Question

1. (#6 on exam 3 review) consider ( f(x)=\frac{x}{x^{2}-9} ) sketch a graph by completing the following steps. (a) locate any asymptotes (if any) and determine the end behavior. (b) find where ( f ) is increasing/decreasing and any local extreme values. (c) find where ( f ) is concave up/concave down and any inflection points. (d) sketch a graph and label the relevant points.

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

Step3: End - behavior

As \(x
ightarrow\infty\), \(f(x)=\frac{x}{x^{2}-9}=\frac{\frac{x}{x^{2}}}{\frac{x^{2}}{x^{2}}-\frac{9}{x^{2}}}=\frac{\frac{1}{x}}{1 - \frac{9}{x^{2}}}
ightarrow0\). As \(x
ightarrow-\infty\), \(f(x)=\frac{x}{x^{2}-9}=\frac{\frac{x}{x^{2}}}{\frac{x^{2}}{x^{2}}-\frac{9}{x^{2}}}=\frac{\frac{1}{x}}{1-\frac{9}{x^{2}}}
ightarrow0\)

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 \(x
eq\pm3\), \(f^\prime(x)<0\) for all \(x
eq\pm3\). 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 \(x
ightarrow\pm\infty\), \(y
ightarrow0\)
(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 \((0,0)\)