Thread: Deriving a quartic equation from inflection points and a local minimum

This is a typical problem - but in reverse. How do I do this? I need to find f(x) = ax^4 + bx^3 - cx^2 -dx - e (solve for a,b,c,d,e). What I'm given are 2 inflection points ((v3)/3, -7.5752) and (-(v3)/3, -.647009) and a local minimum at (1.26255, -11.8696). I hope you can help!

what is (v3)/3?
it must be 0 for having two inflection points that are the same but different in sign.
because for being an inflection point, the second derivative must be zero. which means that 12ax^2+6bx-2c=0
if you substitute (v3)/3 and -(v3)/3 for x and subtract the second equation from the first, you will have: 12b((v3)/3)=0. which indicates whether b or v3 are 0.
I don't have any idea what is v3. Is it another variable?

Sorry. That should've been square root of 3 divided by 3 instead of (v3)/3. It didn't paste well.

Originally Posted by GregPeck

This is a typical problem - but in reverse. How do I do this? I need to find f(x) = ax^4 + bx^3 - cx^2 -dx - e (solve for a,b,c,d,e). What I'm given are 2 inflection points ((v3)/3, -7.5752) and (-(v3)/3, -.647009) and a local minimum at (1.26255, -11.8696). I hope you can help!

(Since a, b, c, d, and e are unknown numbers which can themselves be positive or negative, I see no reason to have those "-"s in the formula- I am going to write it as f(x)= ax^4+ bx^3+ cx^2+ dx+ e.)

Okay, you mean and which, since the y values are given in approximate decimal form, we might as well write as (0.577350, -7.75752) and (-.577350, -.647009).

If anything, this might be overdetermined. That is, there may be too much information.

We need to determine 5 values, a, b, c, d, e so, if anything, that is overdetermined. We know that when x= 0.577350, y= -7.75752, so we must have (0.577350)⁴a+ (0.577350)³b+ (0.577350)^2c+ (0.577350)d+ e= -7.75752. We know that when x= -0.577350, y= -0.647009 so that (-0.577350)⁴a+ (-0.577350)³b+ (-.577350)^2c+ (-0.577350)d+ e= -0.647009. We know that when x= 1.26255, y= -11.8696 so that (1.26255)⁴a+ (1.26255)³b+ (1.26255)^2c+ (1.26255)d+ e= -11.8696. The derivative of f is f'= 4ax^3+ 3bx^2+ 2cx+ d. The derivative will be 0 at a local maximum so we also have 4a(1.26255)³+ 3b(1.26255)²+ 2c(1.26255)+ d= 0. The second derivative of f is f''= 12ax^2+ 6bx+ 2c. The second derivative will be 0 at an inflection point so we have 12a(0.577350)²+ 6b(0.577350)+ 2c= 0 and 12a(-0.577350)²+ 6b(-0.577350)+ 2c= 0.

That is a total of 6 linear equations for the 5 unknown values. Solve any 5 for a, b, c, d, and e and then check to see if they satisfy the sixth equation. If not, there there is NO such function.

I said that if you subtract 12a(0.577350)2+ 6b(0.577350)+ 2c= 0 by 12a(-0.577350)2+ 6b(-0.577350)+ 2c= 0, you will have:
12b(0.577350)=0, which indicates b=0. Now, replace b=0 in HallsofIvy equations. but I think there is something wrong with this problem