# Finding concavity and factoring

I am trying to graph $$y=\frac{x^2}{x^2+3}$$but it is incredibly difficult for me because I am so bad with graphing rules and factoring.

I have come down to concavity and I just can't figure out what to do. I know the second derivative is something close to

$$\frac{6x^4 + 36x^2 +54-24x^4+72x^2}{(x^2+3)^4}$$

No matter how I rewrite it I can't make it something that I can factor or do anything with.

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The expression for the second derivative is incorrect, you should have had $-72x^2$ above. The numerator should be $-18x^4-36x^2+54 = 18 (1-x^2)(x^2+3)$. From this you can see where $y(x)$ is concave and convex. – copper.hat Apr 7 '12 at 2:01
it may help if you notice that $y = 1 - \dfrac{3}{x^2+3}$ – Jason Apr 7 '12 at 2:23

## 2 Answers

I assume you found that $$\frac{dy}{dx}=\frac{6x}{(x^2+3)^2}.\tag{\ast}$$ We want the second derivative. Do not expand anything at this stage, just differentiate. We get, using the Quotient Rule, $$\frac{d^2y}{dx^2}=\frac{(6)(x^2+3)^2-(4x)(x^2+3)(6x)}{(x^2+3)^4}.$$ Again, do not expand, look instead for common factors in the numerator. The numerator is equal to $$(x^2+3)(6)((x^2+3)-4x^2),$$ which simplifies to $(x^2+3)(6)(3-3x^2)$. So $$\frac{d^2y}{dx^2}=\frac{(x^2+3)(6)(3-3x^2)}{(x^2+3)^4}=\frac{(6)(3-3x^2)}{(x^2+3)^3}.$$ The last simplification (cancelling) is not really necessary or even useful, and sometimes can be a bad idea. The reason is that when you use the Quotient Rule, the denominator is naturally a square, so is $\ge 0$, and can usually be forgotten about if we are interested only in the sign of the derivative. So it is usually a good idea to leave the natural denominator alone.

I have not checked whether your expression for the second derivative is correct. But there are two things to consider before expanding. Any calculation carries some risk of error. And after expanding, there is a good chance that you will have to factor in order to find out where your expression is positive, negative, or $0$. After the expansion, the factorization may be far from obvious.

Remark: Scrambling an egg is far easier than unscrambling it. When an expression has structure, it is useful to hang on to that structure as much as possible.

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How did you know not to expand it? – user138246 Apr 7 '12 at 1:40
@Jordan: Remember that the last sentence in your question is about how hard it is to factor things. Why should we expand things out if we need to factor them in the end? – anon Apr 7 '12 at 1:43
@Jordan: Because I have worked out particular examples to classes many times. You too are gathering experience about approaches that work and approaches that often lead to trouble. – André Nicolas Apr 7 '12 at 1:45
@anon Because that is what I always do, I am not sure why. – user138246 Apr 7 '12 at 1:48
@Jordan: Often we can handle things that look complicated only because they have structure. For example, if we are working with $(2x+5)^6$, multiplying out creates a mess; $(2x+5)^6$ is far more tidy, for most purposes far easier to handle. – André Nicolas Apr 7 '12 at 1:51

It might help to remove the $x^2$ in the first numerator before the first differentiation:

$$\frac{d^2}{dx^2}\left(\frac{x^2}{x^2+3}\right)=\frac{d^2}{dx^2}\left(\frac{(x^2+3)-3}{x^2+3}\right)=\frac{d^2}{dx^2}\left(1-\frac{3}{x^2+3}\right) \tag{1}$$

$$=\frac{d}{dx}\left(-3(2x)\frac{-1}{(x^2+3)^2}\right)=\frac{d}{dx}\left(\frac{6x}{(x^2+3)^2}\right) \tag{2}$$

For the rest you can see Andre's answer.

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