Chapter 5, Problem 14RE

a.

To determine

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

Answer to Problem 14RE

The function $y=-x^5+\frac{7}{3}{x}^3+5x^2+4x+2$ is increasing in interval $\left(0.57,1.69\right)$

Explanation of Solution

Given:

The function is

  $y=-x^5+\frac{7}{3}{x}^3+5x^2+4x+2$ .

Calculation:

The function $y=-x^5+\frac{7}{3}{x}^3+5x^2+4x+2$ is increasing when $f^{\prime }\left(x\right)>0$

  $f^{\prime }\left(x\right)=-5x^4+7x^2+10x+4$

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

  $-5x^4+7x^2+10x+4=0$

Below is graph $-5x^4+7x^2+10x+4=0$ to find critical points

  

Therefore ,the critical points are $x=-0.57,1.69$

So , there are three intervals that is $\left(-\infty ,0.57\right),\left(0.57,1.69\right)\text{ and }\left(1.69,\infty \right)$ 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 $\left(-\infty ,0.57\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=-5x^4+7x^2+10x+4\\ f^{\prime }\left(-1\right)=-5{\left(-1\right)}^4+7{\left(-1\right)}^2+10\left(-1\right)+4=-4<0\end{array}$

put $x=1$ to check whether the function is increasing or decreasing in $\left(0.57,1.69\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=-5x^4+7x^2+10x+4\\ f^{\prime }\left(1\right)=-5{\left(1\right)}^4+7{\left(1\right)}^2+10\left(1\right)+4=16>0\end{array}$

put $x=2$ to check whether the function is increasing or decreasing in $\left(1.69,\infty \right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=-5x^4+7x^2+10x+4\\ f^{\prime }\left(2\right)=-5{\left(2\right)}^4+7{\left(2\right)}^2+10\left(2\right)+4=-28<0\end{array}$

Therefore , the function is increasing in interval $\left(0.57,1.69\right)$

Below is the graph of $f^{\prime }\left(x\right)=-5x^4+7x^2+10x+4$

  

From graph of $f^{\prime }\left(x\right)=-5x^4+7x^2+10x+4$ it is clear that the function is increasing in interval $\left(0.57,1.69\right)$ .

b.

To determine

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

Answer to Problem 14RE

The function is decreasing in interval is $\left(-\infty ,0.57\right)\text{ and }\left(1.69,\infty \right)$ .

Explanation of Solution

Given:

The function is

  $y=-x^5+\frac{7}{3}{x}^3+5x^2+4x+2$ .

Calculation:

The function $y=-x^5+\frac{7}{3}{x}^3+5x^2+4x+2$ is decreasing when $f^{\prime }\left(x\right)<0$

  $f^{\prime }\left(x\right)=-5x^4+7x^2+10x+4$

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

  $-5x^4+7x^2+10x+4=0$

Below is graph $-5x^4+7x^2+10x+4=0$ to find critical points

  

Therefore ,the critical points are $x=-0.57,1.69$

So , there are three intervals that is $\left(-\infty ,0.57\right),\left(0.57,1.69\right)\text{ and }\left(1.69,\infty \right)$ 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 $\left(-\infty ,0.57\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=-5x^4+7x^2+10x+4\\ f^{\prime }\left(-1\right)=-5{\left(-1\right)}^4+7{\left(-1\right)}^2+10\left(-1\right)+4=-4<0\end{array}$

put $x=1$ to check whether the function is increasing or decreasing in $\left(0.57,1.69\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=-5x^4+7x^2+10x+4\\ f^{\prime }\left(1\right)=-5{\left(1\right)}^4+7{\left(1\right)}^2+10\left(1\right)+4=16>0\end{array}$

put $x=2$ to check whether the function is increasing or decreasing in $\left(1.69,\infty \right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=-5x^4+7x^2+10x+4\\ f^{\prime }\left(2\right)=-5{\left(2\right)}^4+7{\left(2\right)}^2+10\left(2\right)+4=-28<0\end{array}$

Therefore , the function is decreasing in interval is $\left(-\infty ,0.57\right)\text{ and }\left(1.69,\infty \right)$ .

Below is the graph of $f^{\prime }\left(x\right)=-5x^4+7x^2+10x+4$

  

From graph of $f^{\prime }\left(x\right)=-5x^4+7x^2+10x+4$ it is clear that the function is decreasing in interval is $\left(-\infty ,0.57\right)\text{ and }\left(1.69,\infty \right)$ .

c.

To determine

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

Answer to Problem 14RE

The Function $y=-x^5+\frac{7}{3}{x}^3+5x^2+4x+2$ is concave up in interval $\left(-\infty ,1.079\right)$ .

Explanation of Solution

Given:

The function is $y=-x^5+\frac{7}{3}{x}^3+5x^2+4x+2$ .

