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:
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 .
Just a quick thought… If you want to minimize the x[sup]2[/sup]/(x[sup]3[/sup]-1), Won’t that be the same as maximize the inverse? Look at (x[sup]3[/sup]-1)/x[sup]2[/sup] = x - 1/x[sup]2[/sup] instead!
I see now that this was just some nefarious ploy to get me to waste time in IMHO. Well, I have a special punishment in store for the likes of you.
Guards! Seize him!
I haven’t seen the other thread Firx mentions, so this may be better answered there, but you can factor an (x[sup]3[/sup] + 1) out of that pre-simplified expression. That should leave you with something that’s quadratic on x[sup]3[/sup]. I’m not sure where that 6x[sup]8[/sup] came from, but it shouldn’t be there.
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