Question

Prolly simple enough, but;

given: f'(-1)=2, f'(3)=-1/4, and f'(4)=1

can you determine a suitable (non-piecewise) F(x) that matches?

Answer 1

you are the master at arranging a quadratic given three points.

Answer 2

indeed, but if we forma quadratic with these, doesnt that mean F(x) would be linear? and have the same slope at any given point?

Answer 3

i remember this discussion before, you figured it out in your dreams or something. i always did it the donkey way,

Answer 4

:)

Answer 5

you must have missed your coffee this morning
\(f'\) is quadratic, \(f\) is degree cubic

Answer 6

im thinking in reverse arent i

Answer 7

oh, and the coffee is finally out ;) thnx

Answer 8

i think in reverse at least half the time
the other half i think in neutral,
and the third half i think in third

Answer 9

lol, i have a one track mind that is usually derailed

Answer 10

this question is "prolly simple enough" :D

Answer 11

i had to write it in the teeny vernacular so that the youngins could appreciate it ;)

Answer 12

"youngins" :D

too much appreciated sir.

Answer 13

i spose we could make it more challenging and say that F(x) has to at least be a 5th degree polynomial

Answer 14

Oh "challenging" kinggeorge is enough for that..sir

Answer 15

then we shall keep it at a 3rd degree :)

Answer 16

given 3 points, we can form a setup with 3 coeffs \(c_1,c_2, and ~c_3\) such that the given x values work to zero out the unknown\[y=c_1+c_2(x-x_1)+c_3(x-x_1)(x-x_2)\]

Answer 17

ok ...

Answer 18

you can use the points in any order, but for sake of simplicity; ill just use them as stated such that
\[y=c_1+c_2(x+1)+c_3(x+1)(x-3)\]
when x=-1, y=2
\[2=c_1+c_2(-1+1)+c_3(-1+1)(-1-3)\]
\[2=c_1+0\]

when x=3; y= -1/4

\[-1/4=2+c_2(3+1)+c_3(3+1)(3-3)\]
\[-1/4-2=c_2(4)\]
\[-9/16=c_2\]
etc

Answer 19

from the figure it is clear that f'(x) cannot be linear....... so
let f'(x) be a quadratic, ax^2+bx+c then , f'(x)=ax^2+bx+c
Now, I think we can use three given points to find a,b,c........and after that we can find f(x)