Science:Math Exam Resources/Courses/MATH100/December 2016/Question 08

MATH100 December 2016

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Question 08
Consider the following six graphs on the right hand side. graphs Match the following five functions with the graphs: (i) ${\frac {1}{1-x^{2}}}$ (ii) ${\frac {x}{x^{2}-1}}$ (iii) ${\frac {x^{3}}{2(1-x^{2})}}$ (iv) ${\frac {1}{x^{2}-1}}$ (v) ${\frac {x^{3}}{2(x^{2}-1)}}$

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Hint
Identify the type of asymptotes of the given functions. Two of them have slant asymptotes while three of them have horizontal asymptotes. After you match the functions with slant asymptotes, for the rest of them find their derivatives to see whether there are any extrema and check for the sign of the derivative to help you determine if the functions are increasing/decreasing.

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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. All the five given functions have two vertical asymptotes at $x=1$ and $x=-1$. Functions (i), (ii) and (iv) have a horizontal asymptote at $y=0$. However, functions (iii) and (v) have a slant asymptote which is actually because the degree of the numerator (3) is the degree denominator (2) +1. This means that the graphs for these two functions are (B) and (C). To determine which one is for which function one way is to use long division of $x^{3}$ by $x^{2}-1$ and $x^{3}$ by $1-x^{2}$ to find the equation of the slant asymptote. Without doing any computation, we can see that since in $1-x^{2}$ the leading term has negative coefficient the long division will give a line with negative slope which means that (iii) $\ {\frac {x^{3}}{2(1-x^{2})}}\longrightarrow {\text{slant asymptote with negative slope}}\longrightarrow$ C (v) $\ {\frac {x^{3}}{2(x^{2}-1)}}\longrightarrow {\text{slant asymptote with positive slope}}\longrightarrow$ B For the remaining functions, let's find their derivative to see whether they have any extrema and their decreasing/increasing behavior: (i) $y={\frac {1}{1-x^{2}}}\longrightarrow y'={\frac {2x}{(1-x^{2})^{2}}}=0\longrightarrow x=0$ is a local minimum because for $x<0$ the derivative is negative (decreasing graph) and for $x>0$ the derivative is positive (increasing graph). $\longrightarrow$ F Similarly for (iv) $y={\frac {1}{x^{2}-1}}\longrightarrow y'={\frac {-2x}{(x^{2}-1)^{2}}}=0\longrightarrow x=0$ is a local maximum because for $x<0$ the derivative is positive (increasing graph) and for $x>0$ the derivative is negative (decreasing graph). $\longrightarrow$ A For (ii) $y={\frac {x}{x^{2}-1}}\longrightarrow y'={\frac {(x^{2}-1)-2x^{2}}{(x^{2}-1)^{2}}}={\frac {-1-x^{2}}{(x^{2}-1)^{2}}}\longrightarrow {\text{always negative}}\longrightarrow {\text{decreasing graph}}\longrightarrow$ E.