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specifically, we were to find the points of inflection of the general quartic function: ax^4+bx^3+cx^2+dx+e. so i could get as far as the 2nd derivative and quad form, but then solving for the y value of the point was impossible. sorta

2006-11-14 02:54:59 · 5 answers · asked by ixoye706 5 in Science & Mathematics Mathematics

since there seems to be confusion, my real question is: in mathematical proofs, are you allowed to substitute numbers?

the rest is just the background of what i'm trying to do with it.

2006-11-14 03:13:17 · update #1

5 answers

It depends on what you are wanting to prove. Generally speaking, when you substitute numbers, you are restricting to a specific example. You cannot prove a general result by only looking at specific examples (unless you somehow show those specific cases give *all* the possibilities). However, you *can* show a result is false, if it is false for some specific example, so to show a general result is false, it is enough to 'plug in numbers'.

2006-11-14 04:07:25 · answer #1 · answered by mathematician 7 · 1 0

The quartic equation can be solved generally, though the form is not neat. If you would like to derive it yourself, show that the bx^3 term can be removed (i.e., that you can transform this general equation into one without a cubic term). Working with ax^4+cx^2+dx+e, factor out one solution and then use the quadratic formula for the other.

Cardano was the first to publish a solution to the general quartic, as I recall. Abel showed that the quintic is not solvable in general with addition, multiplication, and root extraction, and that more generally equations of 5th degree or higher cannot be solved in such fashion. Of course numerical methods like Newton's method can find approximate roots, and some special forms can be solved exactly.

2006-11-14 11:04:30 · answer #2 · answered by Charles G 4 · 0 0

Question is not clear. The function is in x variable. You are referring to y. Where does the y come from?

2006-11-14 11:04:44 · answer #3 · answered by openpsychy 6 · 0 0

f' = 4x^3+3bx^2+2cx+d
f'' = 12x^2 + 6bx+2c, why can't you set f''=0 and solve for x, using the quad formula?

2006-11-14 11:04:50 · answer #4 · answered by modulo_function 7 · 0 1

No....u r not allowed

2006-11-14 10:59:06 · answer #5 · answered by Keyur S 2 · 1 2

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