1. (#6 on exam 3 review) consider ( f(x)=\frac{x}{x^{2}-9} ) sketch a graph by completing the following…

Explanation:

Step1: Find vertical asymptotes

Set denominator (x^{2}-9 = 0), then (x^{2}=9), (x=\pm3). So vertical asymptotes are (x = 3) and (x=-3).

Step2: Find horizontal asymptote

Since degree of numerator (n = 1) and degree of denominator (m = 2), and (n<m). By the rule of horizontal asymptotes, (y = 0).

Step3: End - behavior

As (x\rightarrow\infty), (f(x)=\frac{x}{x^{2}-9}=\frac{\frac{x}{x^{2}}}{\frac{x^{2}}{x^{2}}-\frac{9}{x^{2}}}=\frac{\frac{1}{x}}{1 - \frac{9}{x^{2}}}\rightarrow0). As (x\rightarrow-\infty), (f(x)=\frac{x}{x^{2}-9}=\frac{\frac{x}{x^{2}}}{\frac{x^{2}}{x^{2}}-\frac{9}{x^{2}}}=\frac{\frac{1}{x}}{1-\frac{9}{x^{2}}}\rightarrow0)

Step4: Find the first - derivative

Use the quotient rule (y=\frac{u}{v}), (y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}). Here (u = x), (u^\prime=1), (v=x^{2}-9), (v^\prime = 2x). Then (f^\prime(x)=\frac{(x^{2}-9)-x(2x)}{(x^{2}-9)^{2}}=\frac{x^{2}-9 - 2x^{2}}{(x^{2}-9)^{2}}=\frac{-x^{2}-9}{(x^{2}-9)^{2}}=-\frac{x^{2}+9}{(x^{2}-9)^{2}}) Since (x^{2}+9>0) and ((x^{2}-9)^{2}>0) for (x\neq\pm3), (f^\prime(x)<0) for all (x\neq\pm3). So (f(x)) is decreasing on ((-\infty,-3)), ((-3,3)) and ((3,\infty)). There are no local extreme values because (f^\prime(x)) never changes sign.

Step5: Find the second - derivative

Use the quotient rule again. Let (y = f^\prime(x)=-\frac{x^{2}+9}{(x^{2}-9)^{2}}), (u=-(x^{2}+9)), (u^\prime=-2x), (v=(x^{2}-9)^{2}), (v^\prime = 2(x^{2}-9)\times2x = 4x(x^{2}-9)) (f^{\prime\prime}(x)=\frac{-2x(x^{2}-9)^{2}+4x(x^{2}-9)(x^{2}+9)}{(x^{2}-9)^{4}}=\frac{-2x(x^{2}-9)+4x(x^{2}+9)}{(x^{2}-9)^{3}}=\frac{-2x^{3}+18x + 4x^{3}+36x}{(x^{2}-9)^{3}}=\frac{2x^{3}+54x}{(x^{2}-9)^{3}}=\frac{2x(x^{2}+27)}{(x^{2}-9)^{3}}) Set (f^{\prime\prime}(x)=0), then (2x(x^{2}+27)=0) (since (x^{2}+27>0) for all real (x)), (x = 0) Test intervals: For (x\in(-\infty,-3)), let (x=-4), (f^{\prime\prime}(-4)=\frac{2\times(-4)\times((-4)^{2}+27)}{((-4)^{2}-9)^{3}}=\frac{-8\times(16 + 27)}{(16-9)^{3}}=\frac{-8\times43}{343}<0) For (x\in(-3,0)), let (x=-1), (f^{\prime\prime}(-1)=\frac{2\times(-1)\times((-1)^{2}+27)}{((-1)^{2}-9)^{3}}=\frac{-2\times28}{(-8)^{3}}=\frac{-56}{-512}>0) For (x\in(0,3)), let (x = 1), (f^{\prime\prime}(1)=\frac{2\times1\times(1 + 27)}{(1-9)^{3}}=\frac{56}{-512}<0) For (x\in(3,\infty)), let (x = 4), (f^{\prime\prime}(4)=\frac{2\times4\times(16+27)}{(16 - 9)^{3}}=\frac{8\times43}{343}>0)

So (f(x)) is concave down on ((-\infty,-3)) and ((0,3)), concave up on ((-3,0)) and ((3,\infty)). Inflection point at (x = 0), (f(0)=\frac{0}{0 - 9}=0), so inflection point is ((0,0))

Answer:

(a) Vertical asymptotes (x = 3) and (x=-3), horizontal asymptote (y = 0). As (x\rightarrow\pm\infty), (y\rightarrow0) (b) (f(x)) is decreasing on ((-\infty,-3)), ((-3,3)) and ((3,\infty)). No local extreme values (c) Concave down on ((-\infty,-3)) and ((0,3)), concave up on ((-3,0)) and ((3,\infty)). Inflection point ((0,0)) (d) Sketch the graph with vertical asymptotes (x=\pm3), horizontal asymptote (y = 0), decreasing on its domain, concave - down/up as described above and inflection point at ((0,0))