consider the equation below. (if an answer does not exist, enter dne.) f(x) = x / (x^2 + 9) (a) find the…

Explanation:

Step1: Find the first - derivative

Use the quotient rule. If $f(x)=\frac{u}{v}$ where $u = x$ and $v=x^{2}+9$, then $u'=1$ and $v' = 2x$. The quotient rule states that $f'(x)=\frac{u'v - uv'}{v^{2}}$. So $f'(x)=\frac{(x^{2}+9)\times1-x\times(2x)}{(x^{2}+9)^{2}}=\frac{x^{2}+9 - 2x^{2}}{(x^{2}+9)^{2}}=\frac{9 - x^{2}}{(x^{2}+9)^{2}}$.

Step2: Find critical points

Set $f'(x)=0$. Then $\frac{9 - x^{2}}{(x^{2}+9)^{2}}=0$. Since $(x^{2}+9)^{2}>0$ for all real $x$, we solve $9 - x^{2}=0$. Factoring gives $(3 - x)(3 + x)=0$, so $x=-3$ or $x = 3$.

Step3: Determine intervals of increase and decrease

Test intervals $(-\infty,-3)$, $(-3,3)$ and $(3,\infty)$. For $x<-3$, let $x=-4$. Then $f'(-4)=\frac{9-16}{(16 + 9)^{2}}=\frac{-7}{(25)^{2}}<0$, so $f(x)$ is decreasing on $(-\infty,-3)$. For $-3<x<3$, let $x = 0$. Then $f'(0)=\frac{9-0}{(0 + 9)^{2}}=\frac{1}{9}>0$, so $f(x)$ is increasing on $(-3,3)$. For $x>3$, let $x = 4$. Then $f'(4)=\frac{9 - 16}{(16+9)^{2}}=\frac{-7}{(25)^{2}}<0$, so $f(x)$ is decreasing on $(3,\infty)$.

Step4: Find local maximum and minimum

Since $f(x)$ changes from decreasing to increasing at $x=-3$, $f(-3)=\frac{-3}{9 + 9}=-\frac{1}{6}$ is a local minimum. Since $f(x)$ changes from increasing to decreasing at $x = 3$, $f(3)=\frac{3}{9+9}=\frac{1}{6}$ is a local maximum.

Step5: Find the second - derivative

Use the quotient rule on $f'(x)=\frac{9 - x^{2}}{(x^{2}+9)^{2}}$. Let $u = 9 - x^{2}$, $u'=-2x$, $v=(x^{2}+9)^{2}$, and $v'=2(x^{2}+9)\times2x = 4x(x^{2}+9)$. Then $f''(x)=\frac{-2x(x^{2}+9)^{2}-(9 - x^{2})\times4x(x^{2}+9)}{(x^{2}+9)^{4}}=\frac{-2x(x^{2}+9)-4x(9 - x^{2})}{(x^{2}+9)^{3}}=\frac{-2x^{3}-18x-36x + 4x^{3}}{(x^{2}+9)^{3}}=\frac{2x^{3}-54x}{(x^{2}+9)^{3}}=\frac{2x(x^{2}-27)}{(x^{2}+9)^{3}}$.

Step6: Find inflection points

Set $f''(x)=0$. Then $\frac{2x(x^{2}-27)}{(x^{2}+9)^{3}}=0$. Since $(x^{2}+9)^{3}>0$ for all real $x$, we solve $2x(x^{2}-27)=0$. Factoring gives $2x(x - 3\sqrt{3})(x+3\sqrt{3})=0$. So $x=-3\sqrt{3},0,3\sqrt{3}$. $f(-3\sqrt{3})=\frac{-3\sqrt{3}}{27 + 9}=-\frac{\sqrt{3}}{12}$, $f(0)=0$, $f(3\sqrt{3})=\frac{3\sqrt{3}}{27 + 9}=\frac{\sqrt{3}}{12}$. The inflection points are $(-3\sqrt{3},-\frac{\sqrt{3}}{12})$, $(0,0)$ and $(3\sqrt{3},\frac{\sqrt{3}}{12})$. Test intervals for concavity: For $x<-3\sqrt{3}$, let $x=-6$. Then $f''(-6)=\frac{2\times(-6)\times(36 - 27)}{(36 + 9)^{3}}=\frac{-12\times9}{(45)^{3}}<0$. For $-3\sqrt{3}<x<0$, let $x=-3$. Then $f''(-3)=\frac{2\times(-3)\times(9 - 27)}{(9 + 9)^{3}}=\frac{-6\times(-18)}{(18)^{3}}>0$. For $0<x<3\sqrt{3}$, let $x = 3$. Then $f''(3)=\frac{2\times3\times(9 - 27)}{(9 + 9)^{3}}=\frac{6\times(-18)}{(18)^{3}}<0$. For $x>3\sqrt{3}$, let $x = 6$. Then $f''(6)=\frac{2\times6\times(36 - 27)}{(36 + 9)^{3}}=\frac{12\times9}{(45)^{3}}>0$. $f(x)$ is concave up on $(-3\sqrt{3},0)\cup(3\sqrt{3},\infty)$.

Answer:

(a) Increasing interval: $(-3,3)$ Decreasing interval: $(-\infty,-3)\cup(3,\infty)$ (b) Local minimum value: $-\frac{1}{6}$ Local maximum value: $\frac{1}{6}$ (c) Inflection points: $(-3\sqrt{3},-\frac{\sqrt{3}}{12})$, $(0,0)$, $(3\sqrt{3},\frac{\sqrt{3}}{12})$ Concave - up interval: $(-3\sqrt{3},0)\cup(3\sqrt{3},\infty)$