Chapter 5, Problem 15RE

a.

To determine

To find the intervals on which the function is increasing by using analytical method.

Answer to Problem 15RE

The function is increasing in interval $\left(0,0.89\right)$ .

Explanation of Solution

Given:

The function is

  $y=x^{4/5}\left(2-x\right)$ .

Calculation:

The function $y=x^{4/5}\left(2-x\right)$ is increasing when $f^{\prime }\left(x\right)>0$

  $f^{\prime }\left(x\right)=-x^{4/5}+\frac{4}{5}{x}^{-1/5}\left(2-x\right)=\frac{8}{5}{x}^{-1/5}-\frac{9}{5}{x}^{4/5}$

Now put $f^{\prime }\left(x\right)=0$ to find critical points

  $\begin{array}{l}\frac{8}{5}{x}^{-1/5}-\frac{9}{5}{x}^{4/5}=0\\ \frac{1}{5}{x}^{-1/5}\left(8-9x\right)=0\\ \because x^{-1/5}\ne 0\\ \therefore 8-9x=0\\ x=\frac{8}{9}=0.89\end{array}$

So , there are three intervals that is $\left(-\infty ,0\right),\left(0,0.89\right)\text{ and }\left(0.89,\infty \right)$ to check whether $f^{\prime }\left(x\right)>0$ or $f^{\prime }\left(x\right)<0$

put $x=-1$ to check whether the function is increasing or decreasing in $\left(-\infty ,0\right)$

  $f^{\prime }\left(x\right)=\frac{8}{5}{x}^{-1/5}-\frac{9}{5}{x}^{4/5}=\frac{8}{5}{\left(-1\right)}^{-1/5}-\frac{9}{5}{\left(-1\right)}^{4/5}=-3.4<0$

put $x=0.5$ to check whether the function is increasing or decreasing in $\left(0,0.89\right)$

  $f^{\prime }\left(x\right)=\frac{8}{5}{x}^{-1/5}-\frac{9}{5}{x}^{4/5}=\frac{8}{5}{\left(0.5\right)}^{-1/5}-\frac{9}{5}{\left(0.5\right)}^{4/5}=0.80>0$

put $x=1$ to check whether the function is increasing or decreasing in $\left(0.89,\infty \right)$

  $f^{\prime }\left(x\right)=\frac{8}{5}{x}^{-1/5}-\frac{9}{5}{x}^{4/5}=\frac{8}{5}{\left(1\right)}^{-1/5}-\frac{9}{5}{\left(1\right)}^{4/5}=-0.2<0$

Therefore , the function is increasing in interval $\left(0,0.89\right)$

Below is the graph of $f^{\prime }\left(x\right)=\frac{8}{5}{x}^{-1/5}-\frac{9}{5}{x}^{4/5}$

  

From graph of $f^{\prime }\left(x\right)=\frac{8}{5}{x}^{-1/5}-\frac{9}{5}{x}^{4/5}$ it is clear that the function is increasing in interval $\left(0,0.89\right)$ .

b.

To determine

To find the intervals on which the function is decreasing by using analytical method.

Answer to Problem 15RE

The function is decreasing in interval $\left(-\infty ,0\right)$ and $\left(0.89,\infty \right)$

Explanation of Solution

Given:

The function is

  $y=x^{4/5}\left(2-x\right)$ .

Calculation:

The function $y=x^{4/5}\left(2-x\right)$ is decreasing when $f^{\prime }\left(x\right)<0$

  $f^{\prime }\left(x\right)=-x^{4/5}+\frac{4}{5}{x}^{-1/5}\left(2-x\right)=\frac{8}{5}{x}^{-1/5}-\frac{9}{5}{x}^{4/5}$

Now put $f^{\prime }\left(x\right)=0$ to find critical points

  $\begin{array}{l}\frac{8}{5}{x}^{-1/5}-\frac{9}{5}{x}^{4/5}=0\\ \frac{1}{5}{x}^{-1/5}\left(8-9x\right)=0\\ \because x^{-1/5}\ne 0\\ \therefore 8-9x=0\\ x=\frac{8}{9}=0.89\end{array}$

So , there are three intervals that is $\left(-\infty ,0\right),\left(0,0.89\right)\text{ and }\left(0.89,\infty \right)$ to check whether $f^{\prime }\left(x\right)>0$ or $f^{\prime }\left(x\right)<0$

put $x=-1$ to check whether the function is increasing or decreasing in $\left(-\infty ,0\right)$

  $f^{\prime }\left(x\right)=\frac{8}{5}{x}^{-1/5}-\frac{9}{5}{x}^{4/5}=\frac{8}{5}{\left(-1\right)}^{-1/5}-\frac{9}{5}{\left(-1\right)}^{4/5}=-3.4<0$

put $x=0.5$ to check whether the function is increasing or decreasing in $\left(0,0.89\right)$

