Chapter 4, Problem 10RE

$\text{(a)}$

To determine

To find: The interval on that the function is increasing.

Answer to Problem 10RE

The answer: Increasing $[-\sqrt{3},\sqrt{3}]$

Explanation of Solution

Given information:

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

Calculation:

  $\begin{array}{c}y=\frac{x}{x^2+2x+3}\\ y^{\prime }=\frac{1\left(x^2+2x+3\right)-x(2x+2)}{{\left(x^2+2x+3\right)}^2}\\ y^{\prime }=\frac{x^2+2x+3-2x^2-2x}{{\left(x^2+2x+3\right)}^2}\\ y^{\prime }=\frac{3-x^2}{{\left(x^2+2x+3\right)}^2}\end{array}$

  $y\text{'}$ is undefined:

  $\begin{array}{c}{\left(x^2+2x+3\right)}^2=0\\ \to x^2+2x+3=0\\ \to x^2+2x+1=-2\\ \to {(}^x=-2\\ \to \frac{3-x^2}{{\left(x^2+2x+3\right)}^2}=0\\ \to 3-x^2=0\\ \to \pm \sqrt{3}=x\end{array}$

For,

  $(-\infty ,-\sqrt{3}]$

Test

  $x=-2$ ,

So,

  $\begin{array}{c}{y}^{\prime }(-2)=\frac{3-{(-2)}^2}{{\left[{(-2)}^2+2(-2)+3\right]}^2}\\ =\frac{3-4}{{(4-4+3)}^2}\\ =\frac{-1}{3^2}\\ =\frac{-1}{9}<0\end{array}$

For,

  $[-\sqrt{3},\sqrt{3}]$

Test

  $x=0$ ,

So,

  $\begin{array}{c}{y}^{\prime }(0)=\frac{3-0^2}{{\left[{(0)}^2+2(0)+3\right]}^2}\\ =\frac{3}{3^2}\\ =\frac{1}{3}>0\end{array}$

For

  $[\sqrt{3},\infty )$

Test

  $x=2$

So,

  $\begin{array}{c}{y}^{\prime }(2)=\frac{3-2^2}{{\left[2^2+2(2)+3\right]}^2}\\ =\frac{3-4}{{(4+4+3)}^2}\\ =\frac{-1}{{11}^2}\\ =-\frac{1}{121}<0\end{array}$

Therefore, Therefore, the required interval on that the function is increasing $[-\sqrt{3},\sqrt{3}]$

$\text{(b)}$

To determine

To find: The interval on that the function is decreasing

Answer to Problem 10RE

The answer: Decreasing $(-\infty ,-\sqrt{3}]$ and $[\sqrt{3},\infty )$

Explanation of Solution

Given information:

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

Calculation:

  $\begin{array}{c}y=\frac{x}{x^2+2x+3}\\ y^{\prime }=\frac{1\left(x^2+2x+3\right)-x(2x+2)}{{\left(x^2+2x+3\right)}^2}\\ y^{\prime }=\frac{x^2+2x+3-2x^2-2x}{{\left(x^2+2x+3\right)}^2}\\ y^{\prime }=\frac{3-x^2}{{\left(x^2+2x+3\right)}^2}\end{array}$

  $y\text{'}$ is undefined:

  $\begin{array}{c}{\left(x^2+2x+3\right)}^2=0\\ \to x^2+2x+3=0\\ \to x^2+2x+1=-2\\ \to {(}^x=-2\\ \to \frac{3-x^2}{{\left(x^2+2x+3\right)}^2}=0\\ \to 3-x^2=0\\ \to \pm \sqrt{3}=x\end{array}$

For,

  $(-\infty ,-\sqrt{3}]$

Test

  $x=-2$ ,

So,

  $\begin{array}{c}{y}^{\prime }(-2)=\frac{3-{(-2)}^2}{{\left[{(-2)}^2+2(-2)+3\right]}^2}\\ =\frac{3-4}{{(4-4+3)}^2}\\ =\frac{-1}{3^2}\\ =\frac{-1}{9}<0\end{array}$

