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.