Chapter 5, Problem 16RE

a.

To determine

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

Answer to Problem 16RE

The function $y=\frac{5-4x+4x^2-x^3}{x-2}$ is increasing in interval $\left(-\infty ,0.251\right)$.

Explanation of Solution

Given:

The function is

  $y=\frac{5-4x+4x^2-x^3}{x-2}$.

Calculation:

  $f^{\prime }\left(x\right)=\frac{-2x^3+10x^2-16x+3}{{\left(x-2\right)}^2}$

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

  $\begin{array}{l}\frac{-2x^3+10x^2-16x+3}{{\left(x-2\right)}^2}>0\\ \because {\left(x-2\right)}^2\text{ is a positive term}\\ -2x^3+10x^2-16x+3>0\end{array}$ and $x\ne 2$

Below is the graph of $-2x^3+10x^2-16x+3=0$.

  

So , there are three intervals $\left(-\infty ,0.251\right),\left(0.251,2\right)\text{ and }\left(2,\infty \right)$to check the function is increasing or decreasing in these interval

Put $x=0$to check whether function is increasing or decreasing in $\left(-\infty ,0.251\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{-2x^3+10x^2-16x+3}{{\left(x-2\right)}^2}\\ f^{\prime }\left(0\right)=\frac{-2{\left(0\right)}^3+10{\left(0\right)}^2-16\left(0\right)+3}{{\left(0-2\right)}^2}=\frac{3}{4}>0\end{array}$

Function is increasing in interval $\left(-\infty ,0.251\right)$.

Put $x=1$to check whether function is increasing or decreasing in $\left(0.251,2\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{-2x^3+10x^2-16x+3}{{\left(x-2\right)}^2}\\ f^{\prime }\left(1\right)=\frac{-2{\left(1\right)}^3+10{\left(1\right)}^2-16\left(1\right)+3}{{\left(1-2\right)}^2}=-5<0\end{array}$

Function is decreasing in interval $\left(0.251,2\right)$.

Put $x=3$to check whether function is increasing or decreasing in $\left(2,\infty \right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{-2x^3+10x^2-16x+3}{{\left(x-2\right)}^2}\\ f^{\prime }\left(3\right)=\frac{-2{\left(3\right)}^3+10{\left(3\right)}^2-16\left(3\right)+3}{{\left(3-2\right)}^2}=-9<0\end{array}$

Function is decreasing in interval$\left(2,\infty \right)$.

Below is the graph of $f^{\prime }\left(x\right)=\frac{-2x^3+10x^2-16x+3}{{\left(x-2\right)}^2}$

  

From graph also it is clear that the function $y=\frac{5-4x+4x^2-x^3}{x-2}$ is increasing in interval $\left(-\infty ,0.251\right)$.

b.

To determine

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

Answer to Problem 16RE

The function $y=\frac{5-4x+4x^2-x^3}{x-2}$ is decreasing in interval$\left(0.251,2\right)\text{ and }\left(2,\infty \right)$.

Explanation of Solution

Given:

The function is

  $y=\frac{5-4x+4x^2-x^3}{x-2}$.

Calculation:

  $f^{\prime }\left(x\right)=\frac{-2x^3+10x^2-16x+3}{{\left(x-2\right)}^2}$

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

  $\begin{array}{l}\frac{-2x^3+10x^2-16x+3}{{\left(x-2\right)}^2}<0\\ \because {\left(x-2\right)}^2\text{ is a positive term}\\ -2x^3+10x^2-16x+3<0\end{array}$and $x\ne 2$

Below is the graph of $-2x^3+10x^2-16x+3=0$.

  

So , there are three intervals $\left(-\infty ,0.251\right),\left(0.251,2\right)\text{ and }\left(2,\infty \right)$to check the function is increasing or decreasing in these interval

Put $x=0$to check whether function is increasing or decreasing in $\left(-\infty ,0.251\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{-2x^3+10x^2-16x+3}{{\left(x-2\right)}^2}\\ f^{\prime }\left(0\right)=\frac{-2{\left(0\right)}^3+10{\left(0\right)}^2-16\left(0\right)+3}{{\left(0-2\right)}^2}=\frac{3}{4}>0\end{array}$

Function is increasing in interval $\left(-\infty ,0.251\right)$.