For,

  $[-\sqrt{3},\sqrt{3}]$

Test

  $x=0$ ,

So,

  $\begin{array}{c}{y}^{\prime }(0)=\frac{3-0^2}{{\left[{(0)}^2+2(0)+3\right]}^2}\\ =\frac{3}{3^2}\\ =\frac{1}{3}>0\end{array}$

For

  $[\sqrt{3},\infty )$

Test

  $x=2$

So,

  $\begin{array}{c}{y}^{\prime }(2)=\frac{3-2^2}{{\left[2^2+2(2)+3\right]}^2}\\ =\frac{3-4}{{(4+4+3)}^2}\\ =\frac{-1}{{11}^2}\\ =-\frac{1}{121}<0\end{array}$

Therefore, the required interval on that the function is decreasing $(-\infty ,-\sqrt{3}]$ and $[\sqrt{3},\infty )$ .

$(c)$

To determine

To find: The interval on that the function is concave up.

Answer to Problem 10RE

The answer: Concave up in intervals $(-2.584,-0.706)$ and $(3.290,\infty )$

Explanation of Solution

Given information:

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

Calculation:

  $\begin{array}{c}{y}^{\prime }=\frac{3-x^2}{{\left(x^2+2x+3\right)}^2}\\ y^{″}=\frac{{\left(x^2+2x+3\right)}^2(-2x)-\left(3-x^2\right)\left(2\left(x^2+2x+3\right)(2x+2)\right)}{{\left(x^2+2x+3\right)}^4}\\ =\frac{\left(x^2+2x+3\right)(-2x)-\left(3-x^2\right)(2(2x+2))}{{\left(x^2+2x+3\right)}^3}\\ =\frac{2x^3-18x-12}{{\left(x^2+2x+3\right)}^3}\\ =\frac{2\left(x^3-9x-6\right)}{{\left(x^2+2x+3\right)}^3}\\ 0=x^3-9x-6\\ x\approx -2.584\\ \approx -0.706\\ \approx 3.290\end{array}$

Interval

  $(-\infty ,-2.584)$

So,

  $y^{″}(-3)=-1/18<0$ concave down.

Interval

  $(-2.584,-0.706)$

So,

  $y^{″}(-1)=0.5>0$ concave up

Interval

  $(-0.706,3.290)$

So,

  $y^{″}(0)=-4/9<0$ concave down

Interval

  $(3.290,\infty )$

So,

  $y^{″}(4)=44/19683>0$ concave up

Therefore, the required interval on that the function is concave up intervals $(-2.584,-0.706)$ and $(3.290,\infty )$

$(d)$

To determine

To find: The interval on that the function is concave down.

Answer to Problem 10RE

The answer: intervals $(-\infty ,-2.584)$ and $(-0.706,3.290)$ .

Explanation of Solution

Given information:

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

Calculation:

  $\begin{array}{c}{y}^{\prime }=\frac{3-x^2}{{\left(x^2+2x+3\right)}^2}\\ y^{″}=\frac{{\left(x^2+2x+3\right)}^2(-2x)-\left(3-x^2\right)\left(2\left(x^2+2x+3\right)(2x+2)\right)}{{\left(x^2+2x+3\right)}^4}\\ =\frac{\left(x^2+2x+3\right)(-2x)-\left(3-x^2\right)(2(2x+2))}{{\left(x^2+2x+3\right)}^3}\\ =\frac{2x^3-18x-12}{{\left(x^2+2x+3\right)}^3}\\ =\frac{2\left(x^3-9x-6\right)}{{\left(x^2+2x+3\right)}^3}\\ 0=x^3-9x-6\\ x\approx -2.584\\ \approx -0.706\\ \approx 3.290\end{array}$

Interval

  $(-\infty ,-2.584)$

So,

  $y^{″}(-3)=-1/18<0$ concave down.

