THird-Degree

nikchic5

Junior Member
Joined
Feb 16, 2006
Messages
106
Find the third-degree polynomials W such that Q(1)=1, Q'(1)=3, Q''(1)=6, and Q'''(1)=12.

Thank so much to anyone who can help me!!!
 
Q'''(1) = 12, but since this is 3rd degree, Q'''(x)=12
Q(x)=12x+c\displaystyle Q''(x) = 12x + c
Q(1)=6,\displaystyle Q''(1)=6, so: 12+c=6,c=6\displaystyle 12 + c = 6, c = -6
Q(x)=12x6\displaystyle Q''(x) = 12x - 6

Q(x)=6x26x+c\displaystyle Q'(x) = 6x^2 - 6x + c
Q(1)=3,\displaystyle Q'(1) = 3, so: 0+c=3,c=3\displaystyle 0 + c = 3, c = 3
Q(x)=6x26x+3\displaystyle Q'(x) = 6x^2 - 6x + 3

Q(x)=2x33x2+3x+c\displaystyle Q(x)=2x^3 - 3x^2 + 3x + c
Q(1)=1,\displaystyle Q(1) = 1, so: 2+c=1,c=1\displaystyle 2 + c = 1, c=-1
Q(x)=2x33x2+3x1\displaystyle Q(x) = 2x^3 - 3x^2 + 3x - 1
 
It was me who had the error, Gene. You were correct. I fixed mine. You should put your solution back up.
 
I ran out of time before I could go over both of them so I cut mine. Oh well, as long as one method is here. It's nice that we agreed eventually.
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Gene
 
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