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

MATH104 December 2012

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Question 02 (b)
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) concave up? On which intervals is f(x) concave down?

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 where the function is concave up or concave down, we need to check all the intervals between points where the second derivative derivative is undefined and where the derivative is zero. Find these values and check points between all these intervals.

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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. Since we are given the second derivative, it is easy to see that the derivative is 0 at $x=0$. The derivative is undefined when $x^{2}-3=0$ so at the points $x=\pm {\sqrt {3}}$. Let's look at each interval individually. Case 1: $(-\infty ,-{\sqrt {3}})$ On this interval, we have that at the point $x=-2$ that the second derivative is $\displaystyle {\frac {4(-2)((-2)^{2}+9)}{((-2)^{2}-3)^{3}}}=(-8)(13)<0$ and so the function is concave down on this interval. Case 2: $(-{\sqrt {3}},0)$ On this interval, we have that at the point $x=-1$ that the second derivative is $\displaystyle {\frac {4(-1)((-1)^{2}+9)}{((-1)^{2}-3)^{3}}}={\frac {-40}{-8}}=5>0$ and so the function is concave up on this interval. Case 3: $(0,{\sqrt {3}})$ On this interval, we have that at the point $x=1$ that the second derivative is $\displaystyle {\frac {4(1)((1)^{2}+9)}{((1)^{2}-3)^{3}}}={\frac {40}{-8}}=-5<0$ and so the function is concave down on this interval. Case 4: $({\sqrt {3}},\infty )$ On this interval, we have that at the point $x=2$ that the second derivative is $\displaystyle {\frac {4(2)((2)^{2}+9)}{((2)^{2}-3)^{3}}}=(8)(13)>0$ and so the function is concave up on this interval. This completes the problem.

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