Hello,
it's me again, the guy with the Hermite-interpolation.
I now decided to post a concrete task so maybe it's easier to answer.

The following points and the accompanying gradients are given:

x0=0 f(x)=3 f'(x)=0
x1=3 f(x)=3 f'(x)=27
x2=-2 f(x)=-57 f'(x)=72

I now would like to find out a polynomial function with goes through all those mentioned points and the same gradient in those points.

guy.

2. Originally Posted by guuy
Hello,
it's me again, the guy with the Hermite-interpolation.
I now decided to post a concrete task so maybe it's easier to answer.

The following points and the accompanying gradients are given:

x0=0 f(x)=3 f'(x)=0
x1=3 f(x)=3 f'(x)=27
x2=-2 f(x)=-57 f'(x)=72

I now would like to find out a polynomial function with goes through all those mentioned points and the same gradient in those points.

guy.
I don't know anything about that Hermite-something. But I played with your question using what little I know about points, gradients and characteristics of some known equations/functions. And I found an equation/function that you're looking for.

Graphing those ordered pairs on the same x,y plane, I saw that the graph of the unknown function is rising steeply at point (-2,-57), then it level off at (0,3), then goes down after (0,3), and then rises steeply again at (3,3).
That looks like a vertical S-curve. A cubic curve in x.
So I played with:
f(x) = Ax^3 +Bx^2 +Cx +D -----(1)
So,
f'(x) = 3Ax^2 +2Bx +C -----------(2)

At (x=0, f'(x)=0), in (2):
0 = 3A(0) +2B(0) +C
C = 0 --------------------------***
So, (2) becomes
f'(x) = 3Ax^2 +2Bx ------------(2a)

At (x=3, f'(x)=27), in (2a):
27 = 3A(3^2) +2B(3)
9 = 9A +2B --------------------(3)

At (x=-2, f'(x)=72), in (2a):
72 = 3A((-2)^2) +2B(-2)
72 = 12A -4B
18 = 3A -B
B = 3A -18
Plug that into (3),
9 = 9A +2(3A -18)
9 = 9A +6A -36
A = 45/15 = 3 ---------------------------***
So, (2a) becomes f'(x) = 9x^2 +2Bx -----------(2b)

Back to (x=3, f'(x)=27), in (2b):
27 = 9(3^2) +2B(3)
27 = 81 +6B
B = (27-81)/6 = -9 ---------------------***

So (1) becomes f(x) = 3x^3 -9x^2 +D -----(1a)

At (x=0, f(x)=3), in (1a):
3 = 3(0) -9(0) +D
D = 3 ---------------------------------***
So (1a) becomes f(x) = 3x^3 -9x^2 +3 .

And, f(x) = 3x^3 -9x^2 +3 is the function you are looking for.

-----------
I checked that on all the given ordered pairs and it checked allright.