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

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MATH104 December 2011

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Question 02 (d)
Consider the function $f(x)={\frac {x^{3}+x^{2}-2x-3}{x^{2}-3}}$ Its first and second derivatives are given by $f'(x)={\frac {(x^{2}-1)(x^{2}-6)}{(x^{2}-3)^{2}}}$ and $f''(x)={\frac {2x(x^{2}+9)}{(x^{2}-3)^{3}}}$ On which intervals is 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
Consider how concavity of a function relates to the second derivative.

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Solution

In order to find the concavity of the function, we first try to find points where the second derivative is zero or undefined. These points will provide endpoints of intervals for concavity. From part (b) we know that ƒ''(x) is zero exactly when x = 0, and that ƒ''(x) is undefined for x = ±√3 We can now either make a sign table or pick test points between those values to determine whether we are concave up (second derivative positive) or concave down (second derivative negative). For convenience, recall that $f''(x)={\frac {2x(x^{2}+9)}{(x^{2}-3)^{3}}}={\frac {2x(x^{2}+9)}{(x+{\sqrt {3}})^{3}(x-{\sqrt {3}})^{3}}}$ $(-\infty ,-{\sqrt {3}})$ $(-{\sqrt {3}},0)$ $(0,{\sqrt {3}})$ $({\sqrt {3}},\infty )$ $\displaystyle 2x$ $\displaystyle -$ $\displaystyle -$ $\displaystyle +$ $\displaystyle +$ $\displaystyle x^{2}+9$ $\displaystyle +$ $\displaystyle +$ $\displaystyle +$ $\displaystyle +$ $(x+{\sqrt {3}})^{3}$ $\displaystyle -$ $\displaystyle +$ $\displaystyle +$ $\displaystyle +$ $(x-{\sqrt {3}})^{3}$ $\displaystyle -$ $\displaystyle -$ $\displaystyle -$ $\displaystyle +$ $\displaystyle f''(x)$ $\displaystyle -$ $\displaystyle +$ $\displaystyle -$ $\displaystyle +$ Concavity Concave Down Concave Up Concave Down Concave Up Therefore $f(x)$ is concave up on $(-{\sqrt {3}},0)\cup ({\sqrt {3}},\infty )$ and concave down on $(-\infty ,-{\sqrt {3}})\cup (0,{\sqrt {3}})$.

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