What is the Method for Finding a Cubic Function with Specific Zeros?

  • Thread starter Thread starter nanoWatt
  • Start date Start date
  • Tags Tags
    Cubic Function
nanoWatt
Messages
85
Reaction score
2

Homework Statement



From James Stewart's Essential Calculus Early Trancendentals, p.21 #5.

Find an expression for a cubic function f if f(1)=6 and f(-1)=f(0)=f(2)=0

Homework Equations



Used zeros of the function.

The Attempt at a Solution



I understand that the values of x (assuming it's f(x)) which make the value 0 are the zeros of the function. We want these x values to make the whole statement 0, so

since f(-1) = 0, then (x+1) is one of the zeros. Similarly (x-0) and (x-2) are zeros as well.

If I put them together, I get (x+1)(x)(x-2), but then I don't know what to do with the f(1) = 6.

The answer guide gives: f(x) = -3x(x+1)(x-2)

For some reason, the (x-0) just dropped off, and somehow f(1)=6 equates to -3x, or maybe just -3, and the x is from the (x-0).
 
Physics news on Phys.org
Putting the roots together gives you K*(x+1)(x)(x-2) where K is any constant. That still has the right roots. To determine K, put x=1 and adjust K so you get 6.
 
nanoWatt said:
If I put them together, I get (x+1)(x)(x-2), but then I don't know what to do with the f(1) = 6.

Now, f(x) = x(x + 1)(x - 2) is a function that you have arrived at from the fact that x, x + 1 and x - 2 are factors for f(x). However, these three monomials are also factors of.. let's say 2x(x + 1)(x - 2) or 4x(x + 1)(x - 2). Basically, any function of the form: f(x) = ax(x + 1)(x - 2) has those three monomials as it's factors.

so.. can you use the fact that f(1) = 6 to get the value of 'a'?? If done properly, you should get a = -3 which will match your answer...

EDIT:

goddammit.. dick beat me to it.. [no pun intended]
 
Thanks to both of you. Actually, before that was assigned to us, I went through some of the problems on my own, and tried that one. I asked the teacher about it, and she didn't seem to know how to do it. Then she assigned it to us that evening. It was the only problem in this week's homework I've had trouble with.
 
Well, of course. If you don't know how to solve a problem, give it to your students!
 
HallsofIvy said:
Well, of course. If you don't know how to solve a problem, give it to your students!

i'm going to go out and get myself a student...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top