# 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?