Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (f)

MATH104 December 2011

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Question 02 (f)
Consider the function $f(x)={\frac {x^{3}+x^{2}-2x-3}{x^{2}-3}}$ Its first and second derivatives are given by $f'(x)={\frac {(x^{2}-1)(x^{2}-6)}{(x^{2}-3)^{2}}}$ and $f''(x)={\frac {2x(x^{2}+9)}{(x^{2}-3)^{3}}}$ Find any asymptotes of the function f(x) and write their equations.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
How do we identify vertical asymptotes?

Hint 2
The power of the numerator is one greater than the denominator. What does that tell us about horizontal asymptotes? What about slant asymptotes?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Vertical Asymptotes We know that vertical asymptotes occur when the denominator is zero. Here our denominator is $\displaystyle {}x^{2}-3$ which is zero if $\scriptstyle x=\pm {\sqrt {3}}$. Therefore we have vertical asymptotes at $\scriptstyle x=-{\sqrt {3}}$ and $\scriptstyle x={\sqrt {3}}$. To see how the function looks near the asymptotes we could take some limits. However, from part (d) we have that the function is decreasing to the left of $\scriptstyle x=-{\sqrt {3}}$ as well as to the right. Therefore we conclude that $\lim _{x\to -{\sqrt {3}}^{-}}f(x)=-\infty$ while $\lim _{x\to -{\sqrt {3}}^{+}}f(x)=+\infty$ We have the exact same conclusion about the other vertical asymptote at $\scriptstyle x={\sqrt {3}}$ since it has the same decreasing properties as $\scriptstyle x={\sqrt {3}}$. Horizontal or Slant Asymptotes We check for horizontal or slant asymptotes. Since the degree of the numerator is larger than the degree of the denominator then we do not have any horizontal asymptotes. However, the degree of the numerator is exactly one larger than the degree of the denominator and so we do expect slant asymptotes. To figure out the slant asymptote we perform polynomial long division ${\begin{array}{cccccccccccc}&&&&&x&+&1&&&&\\&&&&&\_&\_&\_&\_&\_&\_&\_\\x^{2}&+&0x&-&3)&x^{3}&+&x^{2}&-&2x&-&3\\&&&&&x^{3}&+&0x^{2}&-&3x&&\\&&&&&\_&\_&\_&\_&\_&\_&\_\\&&&&&&&x^{2}&+&x&-&3\\&&&&&&&x^{2}&+&0x&-&3\\&&&&&\_&\_&\_&\_&\_&\_&\_\\&&&&&&&&&x&&\end{array}}$ Therefore we write $\displaystyle {}x^{3}+x^{2}-2x-3=(x^{2}-3)(x+1)+x$. So we get that ${\begin{aligned}{\frac {x^{3}+x^{2}-2x-3}{x^{2}-3}}&={\frac {(x^{2}-3)(x+1)+x}{x^{2}-3}}\\&=x+1+{\frac {x}{x^{2}-3}}\end{aligned}}$ Notice that for the last term, $\lim _{x\to \infty }{\frac {x}{x^{2}-3}}=0.$ Therefore as x gets big we see that ${\frac {x^{3}+x^{2}-2x-3}{x^{2}-3}}$ looks like x+1. If we take x going to $\scriptstyle -\infty$ we see that the limit of the last term still vanishes and it still looks like $x+1$. Therefore, we conclude that $x+1$ is a slant asymptote to the function. Summary Finally then we conclude that the equations for the vertical asymptotes are ${\begin{aligned}x&=-{\sqrt {3}}\\x&={\sqrt {3}}\end{aligned}}$ while the equation of the slant asymptote is $\displaystyle {}y=x+1.$

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