Interval

  $(-2.584,-0.706)$

So,

  $y^{″}(-1)=0.5>0$ concave up

Interval

  $(-0.706,3.290)$

So,

  $y^{″}(0)=-4/9<0$ concave down

Interval

  $(3.290,\infty )$

So,

  $y^{″}(4)=44/19683>0$ concave up

Therefore, the required interval on that the function is intervals $(-\infty ,-2.584)$ and $(-0.706,3.290)$ .

$\text{(e)}$

To determine

To find: The any local extreme value.

Answer to Problem 10RE

The answer: local minimum $(-\sqrt{3},\approx -0.683)$

And Local maximum $(\sqrt{3},\approx 0.183)$

Explanation of Solution

Given information:

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

Calculation:

  $\begin{array}{c}{y}^{\prime }=\frac{x^2+2x+3-x(2x+2)}{{\left(x^2+2x+3\right)}^2}\\ =\frac{x^2+2x+3-2x^2-2x}{{\left(x^2+2x+3\right)}^2}\\ =\frac{-x^2+3}{{\left(x^2+2x+3\right)}^2}\end{array}$

The domain of $y^{\prime }$ is so the only critical points are stationary points,

  $\begin{array}{c}{x}^2=3\\ \Rightarrow x_1=\sqrt{3},\\ x_2=-\sqrt{3}\end{array}$

So,

There is local minimum at $x=-\sqrt{3}$ with the value,

  $\begin{array}{c}y(-\sqrt{3})=\frac{-\sqrt{3}}{3-2\sqrt{3}+3}\\ \approx -0.683\end{array}$

There is local minimum at

  $x=\sqrt{3}$

with the value:

  $\begin{array}{c}y(\sqrt{3})=\frac{\sqrt{3}}{3+2\sqrt{3}+3}\\ \approx 0.183\end{array}$

Therefore, the required value of local minimum $(-\sqrt{3},\approx -0.683)$ and local maximum $(\sqrt{3},\approx 0.183)$ .

$\text{(f)}$

To determine

To find: The inflection point.

Answer to Problem 10RE

The answer: Inflection points are:

  $\begin{array}{c}x\approx -2.584,\\ x\approx -0.706,\\ x\approx 3.290\end{array}$

Explanation of Solution

Given information:

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

Calculation:

  $\begin{array}{c}{y}^{\prime }=\frac{3-x^2}{{\left(x^2+2x+3\right)}^2}\\ y^{″}=\frac{{\left(x^2+2x+3\right)}^2(-2x)-\left(3-x^2\right)\left(2\left(x^2+2x+3\right)(2x+2)\right)}{{\left(x^2+2x+3\right)}^4}\\ =\frac{\left(x^2+2x+3\right)(-2x)-\left(3-x^2\right)(2(2x+2))}{{\left(x^2+2x+3\right)}^3}\\ =\frac{2x^3-18x-12}{{\left(x^2+2x+3\right)}^3}\\ =\frac{2\left(x^3-9x-6\right)}{{\left(x^2+2x+3\right)}^3}\\ 0=x^3-9x-6\\ x\approx -2.584\\ \approx -0.706\\ \approx 3.290\end{array}$

Interval

  $(-\infty ,-2.584)$

So,

  $y^{″}(-3)=-1/18<0$ concave down.

Interval

  $(-2.584,-0.706)$

So,

  $y^{″}(-1)=0.5>0$ concave up

Interval

  $(-0.706,3.290)$

So,

  $y^{″}(0)=-4/9<0$ concave down

Interval

  $(3.290,\infty )$

So,

  $y^{″}(4)=44/19683>0$ concave up

Therefore, the required inflection points of the given equation. are:

  $\begin{array}{c}x\approx -2.584,\\ x\approx -0.706,\\ x\approx 3.290\end{array}$