Science:Math Exam Resources/Courses/MATH110/April 2012/Question 05 (f)

MATH110 April 2012

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Question 05 (f)
Let $\displaystyle f(x)={\frac {2x^{2}}{x^{2}-1}}$ (f) Determine where ƒ is concave up and where it is 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
We determine whether f is concave up or concave down by checking the sign of the second derivative, $f''$.

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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. From the previous question, we know that $f'(x)={\frac {-4x}{(x^{2}-1)^{2}}}$. Using the quotient rule, we calculate the second derivative ${\begin{aligned}f''(x)&={\frac {-4(x^{2}-1)^{2}-(-4x)(2(x^{2}-1)(2x))}{(x^{2}-1)^{4}}}\\&={\frac {-4(x^{2}-1)^{2}+16x^{2}(x^{2}-1)}{(x^{2}-1)^{4}}}\\&={\frac {-4(x^{2}-1)+16x^{2}}{(x^{2}-1)^{3}}}\\&={\frac {12x^{2}+4}{(x^{2}-1)^{3}}}\\\end{aligned}}$ For this fraction, the numerator is always positive so we need only check the sign of the denominator. The denominator is zero when $x=\pm 1$. For $x<-1$, the denominator is positive, for $-1<x<1$ the denominator is negative, and then for $x>1$ the denominator is again positive. Thus we have that the function is concave up on $(-\infty ,-1)\cup (1,\infty )$ and concave down on $(-1,1)$.

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