Question

Suppose I am given 4 points on a graph: (0,10), (4,2), (8,8) and (15,0).
Also, suppose you know that at each of these 4 points, the slope of the function(s) that draws them is zero, i.e. the derivative of this function is 0 at the four points.

Does anyone know a clever way to derive a cubic function that would hit each of these points AND have a derivative=0 at them?

Answer 1

My best guess thus far was to just plug them in as points for a cubic regression in my calculator, but I've no way to restrict the curve of the function so I cant "force" it to be 0 at the given points.

Answer 2

So, im thinking as I want it to be cubic, it will be of the form \[ax^3+bx^2+cx+d\]

and given the points Ive got, I know: f(0)=10 , f(4)=2, f(8)=8 f(15)=0, and since I want the slope to be zero at these points, then f'(0)=f'(4)=f'(8)=f'(15)=0

Answer 3

so using solving for f(0), I have d=10.

since f'(x) = 3ax^2+2bx+c , and given f'(0) = 0 , I think then that c=0. While I continue trying to solve for a and b, does this see the right way to go about this?

Answer 4

http://openstudy.com/study#/updates/5277da18e4b0275c5385dd0c if u can can u help me

Answer 5

a cubic function has at best, 2 points that have a derivative of 0

x^3 derives to 3x^2, and a poly of degree 2 has at most 2 zeros