Put $x=1$to check whether function is increasing or decreasing in $\left(0.251,2\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{-2x^3+10x^2-16x+3}{{\left(x-2\right)}^2}\\ f^{\prime }\left(1\right)=\frac{-2{\left(1\right)}^3+10{\left(1\right)}^2-16\left(1\right)+3}{{\left(1-2\right)}^2}=-5<0\end{array}$

Function is decreasing in interval $\left(0.251,2\right)$.

Put $x=3$to check whether function is increasing or decreasing in $\left(2,\infty \right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{-2x^3+10x^2-16x+3}{{\left(x-2\right)}^2}\\ f^{\prime }\left(3\right)=\frac{-2{\left(3\right)}^3+10{\left(3\right)}^2-16\left(3\right)+3}{{\left(3-2\right)}^2}=-9<0\end{array}$

Function is decreasing in interval $\left(2,\infty \right)$.

Below is the graph of $f^{\prime }\left(x\right)=\frac{-2x^3+10x^2-16x+3}{{\left(x-2\right)}^2}$

  

From graph also it is clear that the function $y=\frac{5-4x+4x^2-x^3}{x-2}$ is decreasing in interval$\left(0.251,2\right)\text{ and }\left(2,\infty \right)$.

c.

To determine

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

Answer to Problem 16RE

The Function $y=\frac{5-4x+4x^2-x^3}{x-2}$is concave up in interval $\left(2,3.71\right)$

Explanation of Solution

Given:

The function is $y=\frac{5-4x+4x^2-x^3}{x-2}$.

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, $y=\frac{5-4x+4x^2-x^3}{x-2}$

First derivative : $y^{\prime }=\frac{-2x^3+10x^2-16x+3}{{\left(x-2\right)}^2}$

Second derivative : $y^{″}=\frac{-2x^3+12x^2-24x+26}{{\left(x-2\right)}^3}$

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

  $\begin{array}{l}\frac{-2x^3+12x^2-24x+26}{{\left(x-2\right)}^3}=0\\ -2x^3+12x^2-24x+26=0,x\ne 2\\ x=3.71,x\ne 2\end{array}$

Therefore , there are four intervals that is $\left(-\infty ,2\right)$,$\left(2,3.71\right)$and $\left(3.71,\infty \right)$

check the value of $f^{″}\left(x\right)$in interval$\left(-\infty ,2\right)$,$\left(2,3.71\right)$and $\left(3.71,\infty \right)$

Now for $x\in \left(-\infty ,2\right)$test for $x=0$

  $\begin{array}{l}{f}^{″}\left(x\right)=\frac{-2x^3+12x^2-24x+26}{{\left(x-2\right)}^3}\\ f^{″}\left(0\right)=\frac{-2{\left(0\right)}^3+12{\left(0\right)}^2-24\left(0\right)+26}{{\left(0-2\right)}^3}=-\frac{26}{8}<0\end{array}$

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

Now for $x\in \left(2,3.71\right)$test for $x=3$

  $\begin{array}{l}{f}^{″}\left(x\right)=\frac{-2x^3+12x^2-24x+26}{{\left(x-2\right)}^3}\\ f^{″}\left(3\right)=\frac{-2{\left(3\right)}^3+12{\left(3\right)}^2-24\left(3\right)+26}{{\left(3-2\right)}^3}=8>0\end{array}$

Therefore, the Function$y=\frac{5-4x+4x^2-x^3}{x-2}$is concave up in interval $\left(2,3.71\right)$

Now for $x\in \left(3.71,\infty \right)$test for $x=4$

  $\begin{array}{l}{f}^{″}\left(x\right)=\frac{-2x^3+12x^2-24x+26}{{\left(x-2\right)}^3}\\ f^{″}\left(4\right)=\frac{-2{\left(4\right)}^3+12{\left(4\right)}^2-24\left(4\right)+26}{{\left(4-2\right)}^3}=-0.75<0\end{array}$

Therefore, the Function $y=\frac{5-4x+4x^2-x^3}{x-2}$is concave down in interval $\left(3.71,\infty \right)$

Below is the graph of$f^{″}\left(x\right)=\frac{-2x^3+12x^2-24x+26}{{\left(x-2\right)}^3}$-

  

From graph it is clear that the Function $y=\frac{5-4x+4x^2-x^3}{x-2}$is concave up in interval $\left(2,3.71\right)$

d.