  $f^{\prime }\left(x\right)=\frac{8}{5}{x}^{-1/5}-\frac{9}{5}{x}^{4/5}=\frac{8}{5}{\left(0.5\right)}^{-1/5}-\frac{9}{5}{\left(0.5\right)}^{4/5}=0.80>0$

put $x=1$ to check whether the function is increasing or decreasing in $\left(0.89,\infty \right)$

  $f^{\prime }\left(x\right)=\frac{8}{5}{x}^{-1/5}-\frac{9}{5}{x}^{4/5}=\frac{8}{5}{\left(1\right)}^{-1/5}-\frac{9}{5}{\left(1\right)}^{4/5}=-0.2<0$

Therefore , the function is decreasing in interval $\left(-\infty ,0\right)$ and $\left(0.89,\infty \right)$ .

Below is the graph of $f^{\prime }\left(x\right)=\frac{8}{5}{x}^{-1/5}-\frac{9}{5}{x}^{4/5}$

  

From graph of $f^{\prime }\left(x\right)=\frac{8}{5}{x}^{-1/5}-\frac{9}{5}{x}^{4/5}$ it is clear that the function is decreasing in interval $\left(-\infty ,0\right)$ and $\left(0.89,\infty \right)$

c.

To determine

To find the intervals on which the function is concave up by using analytical method.

Answer to Problem 15RE

The Function $y=x^{4/5}\left(2-x\right)$ is concave up in interval $\left(-\infty ,-\frac{2}{9}\right)$.

Explanation of Solution

Given:

The function is $y=x^{4/5}\left(2-x\right)$ .

Calculation:

The graph of a twice differentiable function $y=f\left(x\right)$ is

Concave up on any interval where $f^{″}\left(x\right)>0$ and concave down on any interval where $f^{″}\left(x\right)<0$

Since, $y=x^{4/5}\left(2-x\right)$

First derivative : $y^{\prime }=\frac{8}{5}{x}^{-1/5}-\frac{9}{5}{x}^{4/5}$

Second derivative : $y^{″}=-\frac{8}{25}{x}^{-6/5}-\frac{36}{25}{x}^{-1/5}$

Now, put $f^{″}\left(x\right)=0$ to find critical points

  $\begin{array}{l}-\frac{8}{25}{x}^{-6/5}-\frac{36}{25}{x}^{-1/5}=0\\ -\frac{4}{25}{x}^{-6/5}\left(2+9x\right)=0\\ \because x\ne 0\\ \therefore 2+9x=0\\ x=-\frac{2}{9}=-0.23\end{array}$

Therefore , there are three intervals that is $\left(-\infty ,-\frac{2}{9}\right)$ , $\left(-\frac{2}{9},0\right)$ and $\left(0,\infty \right)$

check the value of $f^{″}\left(x\right)$ in interval $\left(-\infty ,-\frac{2}{9}\right)$ , $\left(-\frac{2}{9},0\right)$ and $\left(0,\infty \right)$

Now for $x\in \left(-\infty ,-\frac{2}{9}\right)$ test for $x=-1$

  $\begin{array}{l}{f}^{″}\left(x\right)=-\frac{8}{25}{x}^{-6/5}-\frac{36}{25}{x}^{-1/5}\\ f^{″}\left(-1\right)=-\frac{8}{25}{\left(-1\right)}^{-6/5}-\frac{36}{25}{\left(-1\right)}^{-1/5}=1.12>0\end{array}$

Now for $x\in \left(-\frac{2}{9},0\right)$ test for $x=-0.1$

  $\begin{array}{l}{f}^{″}\left(x\right)=-\frac{8}{25}{x}^{-6/5}-\frac{36}{25}{x}^{-1/5}\\ f^{″}\left(-0.1\right)=-\frac{8}{25}{\left(-0.1\right)}^{-6/5}-\frac{36}{25}{\left(-0.1\right)}^{-1/5}=-2.7<0\end{array}$

Now for $x\in \left(0,\infty \right)$ test for $x=1$

  $\begin{array}{l}{f}^{″}\left(x\right)=-\frac{8}{25}{x}^{-6/5}-\frac{36}{25}{x}^{-1/5}\\ f^{″}\left(1\right)=-\frac{8}{25}{\left(1\right)}^{-6/5}-\frac{36}{25}{\left(1\right)}^{-1/5}=-1.76<0\end{array}$

Therefore, the Function $y=x^{4/5}\left(2-x\right)$ is concave up in interval $\left(-\infty ,-\frac{2}{9}\right)$ .

Below is the graph of $f^{″}\left(x\right)=-\frac{8}{25}{x}^{-6/5}-\frac{36}{25}{x}^{-1/5}$

  

From graph it is clear that, the Function $y=x^{4/5}\left(2-x\right)$ is concave up in interval $\left(-\infty ,-\frac{2}{9}\right)$ .

d.

To determine

To find the intervals on which the function is concave down by using analytical method.

