Science:Math Exam Resources/Courses/MATH104/December 2012/Question 02 (d)

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MATH104 December 2012

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Question 02 (d)
Consider the function $f(x)={\frac {2x^{3}}{3x^{2}-9}}$ Its first and second derivatives are given by $f'(x)={\frac {2x^{2}(x^{2}-9)}{3(x^{2}-3)^{2}}},\qquad {}f''(x)={\frac {4x(x^{2}+9)}{(x^{2}-3)^{3}}}$ Write the equation of all vertical asymptotes (if there are any). Show that $y={\frac {2}{3}}x$ is a slant asymptote.

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Hint
At what values of $x$ does $f(x)$ have possible vertical asymptotes? If you think that $f(x)$ has a vertical asymptote at $x=a$, then what does $\lim _{x\rightarrow a}f(x)$ satisfy? If $y=(2/3)x$ is a slant asymptote, then what does the difference between $f(x)$ and $y=(2/3)x$ become as $x$ gets large? Show this explicity.

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Solution 1

We know that $f(x)$ does not exist where its denominator equals 0: $3x^{2}-9=0\quad \rightarrow \quad x=\pm {\sqrt {3}}.$ So there are possible vertical asymptotes at $\pm {\sqrt {3}}$. To confirm that there is a vertical asymptote at $x={\sqrt {3}}$, we must show that $\lim _{x\rightarrow {\sqrt {3}}^{-}}f(x)=+\infty \,\,{\mbox{or}}\,\,-\infty ,\quad {\mbox{or}}\quad \lim _{x\rightarrow {\sqrt {3}}^{+}}f(x)=+\infty \,\,{\mbox{or}}\,\,-\infty .$ This is easy to show because as $x\rightarrow {\sqrt {3}}$ from the left (from values of $x$ less than ${\sqrt {3}}$), the numerator of $f(x)$ approaches $6{\sqrt {3}}$ and the denominator approaches 0 from the left (negative values). So we know that $\lim _{x\rightarrow {\sqrt {3}}^{-}}f(x)=-\infty .$ Similarly we can show that the following hold: $\lim _{x\rightarrow {\sqrt {3}}^{-}}f(x)=+\infty ,\quad \lim _{x\rightarrow -{\sqrt {3}}^{-}}f(x)=-\infty ,\quad \lim _{x\rightarrow -{\sqrt {3}}^{+}}f(x)=+\infty ,$ Therefore, there are two vertical asymptotes: ${\color {blue}x={\sqrt {3}},x=-{\sqrt {3}}.}$ To find the slant asymptote, we can use long division. ${\begin{array}{rrr}&&2x/3\\&3x^{2}-9&){\overline {2x^{3}+0x}}\\&&{\underline {2x^{3}-6x}}\\&&6x\end{array}}$ Thus $f(x)={\frac {2x}{3}}+{\frac {6x}{3x^{2}-9}}$. Therefore, y = (2/3)x is a slant asymptote of f(x).

Solution 2

Proceed as in solution 1 to find the vertical asymptotes. If $y={\frac {2}{3}}x$ is a slant (oblique) asymptote to $f(x)$ then the difference between $f(x)$ and the slant asymptote has to vanish when x gets very small or very large: $\lim _{x\rightarrow \infty }\left(f(x)-{\frac {2}{3}}x\right)=0,\quad {\mbox{or}}\quad \lim _{x\rightarrow -\infty }\left(f(x)-{\frac {2}{3}}x\right)=0$ We will show that the first equation is true. ${\begin{aligned}\lim _{x\rightarrow \infty }\left(f(x)-{\frac {2}{3}}x\right)&=\lim _{x\rightarrow \infty }\left({\frac {2x^{3}}{3x^{2}-9}}-{\frac {2}{3}}x\right)\\&=\lim _{x\rightarrow \infty }\left({\frac {2x^{3}}{3x^{2}-9}}-{\frac {2x(x^{2}-3)}{3(x^{2}-3)}}\right)\\&=\lim _{x\rightarrow \infty }\left({\frac {2x^{3}}{3x^{2}-9}}-{\frac {2x^{3}-6x}{3x^{2}-9}}\right)\\&=\lim _{x\rightarrow \infty }\left({\frac {6x}{3x^{2}-9}}\right)\\&=\lim _{x\rightarrow \infty }\left({\frac {6/x}{3-9/x^{2}}}\right)\\&=0\end{aligned}}$ Similarly, it can be shown that $\lim _{x\rightarrow -\infty }\left(f(x)-{\frac {2}{3}}x\right)=0.$ Therefore, y = (2/3)x is a slant asymptote of f(x).

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