{{\text{4 - x}}^{\text{2}}}}\end{array}$

The value of $\text{y'}$ depends on the expression ${\text{x(x}}^{\text{2}}\text{ - 6)}$ in the numerator at $\text{x = 0}$ .

So, the value of $\text{y''}$ changes at $\text{x = 0}$

For $\text{-2 < x < 0}$ ,

  $\text{y'' > 0}$

For $\text{0 < x < 2}$ ,

  $\text{y'' < 0}$

Therefore, the interval on which the function is concave down is $\text{(0, 2)}$ .

e.

To determine

To find: To find the local extreme values for the function $\text{y = x}\sqrt{{\text{4 - x}}^{\text{2}}}$ .

Answer to Problem 4RE

The local extreme values have local minimums at $\text{(-}\sqrt{\text{2}}\text{, -2) and (2, 0)}$ , local maximum at $\text{(-2, 0) and (}\sqrt{\text{2}}\text{, 2)}$ .

Explanation of Solution

Given information: The given function is $\text{y = x}\sqrt{{\text{4 - x}}^{\text{2}}}$

Calculation:

The above equation is,

  $\text{y = x}\sqrt{{\text{4 - x}}^{\text{2}}}$

Differentiating the function to get,

  $\begin{array}{c}{\text{f(x)}}^{\text{'}}\text{ = }{\left(\text{x}\sqrt{{\text{4 - x}}^{\text{2}}}\right)}^{\text{'}}\\ {\text{= (x)}}^{\text{'}}\sqrt{{\text{4 - x}}^{\text{2}}}\text{ + x}{\left(\sqrt{{\text{4 - x}}^{\text{2}}}\right)}^{\text{'}}\\ =\sqrt{{\text{4 - x}}^{\text{2}}}\text{ + x}\frac{\text{-2x}\left(\frac{\text{1}}{\text{2}}\right)}{\sqrt{{\text{4 - x}}^{\text{2}}}}\\ =\sqrt{{\text{4 - x}}^{\text{2}}}\text{ - }\frac{{\text{x}}^{\text{2}}}{\sqrt{{\text{4 - x}}^{\text{2}}}}\\ =\frac{{\text{4 - 2x}}^{\text{2}}}{\sqrt{{\text{4 - x}}^{\text{2}}}}\end{array}$

When ${\text{4 - 2x}}^{\text{2}}=0$ ,

  $\begin{array}{c}{\text{f(x)}}^{\text{'}}\text{ = 0}\\ \text{x = }\pm \sqrt{2}\end{array}$

When $\sqrt{{\text{4 - x}}^{\text{2}}}=0$ , ${\text{f(x)}}^{\text{'}}\text{ = 0}$ is undefined

  $\text{x = }\pm 2$

The domain value is $\text{[-2, 2]}$ and the critical value of $\text{x = ±}\sqrt{\text{2}}$

The intervals of the functions are $\text{(-2, -}\sqrt{\text{2}}\text{), (-}\sqrt{\text{2}}\text{, }\sqrt{\text{2}}\text{) and (}\sqrt{\text{2}}\text{, 2)}$

For $\text{(-2, -}\sqrt{\text{2}}\text{)}$ at $\text{x = -1}\text{.5}$

  $\begin{array}{c}{\text{y}}^{\text{'}}\text{(-1}\text{.5) = }\frac{\text{4 - 2(-1}{\text{.5)}}^{\text{2}}}{\sqrt{\text{4 - (-1}{\text{.5)}}^{\text{2}}}}\\ \approx \text{-0}\text{.378 < 0}\end{array}$

For $\text{(-}\sqrt{\text{2}}\text{, }\sqrt{\text{2}}\text{)}$ at $\text{x = 0}$

  $\begin{array}{c}{\text{y}}^{\text{'}}\text{(0) = }\frac{{\text{4 - 2(0)}}^{\text{2}}}{\sqrt{{\text{4 - (0)}}^{\text{2}}}}\\ \text{= 2 > 0}\end{array}$

For $\text{(-}\sqrt{\text{2}}\text{, }\sqrt{\text{2}}\text{)}$ at $\text{x = 1}\text{.5}$

  $\begin{array}{c}{\text{y}}^{\text{'}}\text{(1}\text{.5) = }\frac{\text{4 - 2(1}{\text{.5)}}^{\text{2}}}{\sqrt{\text{4 - (1}{\text{.5)}}^{\text{2}}}}\\ \approx \text{ -0}\text{.378 < 0}\end{array}$

For ${\text{y}}^{\text{'}}\text{ < 0}$ to the right of $\text{x = -2}$ is the local maximum and ${\text{y}}^{\text{'}}$ changes from positive to negative at $\text{x = }\sqrt{2}$ is a local maximum.

For ${\text{y}}^{\text{'}}\text{ < 0}$ to the left of $\text{x = 2}$ is the local minimum and ${\text{y}}^{\text{'}}$ changes from negative to positive at $\text{x = -}\sqrt{2}$ is a local minimum.

