Chapter 5, Problem 10RE

a.

To determine

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

Answer to Problem 10RE

The function $y=\frac{x}{x^2+2x+3}$ is increasing when $x\in \left[-\sqrt{3},\sqrt{3}\right]$

Explanation of Solution

Given:

The function is $y=\frac{x}{x^2+2x+3}$ .

Calculation:

The function is increasing when $f^{\prime }\left(x\right)>0$ .

  $\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{\left(x^2+2x+3\right)\times \frac{d}{dx}\left(x\right)-x\times \frac{d}{dx}\left(x^2+2x+3\right)}{{\left(x^2+2x+3\right)}^2}=\frac{-x^2+3}{{\left(x^2+2x+3\right)}^2}\\ \frac{-x^2+3}{{\left(x^2+2x+3\right)}^2}>0\\ -x^2+3>0\\ x^2-3<0\\ \therefore x\in \left(-\sqrt{3},\sqrt{3}\right)\end{array}$

Therefore, The function $y=\frac{x}{x^2+2x+3}$ is increasing when $x\in \left(-\sqrt{3},\sqrt{3}\right)$

Below is the graph of the function $y=\frac{x}{x^2+2x+3}$

  

From graph it is clear that the function $y=\frac{x}{x^2+2x+3}$ is increasing when $x\in \left[-\sqrt{3},\sqrt{3}\right]$ .

b.

To determine

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

Answer to Problem 10RE

The function $y=\frac{x}{x^2+2x+3}$ is decreasing when $x\in \left(-\infty ,-\sqrt{3}\right]\cup \left[\sqrt{3},\infty \right)$

Explanation of Solution

Given:

The function is $y=\frac{x}{x^2+2x+3}$ .

Calculation:

The function is decreasing when $f^{\prime }\left(x\right)<0$ .

  $\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{\left(x^2+2x+3\right)\times \frac{d}{dx}\left(x\right)-x\times \frac{d}{dx}\left(x^2+2x+3\right)}{{\left(x^2+2x+3\right)}^2}=\frac{-x^2+3}{{\left(x^2+2x+3\right)}^2}\\ \frac{-x^2+3}{{\left(x^2+2x+3\right)}^2}<0\\ -x^2+3<0\\ x^2-3>0\\ \therefore x\in \left(-\infty ,-\sqrt{3}\right]\cup \left[\sqrt{3},\infty \right)\end{array}$

Therefore, the function $y=\frac{x}{x^2+2x+3}$ is decreasing when $x\in \left(-\infty ,-\sqrt{3}\right]\cup \left[\sqrt{3},\infty \right)$

Below is the graph of the function $y=\frac{x}{x^2+2x+3}$

  

From graph it is clear that , the function $y=\frac{x}{x^2+2x+3}$ is decreasing when $x\in \left(-\infty ,-\sqrt{3}\right]\cup \left[\sqrt{3},\infty \right)$

c.

To determine

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

Answer to Problem 10RE

The function $y=\frac{x}{x^2+2x+3}$ is concave up in interval $\left(-2.58,-0.705\right)$ and $\left(3.28,\infty \right)$

Explanation of Solution

Given:

The function is $y=\frac{x}{x^2+2x+3}$ .

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, $y=\frac{x}{x^2+2x+3}$

First derivative : $y^{\prime }=\frac{-x^2+3}{{\left(x^2+2x+3\right)}^2}$

Second derivative : $y^{″}=\frac{2\left(x^3-9x-6\right)}{{\left(x^2+2x+3\right)}^3}$

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

  $\begin{array}{l}\frac{2\left(x^3-9x-6\right)}{{\left(x^2+2x+3\right)}^3}=0\\ \because {\left(x^2+2x+3\right)}^3\text{ is always positive}\\ \therefore x^3-9x-6=0\\ x=3.28,-0.705,-2.58\end{array}$

Domain of $y=\frac{x}{x^2+2x+3}$ is $\left(-\infty ,\infty \right)$

Therefore , there are four intervals that is $\left(-\infty ,-2.58\right)$ , $\left(-2.58,-0.705\right)$ , $\left(-0.705,3.28\right)$ and $\left(3.28,\infty \right)$

check the value of $f^{″}\left(x\right)$ in interval $\left(-\infty ,-2.58\right)$ , $\left(-2.58,-0.705\right)$ , $\left(-0.705,3.28\right)$ and $\left(-0.705,3.28\right)$

Now for $x\in \left(-2.58,-0.705\right)$ test for $x=-1$

  $\begin{array}{l}{f}^{″}\left(x\right)=\frac{2\left(x^3-9x-6\right)}{{\left(x^2+2x+3\right)}^3}\\ f^{″}\left(-3\right)=\frac{2\left({\left(-1\right)}^3-9\left(-1\right)-6\right)}{{\left({\left(-1\right)}^2+2\left(-1\right)+3\right)}^3}=\frac{1}{2}>0\end{array}$

Therefore, the Function $y=\frac{x}{x^2+2x+3}$ is concave up in interval $\left(-2.58,-0.705\right)$

Now for $x\in \left(3.28,\infty \right)$ test for $x=4$

  $\begin{array}{l}{f}^{″}\left(x\right)=\frac{2\left(x^3-9x-6\right)}{{\left(x^2+2x+3\right)}^3}\\ f^{″}\left(0\right)=\frac{2\left({\left(4\right)}^3-9\left(4\right)-6\right)}{{\left({\left(4\right)}^2+2\left(4\right)+3\right)}^3}=\frac{44}{19683}<0\end{array}$

Therefore, the Function $y=\frac{x}{x^2+2x+3}$ is concave up in interval $\left(3.28,\infty \right)$

Below is the graph of the function $y=\frac{x}{x^2+2x+3}$

  

From graph it is clear that , the function $y=\frac{x}{x^2+2x+3}$ is concave up in interval $\left(-2.58,-0.705\right)$ and $\left(3.28,\infty \right)$ .

d.

