Science:Math Exam Resources/Courses/MATH104/December 2016/Question 11 (a)

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MATH104 December 2016

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Question 11 (a)
Consider the function $f(x)={\frac {x^{2}-2}{(x-2)^{2}}}$ , and its derivatives $f'(x)={\frac {4(1-x)}{(x-2)^{3}}},~~~~~f''(x)={\frac {4(2x\color {red}-\color {black}1)}{(x-2)^{4}}}.$ Note: The expression for the second derivative given in the exam PDF is incorrect; the correct expression is given here. (a) Find all the asymptotes of ${\textstyle f(x)}$ .

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
To find vertical asymptotes, consider the points at which the denominator of $f$ vanishes. To get the horizontal ones, take limit on $f$ as $x$ goes to $\pm \infty$ .

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Solution
First, we find vertical asymptote(s) of $f$ . Since the function $f$ is rational and its denominator is zero at $x=2$ , it is a candidate for the vertical asymptote. To check this, we evaluate the limit of $f$ as $x$ approaches to $2$ from the left and the right. Note that for $x$ close to $2$ , the numerator is close to $2$ , because of $\lim _{x\to 2}x^{2}-2=2$ . Since the denominator $(x-2)^{2}$ is always non-negative, we get $\lim _{x\to 2^{-}}f(x)=\lim _{x\to 2^{-}}{\frac {x^{2}-2}{(x-2)^{2}}}=\lim _{x\to 2^{-}}(x^{2}-2)\cdot {\frac {1}{(x-2)^{2}}}=+\infty ,\quad \lim _{x\to 2^{+}}f(x)=+\infty .$ Therefore, we justify $x=2$ is the vertical asymptote of $g$ Now, we consider horizontal asymptote(s) of $f$ . For this purpose, it is enough to evaluate $\lim _{x\to \pm \infty }f(x)$ ; $\lim _{x\to \pm \infty }f(x)=\lim _{x\to \pm \infty }{\frac {x^{2}-2}{(x-2)^{2}}}=\lim _{x\to \pm \infty }{\frac {x^{2}-2}{x^{2}-4x+4}}=\lim _{x\to \pm \infty }{\frac {1-{\frac {2}{x^{2}}}}{1-{\frac {4}{x}}+{\frac {4}{x^{2}}}}}=1$ This implies that $y=1$ is the horizontal asymptote of $f$ . The asymptotes of $f$ : $\color {blue}x=2,y=1$ .