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

MATH104 December 2011

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Question 02 (a)
Consider the function $f(x)={\frac {x^{3}+x^{2}-2x-3}{x^{2}-3}}.$ Its first and second derivatives are given by $f'(x)={\frac {(x^{2}-1)(x^{2}-6)}{(x^{2}-3)^{2}}},\qquad f''(x)={\frac {2x(x^{2}+9)}{(x^{2}-3)^{3}}}.$ Find all $x$ such that $f'(x)=0$ or $f'(x)$ does not exist.

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
Try setting the numerator and the denominator to 0. What does a value of 0 in the numerator correspond to? How about when the denominator is 0?

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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. To determine where f'(x)=0, we look at the numerator of f'(x) and note that f'(x)=0 when the numerator is zero, i.e. $(x^{2}-1)(x^{2}-6)=0,$ which implies x2-1=0 or x2-6=0. Solving each equation and noting that both positive and negative solutions are valid, we obtain $f'(x)=0{\text{ if and only if }}x=\pm 1{\text{ or }}x=\pm {\sqrt {6}}.$ To determine where f'(x) does not exist, we set the denominator to 0: $x^{2}-3=0.$ So, f'(x) does not exist when $x=\pm {\sqrt {3}}$. Remark: Note that the denominator and numerator are never simultaneously 0. If they were, say at x=a, we would have to take the limit of f'(x) for $x\rightarrow a$ to determine if f'(x) did not exist or was zero (or another finite number).

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