Answer to Problem 15RE

The Function $y=x^{4/5}\left(2-x\right)$ is concave down in interval $\left(-\frac{2}{9},0\right)$ and $\left(0,\infty \right)$

Explanation of Solution

Given:

The function is $y=x^{4/5}\left(2-x\right)$ .

Calculation:

The graph of a twice differentiable function $y=f\left(x\right)$ is

Concave up on any interval where $f^{″}\left(x\right)>0$ and concave down on any interval where $f^{″}\left(x\right)<0$

Since, $y=x^{4/5}\left(2-x\right)$

First derivative : $y^{\prime }=\frac{8}{5}{x}^{-1/5}-\frac{9}{5}{x}^{4/5}$

Second derivative : $y^{″}=-\frac{8}{25}{x}^{-6/5}-\frac{36}{25}{x}^{-1/5}$

Now, put $f^{″}\left(x\right)=0$ to find critical points

  $\begin{array}{l}-\frac{8}{25}{x}^{-6/5}-\frac{36}{25}{x}^{-1/5}=0\\ -\frac{4}{25}{x}^{-6/5}\left(2+9x\right)=0\\ \because x\ne 0\\ \therefore 2+9x=0\\ x=-\frac{2}{9}=-0.23\end{array}$

Therefore , there are three intervals that is $\left(-\infty ,-\frac{2}{9}\right)$ , $\left(-\frac{2}{9},0\right)$ and $\left(0,\infty \right)$

check the value of $f^{″}\left(x\right)$ in interval $\left(-\infty ,-\frac{2}{9}\right)$ , $\left(-\frac{2}{9},0\right)$ and $\left(0,\infty \right)$

Now for $x\in \left(-\infty ,-\frac{2}{9}\right)$ test for $x=-1$

  $\begin{array}{l}{f}^{″}\left(x\right)=-\frac{8}{25}{x}^{-6/5}-\frac{36}{25}{x}^{-1/5}\\ f^{″}\left(-1\right)=-\frac{8}{25}{\left(-1\right)}^{-6/5}-\frac{36}{25}{\left(-1\right)}^{-1/5}=1.12>0\end{array}$

Now for $x\in \left(-\frac{2}{9},0\right)$ test for $x=-0.1$

  $\begin{array}{l}{f}^{″}\left(x\right)=-\frac{8}{25}{x}^{-6/5}-\frac{36}{25}{x}^{-1/5}\\ f^{″}\left(-0.1\right)=-\frac{8}{25}{\left(-0.1\right)}^{-6/5}-\frac{36}{25}{\left(-0.1\right)}^{-1/5}=-2.7<0\end{array}$

Now for $x\in \left(0,\infty \right)$ test for $x=1$

  $\begin{array}{l}{f}^{″}\left(x\right)=-\frac{8}{25}{x}^{-6/5}-\frac{36}{25}{x}^{-1/5}\\ f^{″}\left(1\right)=-\frac{8}{25}{\left(1\right)}^{-6/5}-\frac{36}{25}{\left(1\right)}^{-1/5}=-1.76<0\end{array}$

Therefore, the Function $y=x^{4/5}\left(2-x\right)$ is concave down in interval $\left(-\frac{2}{9},0\right)$ and $\left(0,\infty \right)$

Below is the graph of $f^{″}\left(x\right)=-\frac{8}{25}{x}^{-6/5}-\frac{36}{25}{x}^{-1/5}$

  

From graph it is clear that, the Function $y=x^{4/5}\left(2-x\right)$ is concave down in interval $\left(-\frac{2}{9},0\right)$ and $\left(0,\infty \right)$ .

e.

To determine

To find any local extreme values.

Answer to Problem 15RE

The function $y=x^{4/5}\left(2-x\right)$ has local maxima at point $\left(0.88,1.01\right)$ and local minima at $\left(0,0\right)$

Explanation of Solution

Given:

The function is $y=x^{4/5}\left(2-x\right)$ .

Calculation:

Graph of $y=x^{4/5}\left(2-x\right)$ is given below

  

From graph it is clear that the function $y=x^{4/5}\left(2-x\right)$ has local maxima at point $\left(0.88,1.01\right)$ and local minima at $\left(0,0\right)$ .

f.

To determine

To find inflections points.

Answer to Problem 15RE

The inflection point is at $x=-\frac{2}{9}$ .

Explanation of Solution

Given:

The function is $y=x^{4/5}\left(2-x\right)$ .

Calculation:

Inflection point of any function is a point where the graph of function has a tangent line and where the concavity changes.

Since, $y=x^{4/5}\left(2-x\right)$ changes concavity in interval $\left(-\infty ,-\frac{2}{9}\right)$ , $\left(-\frac{2}{9},0\right)$ and $\left(0,\infty \right)$ , and remains concave down in interval $\left(-\frac{2}{9},0\right)$ and $\left(0,\infty \right)$ .

Therefore, the inflection point is at $x=-\frac{2}{9}$ .