Algebra/calculus help...

In My Humble Opinion
1

OK I have this ridiculous algebra problem that’s resulted from a relatively simple calculus problem (graph function with min/max, concavity, etc…). Well, I took the second derivative of the function (x^2)/(x^3 + 1) and got up to this:

(x^3 + 1)^2 (-4x^3 + 2) - 6x^2 (x^4 + 2x)(x^3 + 1)

Now I have to solve for x with the function equal to zero. My classmates and I have spent a lot of time on this and the furthest any of us have gotten is
2x^9 + 6x^8 - 2x^6 - 20x^3 + 2 = 0

This is actually part of a take home test that we’ve been working on for quite a while. I know that I can look at the graph and see where the inflection points are but we need to show work for this. If anyone could help us out with either part of the problem it would be greatly appreciated, though some of you geniuses may be better off starting with the pre-simplification part because in all likelyhood we messed up somewhere on the way .

2

I believe this is the wrong forum…

3

Calculus! Booooo!! You will get much better help than me when this gets flipped over to GQ.

I took calculus last spring, and I promise to chime in later when I can borrow my friends book to take a look see to make sure I’m right.

If you haven’t already, double check to make sure that you’re using the proper rules with regards to how the function is formatted. IIRC, there’s a lot of goofy variations and combinations out there and sometimes it’s tough to get the right one.

4

This really belongs in GQ, I think, but…I don’t think I can pass up a tasty morsel of calculus such as this.

A general form of the quotient rule, where A and B are functions of x, and d denotes defferentiation, is:

f(x) = A/B

f’(x) = (BdA - AdB)/(B[sup]2[/sup])

and

f’’(x) = {B[sup]2[/sup][d(BdA - AdB)] - 2BdB(BdA - AdB)}/(B[sup]4[/sup])

Which can be simplified to:

f’’(x) = [d(Bda - AdB)/(B[sup]2[/sup])] - [2dB(BdA - AdB)/(B[sup]3[/sup])]

So now all we have to do is use the given function, f(x) = (x[sup]2[/sup])/(x[sup]3[/sup] + 1). In other words, A from our general example is now x[sup]2[/sup], and B is now x[sup]3[/sup] + 1. So, dA is 2x, and dB is 3x[sup]2[/sup], and we can substitute.

f(x) = (x[sup]2[/sup])/(x[sup]3[/sup] + 1)

f’(x) = (2x * (x[sup]3[/sup] + 1) - (x[sup]2[/sup] * 3x[sup]2[/sup]))/((x[sup]3[/sup] + 1)[sup]2[/sup])

= (2x - x[sup]4[/sup])/((x[sup]3[/sup] + 1)[sup]2[/sup])

and

f’’(x) = [((x[sup]3[/sup] + 1)[sup]2[/sup]) * (2 - 4x[sup]3[/sup]) - 2 * 3x[sup]2[/sup] * (x[sup]3[/sup] + 1) * (2x - x[sup]4[/sup])]/((x[sup]3[/sup] + 1)[sup]4[/sup])

Which, simplified, is:

f’’(x) = [(2 - 4x[sup]3[/sup])/((x[sup]3[/sup] + 1)[sup]2[/sup])] - [(6x[sup]2[/sup] * (2x - x[sup]4[/sup]))/((x[sup]3[/sup] + 1)[sup]3[/sup])]

If we then set f’’(x) equal to zero, we have:

[(2 - 4x[sup]3[/sup])/((x[sup]3[/sup] + 1)[sup]2[/sup])] - [(6x[sup]2[/sup] * (2x - x[sup]4[/sup]))/((x[sup]3[/sup] + 1)[sup]3[/sup])] = 0

or,

(x[sup]3[/sup] + 1) * (2 - 4x[sup]3[/sup]) - 6x[sup]2[/sup] * (2x - x[sup]4[/sup]) = 0

Which is just about where you were when you got here. (I really couldn’t control myself.) However, this can be further simplified by multiplying and adding like terms:

(2x[sup]3[/sup] - 4x[sup]6[/sup] + 2 - 4x[sup]3[/sup]) - (12x[sup]3[/sup] - 6x[sup]6[/sup]) = 0

or,

2(x[sup]6[/sup] - 7x[sup]3[/sup] + 1) = 0

But we can just drop the two. Letting y = x[sup]3[/sup], we have:

y[sup]2[/sup] - 7y + 1 = 0

A nice quadratic! Using the quadratic formula, which I don’t really feel like talking about, we get:

y = (7/2) + or - ((3/2) * 5sup[/sup])

and, since y = x[sup]3[/sup],

x = [(7/2) + or - ((3/2) * 5sup[/sup])]sup[/sup]

These values can easily be checked with a graphing calculator, and, having checked, I declare them accurate.

In place of a fee, I will accept co-authorship credit on your test, and a subsequent fraction of all the royalties derived therefrom.

5

Thanks a lot. It’s not a hard problem at all, but that algebra is a pain in the ass. I’m lucky that you have mad skills. That’s 20 points on the test that you helped me get. One more thing. I had another problem where I kept plugging numbers to the left and right of critical points into the original function to find whether the function in those areas was increasing or decreasing, but I was getting some wrong answers (I was comparing my results with a graph of the function on a TI-86). Then I started plugging the numbers into the derivative of the function and everything was coming out right. I remember being tought to plug the numbers into the original function, but I haven’t seen this stuff in over a year. Which is it?

6

Posting the same thread in two different forums is called cross-posting and is not allowed here. If you suspect you have posted in the wrong forum, please e-mail a moderator to move it for you.

bibliophage
moderator GQ