To determine

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

Answer to Problem 16RE

The Function $y=\frac{5-4x+4x^2-x^3}{x-2}$is concave down in interval $\left(-\infty ,2\right)$and $\left(3.71,\infty \right)$

Explanation of Solution

Given:

The function is $y=\frac{5-4x+4x^2-x^3}{x-2}$.

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, $y=\frac{5-4x+4x^2-x^3}{x-2}$

First derivative : $y^{\prime }=\frac{-2x^3+10x^2-16x+3}{{\left(x-2\right)}^2}$

Second derivative : $y^{″}=\frac{-2x^3+12x^2-24x+26}{{\left(x-2\right)}^3}$

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

  $\begin{array}{l}\frac{-2x^3+12x^2-24x+26}{{\left(x-2\right)}^3}=0\\ -2x^3+12x^2-24x+26=0,x\ne 2\\ x=3.71,x\ne 2\end{array}$

Therefore , there are three intervals that is $\left(-\infty ,2\right)$,$\left(2,3.71\right)$and $\left(3.71,\infty \right)$

check the value of $f^{″}\left(x\right)$in interval $\left(-\infty ,2\right)$,$\left(2,3.71\right)$and $\left(3.71,\infty \right)$

Now for $x\in \left(-\infty ,2\right)$test for $x=0$

  $\begin{array}{l}{f}^{″}\left(x\right)=\frac{-2x^3+12x^2-24x+26}{{\left(x-2\right)}^3}\\ f^{″}\left(0\right)=\frac{-2{\left(0\right)}^3+12{\left(0\right)}^2-24\left(0\right)+26}{{\left(0-2\right)}^3}=-\frac{26}{8}<0\end{array}$

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

Now for $x\in \left(2,3.71\right)$test for $x=3$

  $\begin{array}{l}{f}^{″}\left(x\right)=\frac{-2x^3+12x^2-24x+26}{{\left(x-2\right)}^3}\\ f^{″}\left(3\right)=\frac{-2{\left(3\right)}^3+12{\left(3\right)}^2-24\left(3\right)+26}{{\left(3-2\right)}^3}=8>0\end{array}$

Therefore, the Function $y=\frac{5-4x+4x^2-x^3}{x-2}$is concave up in interval $\left(2,3.71\right)$

Now for $x\in \left(3.71,\infty \right)$test for $x=4$

  $\begin{array}{l}{f}^{″}\left(x\right)=\frac{-2x^3+12x^2-24x+26}{{\left(x-2\right)}^3}\\ f^{″}\left(4\right)=\frac{-2{\left(4\right)}^3+12{\left(4\right)}^2-24\left(4\right)+26}{{\left(4-2\right)}^3}=-0.75<0\end{array}$

Therefore, the Function $y=\frac{5-4x+4x^2-x^3}{x-2}$is concave down in interval $\left(3.71,\infty \right)$

Below is the graph of$f^{″}\left(x\right)=\frac{-2x^3+12x^2-24x+26}{{\left(x-2\right)}^3}$-

  

From graph also it is clear that the Function $y=\frac{5-4x+4x^2-x^3}{x-2}$is concave down in interval $\left(-\infty ,2\right)$and $\left(3.71,\infty \right)$.

e.

To determine

To find any local extreme values.

Answer to Problem 16RE

The function $y=\frac{5-4x+4x^2-x^3}{x-2}$has local maxima at point $\left(0.251,-2.4173\right)$.

Explanation of Solution

Given:

The function is$y=\frac{5-4x+4x^2-x^3}{x-2}$.

Calculation:

Graph of $y=\frac{5-4x+4x^2-x^3}{x-2}$ is given below

  

From graph it is clear that the function $y=\frac{5-4x+4x^2-x^3}{x-2}$has local maxima at point $\left(0.251,-2.4173\right)$.

f.

To determine

To find inflections points.

Answer to Problem 16RE

The inflection point is at $x=3.71$.

Explanation of Solution

Given:

The function is $y=\frac{5-4x+4x^2-x^3}{x-2}$.

Calculation:

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

Since, $y=\frac{5-4x+4x^2-x^3}{x-2}$

changes concavity in interval $\left(-\infty ,2\right)$,$\left(2,3.71\right)$and $\left(3.71,\infty \right)$

Therefore, theinflection point is at $x=3.71$.