To determine

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

Answer to Problem 10RE

The function $y=\frac{x}{x^2+2x+3}$ is concave down in interval $\left(-\infty ,-2.58\right)$ and $\left(-0.705,3.28\right)$ .

Explanation of Solution

Given:

The function is $y=\frac{x}{x^2+2x+3}$ .

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, $y=\frac{x}{x^2+2x+3}$

First derivative : $y^{\prime }=\frac{-x^2+3}{{\left(x^2+2x+3\right)}^2}$

Second derivative : $y^{″}=\frac{2\left(x^3-9x-6\right)}{{\left(x^2+2x+3\right)}^3}$

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

  $\begin{array}{l}\frac{2\left(x^3-9x-6\right)}{{\left(x^2+2x+3\right)}^3}=0\\ \because {\left(x^2+2x+3\right)}^3\text{ is always positive}\\ \therefore x^3-9x-6=0\\ x=3.28,-0.705,-2.58\end{array}$

Domain of $y=\frac{x}{x^2+2x+3}$ is $\left(-\infty ,\infty \right)$

Therefore , there are four intervals that is $\left(-\infty ,-2.58\right)$ , $\left(-2.58,-0.705\right)$ , $\left(-0.705,3.28\right)$ and $\left(3.28,\infty \right)$

check the value of $f^{″}\left(x\right)$ in interval $\left(-\infty ,-2.58\right)$ , $\left(-2.58,-0.705\right)$ , $\left(-0.705,3.28\right)$ and $\left(-0.705,3.28\right)$

Now for $x\in \left(-\infty ,-2.58\right)$ test for $x=-3$

  $\begin{array}{l}{f}^{″}\left(x\right)=\frac{2\left(x^3-9x-6\right)}{{\left(x^2+2x+3\right)}^3}\\ f^{″}\left(-3\right)=\frac{2\left({\left(-3\right)}^3-9\left(-3\right)-6\right)}{{\left({\left(-3\right)}^2+2\left(-3\right)+3\right)}^3}=-\frac{1}{18}<0\end{array}$

Therefore, the Function $y=\frac{x}{x^2+2x+3}$ is concave down in interval $\left(-\infty ,-2.58\right)$

Now for $x\in \left(-0.705,3.28\right)$ test for $x=1$

  $\begin{array}{l}{f}^{″}\left(x\right)=\frac{2\left(x^3-9x-6\right)}{{\left(x^2+2x+3\right)}^3}\\ f^{″}\left(0\right)=\frac{2\left({\left(0\right)}^3-9\left(0\right)-6\right)}{{\left({\left(0\right)}^2+2\left(0\right)+3\right)}^3}=-\frac{4}{9}<0\end{array}$

Therefore, the Function $y=\frac{x}{x^2+2x+3}$ is concave down in interval $\left(-0.705,3.28\right)$

Below is the graph of the function $y=\frac{x}{x^2+2x+3}$

  

From graph it is clear that , the function $y=\frac{x}{x^2+2x+3}$ is concave down in interval $\left(-\infty ,-2.58\right)$ and $\left(-0.705,3.28\right)$ .

e.

To determine

To find any local extreme values.

Answer to Problem 10RE

  $y=\frac{x}{x^2+2x+3}$ has local maximum value $-0.683$ at $x=\sqrt{3}$ and local minimum value $0.183$ at $x=-\sqrt{3}$ point .

Explanation of Solution

Given:

The function is $y=\frac{x}{x^2+2x+3}$ .

Calculation:

Graph of $y=\frac{x}{x^2+2x+3}$ is given below

  

From graph it is clear that $y=\frac{x}{x^2+2x+3}$ has local maximum value $-0.683$ at $x=\sqrt{3}$ and local minimum value $0.183$ at $x=-\sqrt{3}$ point .

f.

To determine

To find inflections points.

Answer to Problem 10RE

There are three inflection point $x=-2.58,x=-0.705\text{ and }x=3.28$ .

Explanation of Solution

Given:

The function is $y=\frac{x}{x^2+2x+3}$ .

Calculation:

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

Since, $y=\frac{x}{x^2+2x+3}$ changes concavity in interval $\left(-\infty ,-2.58\right)$ , $\left(-2.58,-0.705\right)$ , $\left(-0.705,3.28\right)$ and $\left(-0.705,3.28\right)$

Therefore, there are three inflection point $x=-2.58,x=-0.705\text{ and }x=3.28$ .