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

MATH104 December 2016

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Question 11 (b)

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. (b) Find the intervals on which ${\textstyle f(x)}$ is increasing or decreasing and classify the local extreme values.

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
If $f'>0$ on a interval $(a,b)$, then $f$ is increasing on the interval. On the other hand, if $f'<0$ on a interval $(c,d)$, then $f$ is decreasing on it.

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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. As in the Hint, if $f'>0$ on a interval $(a,b)$, then $f$ is increasing on the interval. On the other hand, if $f'<0$ on a interval $(c,d)$, then $f$ is decreasing on it. Since the derivative of $f$ is given in the question as follows $f'(x)={\frac {4(1-x)}{(x-2)^{3}}}$, we see that the numerator is 0 for $x=1$, and the denominator is 0 for $x=2$. These are our critical points, i.e. the only places where the might change sign. We first determine the sign of the derivative, based on the sign of the numerator and the denominator; - $(-\infty ,1)$ $(1,2)$ $(2,\infty )$ $4(1-x)$ $+$ $-$ $-$ $(x-2)^{3}$ $-$ $-$ $+$ $f'(x)={\frac {4(1-x)}{(x-2)^{3}}}$ $-$ $+$ $-$ Then, this implies that $f$ is increasing on ${\textstyle (1,2)}$ and decreasing on ${\textstyle (-\infty ,1),(2,\infty )}$. Since $f'(1)=0$ and $f''(1)>0$, we have a local mininum at the point ${\textstyle (1,-1)}$. On the other hand, $f$ is not defined on $x=2$ (recall that $x=2$ is actually a vertical asymptote from part (a)), a local extremum doesn't occur at $x=2$. Answer: $\color {blue}{\text{Increasing on }}(1,2),{\text{ decreasing on }}(-\infty ,1),(2,\infty ),{\text{ local minimum at }}(1,-1)$