To determine the $\text{y}$ coordinates,

  $\begin{array}{c}\text{y(-2) = -2}\sqrt{{\text{4-(-2)}}^{\text{2}}}\\ \text{= 0}\\ \text{y(-}\sqrt{\text{2}}\text{) = -}\sqrt{\text{2}}\sqrt{\text{4 -(-}\sqrt{\text{2}}{\text{)}}^{\text{2}}}\\ \text{= -2}\\ \text{y(}\sqrt{\text{2}}\text{) = }\sqrt{\text{2}}\sqrt{{\text{4 - (2)}}^{\text{2}}}\\ \text{= 2}\\ \text{y(2) = 2}\sqrt{{\text{4 - (2)}}^{\text{2}}}\\ \text{= 0}\end{array}$

Therefore, the local extreme values have local minimums at $\text{(-}\sqrt{\text{2}}\text{, -2) and (2, 0)}$ , local maximum at $\text{(-2, 0) and (}\sqrt{\text{2}}\text{, 2)}$ .

f.

To determine

To find: To find the inflection points for the function $\text{y = x}\sqrt{{\text{4 - x}}^{\text{2}}}$ .

Answer to Problem 4RE

The function has the inflection point at $\text{(0, 0)}$ .

Explanation of Solution

Given information: The given function is $\text{y = x}\sqrt{{\text{4 - x}}^{\text{2}}}$

Calculation:

The above equation is,

  $\text{y = x}\sqrt{{\text{4 - x}}^{\text{2}}}$

Differentiating the function to get,

  $\begin{array}{c}{\text{f(x)}}^{\text{'}}\text{ = }{\left(\text{x}\sqrt{{\text{4 - x}}^{\text{2}}}\right)}^{\text{'}}\\ {\text{= (x)}}^{\text{'}}\sqrt{{\text{4 - x}}^{\text{2}}}\text{ + x}{\left(\sqrt{{\text{4 - x}}^{\text{2}}}\right)}^{\text{'}}\\ =\sqrt{{\text{4 - x}}^{\text{2}}}\text{ + x}\frac{\text{-2x}\left(\frac{\text{1}}{\text{2}}\right)}{\sqrt{{\text{4 - x}}^{\text{2}}}}\\ =\sqrt{{\text{4 - x}}^{\text{2}}}\text{ - }\frac{{\text{x}}^{\text{2}}}{\sqrt{{\text{4 - x}}^{\text{2}}}}\\ =\frac{{\text{4 - 2x}}^{\text{2}}}{\sqrt{{\text{4 - x}}^{\text{2}}}}\end{array}$

Using quotient and chain rule, find $\text{y''}$

  $\begin{array}{c}{\text{y}}^{\text{''}}\text{ = }\frac{{\left({\text{4 - 2x}}^{\text{2}}\right)}^{\text{'}}\sqrt{{\text{4 - x}}^{\text{2}}}\text{ - }\left({\text{4 - 2x}}^{\text{2}}\right)\text{×}{\left(\sqrt{{\text{4 - x}}^{\text{2}}}\right)}^{\text{'}}}{{\left(\sqrt{{\text{4 - x}}^{\text{2}}}\right)}^{\text{2}}}\\ \text{=}\frac{\text{-4x}\sqrt{{\text{4 - x}}^{\text{2}}}\text{ - }\left({\text{4 - 2x}}^{\text{2}}\right)\text{×}\frac{\text{1}}{\text{2}\sqrt{{\text{4 - x}}^{\text{2}}}}\text{(-2x)}}{{\left(\sqrt{{\text{4 - x}}^{\text{2}}}\right)}^{\text{2}}}\\ \text{=}\frac{\text{-4x}\sqrt{{\text{4 - x}}^{\text{2}}}\text{+}\frac{\text{x}\left({\text{4 - 2x}}^{\text{2}}\right)}{\sqrt{{\text{4 - x}}^{\text{2}}}}}{{\text{4 - x}}^{\text{2}}}\\ \text{=}\frac{\text{-4x}\left({\text{4 - x}}^{\text{2}}\right)\text{+ x}\left({\text{4 - 2x}}^{\text{2}}\right)}{\left({\text{4 - x}}^{\text{2}}\right)\sqrt{{\text{4 - x}}^{\text{2}}}}\\ \text{=}\frac{{\text{-16x + 4x}}^{\text{3}}{\text{ + 4x - 2x}}^{\text{3}}}{\left({\text{4 - x}}^{\text{2}}\right)\sqrt{{\text{4 - x}}^{\text{2}}}}\\ \text{=}\frac{{\text{2x}}^{\text{3}}\text{ - 12x}}{\left({\text{4 - x}}^{\text{2}}\right)\sqrt{{\text{4 - x}}^{\text{2}}}}\\ \text{=}\frac{\text{2x}\left({\text{x}}^{\text{2}}\text{ - 6}\right)}{\left({\text{4 - x}}^{\text{2}}\right)\sqrt{{\text{4 - x}}^{\text{2}}}}\end{array}$

The value of $\text{y'}$ depends on the expression ${\text{x(x}}^{\text{2}}\text{ - 6)}$ in the numerator at $\text{x = 0}$ .

So, the value of $\text{y''}$ changes at $\text{x = 0}$

For $\text{-2 < x < 0}$ ,

  $\text{y'' > 0}$

For $\text{0 < x < 2}$ ,

  $\text{y'' < 0}$

Since $\text{y''}$ changes its sign when passing through zero, there is an inflection at $\text{x = 0}$ .

Therefore, the function has the inflection point at $\text{(0, 0)}$ .