for the following function a) give the coordinates of any critical points and classify each point as a…

Explanation:

Step1: Find the first - derivative

We use the quotient rule. If $y = \frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = x$, $u'=1$, $v=x^{2}+25$, $v' = 2x$. So $y'=\frac{(x^{2}+25)\times1-x\times(2x)}{(x^{2}+25)^{2}}=\frac{x^{2}+25 - 2x^{2}}{(x^{2}+25)^{2}}=\frac{25 - x^{2}}{(x^{2}+25)^{2}}$.

Step2: Find critical points

Set $y' = 0$. Then $\frac{25 - x^{2}}{(x^{2}+25)^{2}}=0$. Since $(x^{2}+25)^{2}>0$ for all real $x$, we solve $25 - x^{2}=0$. So $x=\pm5$. When $x = 5$, $y=\frac{5}{5^{2}+25}=\frac{5}{50}=\frac{1}{10}$; when $x=-5$, $y=\frac{-5}{(-5)^{2}+25}=-\frac{1}{10}$. The critical points are $(5,\frac{1}{10})$ and $(-5,-\frac{1}{10})$.

Step3: Classify critical points

We use the second - derivative test. First, find the second - derivative. Using the quotient rule on $y'=\frac{25 - x^{2}}{(x^{2}+25)^{2}}$, where $u = 25 - x^{2}$, $u'=-2x$, $v=(x^{2}+25)^{2}$, $v'=2(x^{2}+25)\times2x = 4x(x^{2}+25)$. Then $y''=\frac{-2x(x^{2}+25)^{2}-(25 - x^{2})\times4x(x^{2}+25)}{(x^{2}+25)^{4}}=\frac{-2x(x^{2}+25)-4x(25 - x^{2})}{(x^{2}+25)^{3}}=\frac{-2x^{3}-50x-100x + 4x^{3}}{(x^{2}+25)^{3}}=\frac{2x^{3}-150x}{(x^{2}+25)^{3}}$. When $x = 5$, $y''=\frac{2\times5^{3}-150\times5}{(5^{2}+25)^{3}}=\frac{250 - 750}{(50)^{3}}<0$, so $(5,\frac{1}{10})$ is a relative maximum. When $x=-5$, $y''=\frac{2\times(-5)^{3}-150\times(-5)}{((-5)^{2}+25)^{3}}=\frac{-250 + 750}{(50)^{3}}>0$, so $(-5,-\frac{1}{10})$ is a relative minimum.

Step4: Find intervals of increase and decrease

Set up a sign - chart for $y'$. The critical points are $x=-5$ and $x = 5$. Test intervals: $(-\infty,-5)$: Let $x=-6$, $y'=\frac{25-(-6)^{2}}{((-6)^{2}+25)^{2}}=\frac{25 - 36}{(36 + 25)^{2}}<0$, so the function is decreasing on $(-\infty,-5)$. $(-5,5)$: Let $x = 0$, $y'=\frac{25-0}{(0 + 25)^{2}}>0$, so the function is increasing on $(-5,5)$. $(5,\infty)$: Let $x = 6$, $y'=\frac{25-6^{2}}{(6^{2}+25)^{2}}=\frac{25 - 36}{(36 + 25)^{2}}<0$, so the function is decreasing on $(5,\infty)$.

Step5: Find points of inflection

Set $y'' = 0$. Then $\frac{2x^{3}-150x}{(x^{2}+25)^{3}}=0$. Since $(x^{2}+25)^{3}>0$ for all real $x$, we solve $2x^{3}-150x=2x(x^{2}-75)=0$. So $x = 0,\pm5\sqrt{3}$. When $x = 0$, $y = 0$; when $x = 5\sqrt{3}$, $y=\frac{5\sqrt{3}}{(5\sqrt{3})^{2}+25}=\frac{5\sqrt{3}}{75 + 25}=\frac{\sqrt{3}}{20}$; when $x=-5\sqrt{3}$, $y=-\frac{\sqrt{3}}{20}$. The points of inflection are $(0,0),(5\sqrt{3},\frac{\sqrt{3}}{20}),(-5\sqrt{3},-\frac{\sqrt{3}}{20})$.

Step6: Find intervals of concavity

Set up a sign - chart for $y''$. The inflection points are $x=-5\sqrt{3},0,5\sqrt{3}$. Test intervals: $(-\infty,-5\sqrt{3})$: Let $x=-10$, $y''=\frac{2\times(-10)^{3}-150\times(-10)}{((-10)^{2}+25)^{3}}=\frac{-2000 + 1500}{(100 + 25)^{3}}<0$, so the function is concave down on $(-\infty,-5\sqrt{3})$. $(-5\sqrt{3},0)$: Let $x=-5$, $y''=\frac{2\times(-5)^{3}-150\times(-5)}{((-5)^{2}+25)^{3}}>0$, so the function is concave up on $(-5\sqrt{3},0)$. $(0,5\sqrt{3})$: Let $x = 5$, $y''=\frac{2\times5^{3}-150\times5}{(5^{2}+25)^{3}}<0$, so the function is concave down on $(0,5\sqrt{3})$. $(5\sqrt{3},\infty)$: Let $x = 10$, $y''=\frac{2\times10^{3}-150\times10}{(10^{2}+25)^{3}}=\frac{2000 - 1500}{(100 + 25)^{3}}>0$, so the function is concave up on $(5\sqrt{3},\infty)$.

Step7: Sketch the graph

Plot the critical points, points of inflection, and use the information about intervals of increase/decrease and concavity to sketch the graph.

Answer:

a) Critical points: $(5,\frac{1}{10})$ (relative maximum), $(-5,-\frac{1}{10})$ (relative minimum). b) Increasing on $(-5,5)$, decreasing on $(-\infty,-5)\cup(5,\infty)$. c) Points of inflection: $(0,0),(5\sqrt{3},\frac{\sqrt{3}}{20}),(-5\sqrt{3},-\frac{\sqrt{3}}{20})$. d) Concave up on $(-5\sqrt{3},0)\cup(5\sqrt{3},\infty)$, concave down on $(-\infty,-5\sqrt{3})\cup(0,5\sqrt{3})$. e) Sketch the graph using the above - found information.