Finding a function

1. Dec 10, 2008

Olle.s

1. The problem statement, all variables and given/known data
The function goes through the points -3,1 -1,3 and 1,4
f'(-3)=0 and f'(1)=0
f''(x)>0 when x<-1 and f''(x)<0 when x>-1
What is the function?
2. Relevant equations

3. The attempt at a solution
I take it this is a third degree equation?
f(x) = ax^3 + bx^2 + cx + d

When I plug in the values I get
From derivative
0 =27a– 6b+ c
0 =3a+ 2b+ c
From f(x)
1 =-27a +9b− 3c+ d
3 =-a+ b− c+ d
4 =a+ b+ c+ d

I substitute and get rid of c and d and then get
a=-5/48
and from this i get
b=-0.3125 c= 0.9375 d ≈ 3.48

but this is wrong, as the graph doesn't go through -3,1
What did I do wrong? ?
Thanks for help

2. Dec 10, 2008

tiny-tim

Welcome to PF!

Hi Olle.s! Welcome to PF!
erm … you had five equations and only four variables!

Try a fourth degree equation.

3. Dec 10, 2008

Defennder

Hmm, why did you assume it was a 3rd degree polynomial? There doesn't appear to be any soution to this system of equations; you can check it online via a linear solver:

4. Dec 10, 2008

gabbagabbahey

Re: Welcome to PF!

I don't think a fourth degree equation will work; the fact that $f''(x)<0 \text{ if } x>-1$ and $f''(x)>0\text{ if } x<-1$ says to me that the graph of f(x) has only one inflection point, and its at x=-1. A quartic won't accomplish that.

5. Dec 10, 2008

tiny-tim

oops!

oooh … I never noticed they were both at -1.

hmmm … that gives a sixth equation, f''(-1) = 0 (or ∞?) …

so maybe a fifth degree equation will do it?

Thanks, gabbagabbahey! It's a good job other people check up on me!

6. Dec 10, 2008

gabbagabbahey

Are you sure it doesn't pass through (-1,2) instead?

7. Dec 10, 2008

gabbagabbahey

Re: oops!

Actually I don't think there is a solution to the problem written as-is (if it passes through (-1,2) instead of (-1,3) then it's a different story)

Try sketching a graph of the function using the fact that it must be concave up for x<-1 and concave down for x>-1....it looks almost cubic, but a little fishy at x=-1.

-EDIT- I just realized that the problem doesn't claim that f is continuous....that changes things a little

8. Dec 10, 2008

tiny-tim

hmm … never thought of sketching it …

yes, (-1,2) does look a lot more likely!
uhh? what's wrong with looking fishy??

9. Dec 10, 2008

gabbagabbahey

The accompanying golden hue distracts one from the mathematical tasks at hand of course. :uhh: (-grasps at straws in search of witticisms-)

10. Dec 10, 2008

tiny-tim

i'll take that as a compliment!

11. Dec 10, 2008

Olle.s

Hi,
I'm certain it's -1, 3,
yeah I've sketched it, it's a third degree function with a negative value in front of x^3, I think.
And yeah, I also had some problems when juggling with the variables..
But perhaps there's a typo somewhere in the question that makes it unsolvable?
Because I think this should be failry easy!

12. Dec 10, 2008

gabbagabbahey

Well; it's not unsolvable. The function may or may not be continuous.

Given that you are told f''(x)<0 for all x>-1 and f''(0) for all x<-1, does that tell you that f''(x) exists everywhere? If a function is twice differentiable everywhere, then can you say whether or not it is continuous?
What places, if any might f(x) be discontinuous?

13. Dec 11, 2008

Olle.s

Hm... I'm not sure I follow you.
Why is the graph discontinuous?

14. Dec 11, 2008

gabbagabbahey

I haven't said that it is.

Is there any reason to assume otherwise though?