5. -/1 points details my notes tanapcalc10 4.2.042.ep. consider the following function. g(x) = x / (2 + x^2)…

Explanation:

Step1: Find the first - derivative using quotient rule

The quotient rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. For $g(x)=\frac{x}{2 + x^{2}}$, where $u = x$, $u^\prime=1$, $v = 2 + x^{2}$, and $v^\prime = 2x$. Then $g^\prime(x)=\frac{1\cdot(2 + x^{2})-x\cdot(2x)}{(2 + x^{2})^{2}}=\frac{2 + x^{2}-2x^{2}}{(2 + x^{2})^{2}}=\frac{2 - x^{2}}{(2 + x^{2})^{2}}$.

Step2: Find the second - derivative using quotient rule

Let $u = 2 - x^{2}$, $u^\prime=-2x$, $v=(2 + x^{2})^{2}$, and $v^\prime = 2(2 + x^{2})\cdot2x=4x(2 + x^{2})$. Then $g^{\prime\prime}(x)=\frac{-2x\cdot(2 + x^{2})^{2}-(2 - x^{2})\cdot4x(2 + x^{2})}{(2 + x^{2})^{4}}=\frac{-2x(2 + x^{2})[(2 + x^{2})+2(2 - x^{2})]}{(2 + x^{2})^{4}}=\frac{-2x(2 + x^{2})(2 + x^{2}+4 - 2x^{2})}{(2 + x^{2})^{4}}=\frac{-2x(6 - x^{2})}{(2 + x^{2})^{3}}$.

Step3: Find intervals of concavity

Set $g^{\prime\prime}(x)=0$. Then $-2x(6 - x^{2})=0$, which gives $x = 0$, $x=\sqrt{6}$, and $x =-\sqrt{6}$. Test the intervals: For $x<-\sqrt{6}$, let $x=-3$. Then $g^{\prime\prime}(-3)=\frac{-2(-3)(6 - 9)}{(2 + 9)^{3}}=\frac{6\times(-3)}{11^{3}}<0$. For $-\sqrt{6}<x<0$, let $x=-1$. Then $g^{\prime\prime}(-1)=\frac{-2(-1)(6 - 1)}{(2 + 1)^{3}}=\frac{2\times5}{27}>0$. For $0<x<\sqrt{6}$, let $x = 1$. Then $g^{\prime\prime}(1)=\frac{-2\times1\times(6 - 1)}{(2 + 1)^{3}}=\frac{-10}{27}<0$. For $x>\sqrt{6}$, let $x = 3$. Then $g^{\prime\prime}(3)=\frac{-2\times3\times(6 - 9)}{(2+9)^{3}}=\frac{-6\times(-3)}{11^{3}}>0$.

Answer:

$g^\prime(x)=\frac{2 - x^{2}}{(2 + x^{2})^{2}}$ $g^{\prime\prime}(x)=\frac{-2x(6 - x^{2})}{(2 + x^{2})^{3}}$ concave upward: $(-\sqrt{6},0)\cup(\sqrt{6},\infty)$ concave downward: $(-\infty,-\sqrt{6})\cup(0,\sqrt{6})$