If you have two arbitrary points on the x-y plane, each with

Plot the derivatives and find a function that goes through the two derivative values. Take the integral of this function and find c such that the integral equation goes through one of the original points given.

>>8010463
f'(x) = (f'(a)-f'(b))/(a-b) (x-a) + f'(a)

>>8010463
>Plot the derivatives and find a function that goes through the two derivative values.
But that's just restating the problem. What is the function?

>>8010474
Meant to reply to>>8010470

>>8010474
No it's not. The problem is to graph a function f(x) with the slopes given. Graph the slopes as values of a new function, f'(x). The simplest function through two points is a line. So define f'(x) as a line which goes through f'(a) and f'(b). This is >>8010472

Now we take the integral of f'(x)

>>8010463
You want to use Newton's divided differences interpolation polynomial. To include derivatives, repeat the points and plug in the derivate f'(xi) for (f(xi)-f(xi))/(xi-xi)
https://en.wikipedia.org/wiki/Newton_polynomial

>>8010463
f(x) = f(a)+(x-a)((x+a-2b)f'(a)-(x-a)f'(b))/(2(a-b))

>>8010463
It can be pretty much anything really, it could be a squiggly wiggly, it could morph into a sign wave.

However, from your description, you probably want to use a cubic spline.

I would suggest reading up on C1 continuity though:
https://en.wikipedia.org/wiki/Smoothness

>>8010490

y(x) = y1 + y'1 (x-x1) + [(y2-y1)/(x2-x1) - y'1]/(x2-x1) (x-x1)^2 + [y'2+y'1 - 2(y2-y1)/(x2-x1)]/(x2-x1)^2 (x-x1)^2(x-x2)

>>8010510
Mine's better.

f(x) = f(a)+(x-a)((x+a-2b)f'(a)-(x-a)f'(b))/(2(a-b))

File: 1450771622829.png (300KB, 469x540px) Image search: [Google]

300KB, 469x540px

>>8010514
Well, n-now it is:
[eqn]f(x) = f(a) + (x-a) \frac{(x + a - 2b) f'(a) - (x-a) f'(b)}{2(a - b)}[/eqn]

>>8010463
I'm pretty sure there are at least continuum many functions that go through points A and B and have the specified derivatives at those points.

>>8010685
You're looking at the inverse image of a point under a linear transformation from an infinite dimensional space to a four dimensional one. Why are you only pretty sure?

>>8010692
Because there are definitely people who know more about functions than me on this board, so I may be proven wrong.

All functions f(x) that

f(a)=A
f(b)=B
f'(a)=C
f'(b)=D

>>8010472
>>8010497
>>8010510
>>8010514
>>8010522
So I just tried this and it's wrong.

>>8010463
Catmull-Rom interpolation would work wouldn't it?
Bezier would work too I guess.
Either way you'd just need to generate points along the slope of each

>>8012023
Yeah, I'm going with numerically approximated Bezier curves. Thanks, though,

>>8010463
People are really overthinking this. Given the following data points:

Point 1: (x0, y0) with slope m0
Point 2: (x1, y1) with slope m1

You need a function like:

f'(x)= m0*g0(x) + m1*g1(x) with g0 = 1 at x0 and 0 at x1. Similarly g1 = 1 at x1 and 0 at x0.

g0(x) = (x-x1)/(x0-x1) works for the first part.
g1(x) = (x-x0)/(x1-x0)works fro the second part.

Putting this together:

f'(x) = m0*(x-x1)/(x0-x1) + m1*(x-x0)/(x1-x0)

Integrate this and you get a nice polynomial of degree 2.You can extrapolate this to any number of points and slopes.

Note this is not a unique solution, just a polynomial solution that works. There are an infinite number of solution many of which may not be explicitly expressible in a closed form

>>8010463
Use Newton's Interpolating Polynomial and a difference table