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

MATH104 December 2016

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Question 11 (d)
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. (d) Sketch the graph of $y=f(x)$.

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 produce an accurate sketch of a given function $f$, we may consider the following steps: Find the domain of $f$. Generally, we assume that the domain is the entire real line then find restrictions, such as where a denominator is $0$ or where negatives appear under the radical. Find the critical values of $f$. Find the possible points of inflection of $f$. Find the location of any vertical asymptotes of $f$ (usually done in conjunction with item 1 above). Consider the limits $\lim _{x\to -\infty }f(x)$ and $\lim _{x\to \infty }f(x)$ to determine the end behaviour of the function. Create a number line that includes all critical points, possible points of inflection, and locations of vertical asymptotes. For each interval created, determine whether $f$ is increasing or decreasing, concave up or down. Evaluate $f$ at each critical point and possible point of inflection. Plot these points on a set of axes. Connect theses points with curves exhibiting the proper concavity. Sketch asymptotes and $x$ and $y$ intercepts where applicable.

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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. Following the hint, we have The domain of $f$ is $(-\infty ,2),(2,\infty )$ since the denominator of $f$ is $0$ when $x=2$. From (b), we know that the function $f$ has a local minimum at $x=1$. The minimum point is $(1,-1).$ From (c), we know that $({\frac {1}{2}},-{\frac {7}{9}})$ is the only inflection point. From (a), we know that the vertical asymptote of $f$ is $x=2.$ By considering the limits $\lim _{x\to -\infty }f(x)$ and $\lim _{x\to \infty }f(x)$, we know that the horizontal asymptote is $y=1$. Details can be found in solution to part (a). From (b), we know that $f$ is increasing on $(1,2)$ and decreasing on $(-\infty ,1),(2,\infty ).$ From (c), we know that the function is concave up on $({\frac {1}{2}},2),(2,\infty )$ and concave down on $(-\infty ,{\frac {1}{2}}).$ Label the critical point $(1,-1)$ , the inflection point $({\frac {1}{2}},-{\frac {7}{9}})$ and the vertical asymptote $x=2$on the real line. Connect these points with curves exhibiting the proper concavity.