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, $y=-x^5+\frac{7}{3}{x}^3+5x^2+4x+2$

First derivative : $y^{\prime }=-5x^4+7x^2+10x+4$

Second derivative : $y^{″}=-20x^3+14x+10$

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

  $\begin{array}{l}-20x^3+14x+10=0\\ x=1.07\end{array}$

Therefore , there are two intervals that is $\left(-\infty ,1.079\right)$ and $\left(1.079,\infty \right)$

check the value of $f^{″}\left(x\right)$ in interval $\left(-\infty ,1.079\right)$ and $\left(1.079,\infty \right)$

Now for $x\in \left(-\infty ,1.079\right)$ test for $x=0$

  $\begin{array}{l}{f}^{″}\left(x\right)=-20x^3+14x+10\\ f^{″}\left(0\right)=-20{\left(0\right)}^3+14\left(0\right)+10=10>0\end{array}$

Now for $x\in \left(1.079,\infty \right)$ test for $x=2$

  $\begin{array}{l}{f}^{″}\left(x\right)=-20x^3+14x+10\\ f^{″}\left(2\right)=-20{\left(2\right)}^3+14\left(2\right)+10=-122<0\end{array}$

Therefore, the Function $y=-x^5+\frac{7}{3}{x}^3+5x^2+4x+2$ is concave up in interval $\left(-\infty ,1.079\right)$ .

Below is the graph of $f^{″}\left(x\right)=-20x^3+14x+10$

  

From graph it is clear that, the Function $y=-x^5+\frac{7}{3}{x}^3+5x^2+4x+2$ is concave up in interval $\left(-\infty ,1.079\right)$ .

d.

To determine

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

Answer to Problem 14RE

The Function $y=-x^5+\frac{7}{3}{x}^3+5x^2+4x+2$ is concave down in interval $\left(1.079,\infty \right)$ .

Explanation of Solution

Given:

The function is $y=-x^5+\frac{7}{3}{x}^3+5x^2+4x+2$ .

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, $y=-x^5+\frac{7}{3}{x}^3+5x^2+4x+2$

First derivative : $y^{\prime }=-5x^4+7x^2+10x+4$

Second derivative : $y^{″}=-20x^3+14x+10$

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

  $\begin{array}{l}-20x^3+14x+10=0\\ x=1.07\end{array}$

Therefore , there are two intervals that is $\left(-\infty ,1.079\right)$ and $\left(1.079,\infty \right)$

check the value of $f^{″}\left(x\right)$ in interval $\left(-\infty ,1.079\right)$ and $\left(1.079,\infty \right)$

Now for $x\in \left(-\infty ,1.079\right)$ test for $x=0$

  $\begin{array}{l}{f}^{″}\left(x\right)=-20x^3+14x+10\\ f^{″}\left(0\right)=-20{\left(0\right)}^3+14\left(0\right)+10=10>0\end{array}$

Now for $x\in \left(1.079,\infty \right)$ test for $x=2$

  $\begin{array}{l}{f}^{″}\left(x\right)=-20x^3+14x+10\\ f^{″}\left(2\right)=-20{\left(2\right)}^3+14\left(2\right)+10=-122<0\end{array}$

Therefore, the Function $y=-x^5+\frac{7}{3}{x}^3+5x^2+4x+2$ is concave down in interval $\left(1.079,\infty \right)$ .

Below is the graph of $f^{″}\left(x\right)=-20x^3+14x+10$

  

From graph it is clear that, the Function $y=-x^5+\frac{7}{3}{x}^3+5x^2+4x+2$ is concave down in interval $\left(1.079,\infty \right)$ .

e.

To determine

To find any local extreme values.

Answer to Problem 14RE

The function $y=-x^5+\frac{7}{3}{x}^3+5x^2+4x+2$ has local maxima at point $\left(1.5,16.4\right)$ and local minima at $\left(-0.48,1.10\right)$ .

Explanation of Solution

Given:

The function is $y=-x^5+\frac{7}{3}{x}^3+5x^2+4x+2$ .

Calculation:

Graph of $y=-x^5+\frac{7}{3}{x}^3+5x^2+4x+2$ is given below

  

From graph it is clear that the function $y=-x^5+\frac{7}{3}{x}^3+5x^2+4x+2$ has local maxima at point $\left(1.5,16.4\right)$ and local minima at $\left(-0.48,1.10\right)$ .

f.

To determine

To find inflections points.

Answer to Problem 14RE

The inflection point is at $x=1.079$ .

Explanation of Solution

Given:

The function is $y=-x^5+\frac{7}{3}{x}^3+5x^2+4x+2$ .

Calculation:

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

Since, $y=-x^5+\frac{7}{3}{x}^3+5x^2+4x+2$ changes concavity in interval $\left(-\infty ,1.079\right)$ and $\left(1.079,\infty \right)$ .

Therefore, the inflection point is at $x=1.079$ .