THird-Degree

[nikchic5]

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!!!

 

[daon]

Q'''(1) = 12, but since this is 3rd degree, Q'''(x)=12
\(\displaystyle Q''(x) = 12x + c\)
\(\displaystyle Q''(1)=6,\) so: \(\displaystyle 12 + c = 6, c = -6\)
\(\displaystyle Q''(x) = 12x - 6\)

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

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

 

[daon]

It was me who had the error, Gene. You were correct. I fixed mine. You should put your solution back up.

 

[Gene]

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