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

MATH104 December 2012

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q1 (g)  •  Q1 (h)  •  Q1 (i)  •  Q1 (j)  •  Q1 (k)  •  Q1 (l)  •  Q1 (m)  •  Q1 (n)  •  Q1 (o)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q2 (d)  •  Q2 (e)  •  Q3  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q5 (a)  •  Q5 (b)  •  Q6 (a)  •  Q6 (b)  •  Q6 (c)  •  Q6 (d)  •  

   Other MATH104 Exams

  •  December 2014  •  December 2011  •  December 2010  •  December 2012  •  December 2013  •  December 2015  •  December 2016  •  

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.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

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.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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).

Click here for similar questions

MER QGH flag, MER QGQ flag, MER QGS flag, MER RT flag, MER Tag Asymptote, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag