Science:Math Exam Resources/Courses/MATH104/December 2012/Question 02 (a)

MATH104 December 2012

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Question 02 (a)
Consider the function $f(x)={\frac {2x^{3}}{3x^{2}-9}}$ Its first and second derivatives are given by $f'(x)={\frac {2x^{2}(x^{2}-9)}{3(x^{2}-3)^{2}}},\qquad {}f''(x)={\frac {4x(x^{2}+9)}{(x^{2}-3)^{3}}}$ On which intervals if f(x) increasing? On which intervals is f(x) decreasing?

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
Remember that to find intervals of increase and decrease, we need to check all the intervals between points where the derivative is undefined and where the derivative is zero. Find these values and check points between all these intervals.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. Since we are given the derivative, it is easy to see that the values of the derivative where it is equal to zero or undefined. The derivative is equal to zero where $x=0$ and where $x^{2}-9=0$, so at $x=\pm 3$. The derivative is undefined where $x^{2}-3=0$ so at the points $x=\pm {\sqrt {3}}$. Notice that the derivative's denominator is always positive since it is a positive number times a square. So when checking for sign changes, we need only to look at the derivative's numerator, that is, the function $\displaystyle 2x^{2}(x^{2}-9)$ (note we could check the denominator as well, but this trick saves a lot of valuable time on an exam!). Let's look at each interval individually. Case 1: $(-\infty ,-3)$ On this interval, we have that at the point $x=-4$ that the numerator is $\displaystyle 2(-4)^{2}((-4)^{2}-9)=224>0$ and so the function is increasing on this interval. Case 2: $(-3,-{\sqrt {3}})$ On this interval, we have that at the point $x=-2$ that the numerator is $\displaystyle 2(-2)^{2}((-2)^{2}-9)=-40<0$ and so the function is decreasing on this interval. Case 3: $(-{\sqrt {3}},0)$ On this interval, we have that at the point $x=-1$ that the numerator is $\displaystyle 2(-1)^{2}((-1)^{2}-9)=-16<0$ and so the function is decreasing on this interval. Case 4: $(0,{\sqrt {3}})$ On this interval, we have that at the point $x=1$ that the numerator is $\displaystyle 2(1)^{2}((1)^{2}-9)=-16<0$ and so the function is decreasing on this interval. Case 5: $({\sqrt {3}},3)$ On this interval, we have that at the point $x=2$ that the numerator is $\displaystyle 2(2)^{2}((2)^{2}-9)=-40<0$ and so the function is decreasing on this interval. Case 6: $(3,\infty )$ On this interval, we have that at the point $x=4$ that the numerator is $\displaystyle 2(4)^{2}((4)^{2}-9)=224>0$ and so the function is increasing on this interval. This completes the problem.

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