Science:Math Exam Resources/Courses/MATH110/April 2019/Question 05 (c)

MATH110 April 2019

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Question 05 (c)
Let $f(x)={\frac {x^{3}}{x^{2}+1}}$. Find the interval(s) on which f(x) 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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Recall that a function f(x) is concave down at all points where its second derivative is negative; that is, f ′′(x) < 0. Computing f ′(x) is straightforward, but computing f ′′(x) will require more effort. The final expression is not that messy!

Hint 2
After simplification, the second derivative is: $f''(x)=-{\frac {2x(x^{2}-3)}{(x^{2}+1)^{3}}}.$

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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. Recall that a function f(x) is concave down at all points where its second derivative is negative; that is, f ′′(x) < 0. From part (a), we know that $f'(x)={\frac {x^{4}+3x^{2}}{(x^{2}+1)^{2}}}.$ Using the quotient rule again, we find that $f''(x)={\frac {(x^{4}+3x^{2})'(x^{2}+1)^{2}-(x^{4}+3x^{2})((x^{2}+1)^{2})'}{(x^{2}+1)^{4}}}={\frac {(4x^{3}+6x)(x^{2}+1)^{2}-(x^{4}+3x^{2})(2(x^{2}+1)\cdot 2x)}{(x^{2}+1)^{4}}}.$ This simplifies to $f''(x)={\frac {(x^{2}+1)\cdot \left(x(4x^{2}+6)(x^{2}+1)-4x(x^{4}+3x^{2})\right)}{(x^{2}+1)^{4}}}={\frac {x(6-2x^{2})}{(x^{2}+1)^{3}}}=-{\frac {2x(x^{2}-3)}{(x^{2}+1)^{3}}}.$ The denominator is always positive (by the same reasoning as in the solution to part (b)), so f "(x) will be negative whenever the numerator is negative, in symbols, $f''(x)<0\,{\text{ whenever }}\,x(x^{2}-3)<0.$ To figure out for which values of x the inequality $x(x^{2}-3)<0$ holds, factor the expression as $x(x^{2}-3)=x(x-{\sqrt {3}})(x+{\sqrt {3}}).$ We do this because it's easier to figure out for what x each of $x$, $x-{\sqrt {3}}$ or $x+{\sqrt {3}}$ is negative and we know that a product of three numbers is negative if one of them is negative and the other two are positive or if all three are negative. Therefore $x(x-{\sqrt {3}})(x+{\sqrt {3}})<0$ if $\color {blue}x<-{\sqrt {3}}$ or if $x$ is in the interval $\color {blue}(0,{\sqrt {3}})$, in which case $x-{\sqrt {3}}<0$ and $x(x+{\sqrt {3}})$ is positive.