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4xloe

Member

Registered: 2008-12-10

Posts: 1

Finding the function

I know that
the graph goes through the points -3,1 -1,3 and 1,4
has f'(x)= 0 at x=-1 and x=1
and f''(x)<0 for x>-1 and f''(x)>0 for x<-1

I guess it's a third degree equation?
like this? f(x)=ax³+bx²+cx+d                 

If i plug in the numbers i get
-27a+9b−3c+d=1  3=-a+b−c+d   and     4=a+b+c+d 
and
0=27a–6b+c   0=3a+2b+c   
Now i substitue, right? After substitutions i find that b = -0.3125 and a = -5/48
With these two values i can find c and d, but i think something has gone wrong..
The graph i get doesnt go through -3,1:|

help? what am I doing wrong?

Last edited by 4xloe (2008-12-10 00:38:46)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Finding the function

I think your equation from f'(-1)=0 should read  3a - 2b + c = 0.

But the main reason this isn't working is that there are five simultaneous equations that you need to solve, but by trying to get a cubic you've only given yourself 4 variables.

If you try making f(x) quartic (ie. f(x) = ax^4 + bx^3 + cx^2 + dx + e), then you'll be able to find a, b, c, d, e that satisfy the conditions on f and f'.
f'' might still mess it up though, in which case you'll have to bring higher-order terms in again.

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It wanted to be normal.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Finding the function

cannot be a quartic (if it’s a polynomial).

f''(x)<0 for x>-1 and f''(x)>0 for x<-1

implies that

must be an odd-degree polynomial. Hence

itself must be of odd degree.

In fact,

must be of degree at least 7. For suppose

is quintic. Then

is a cubic of the form

where

and

. Thus

the sum of the coefficients of the odd-powered

-terms is 0,

which is totally impossible.

Hence

must at the very least be a septic polynomial. Good luck finding it, dude.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Finding the function

You can't even cheat with a piecewise function, since it has to be differentiable everywhere.
Maybe if we give it some antiseptic it'll go away.

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Why did the vector cross the road?
It wanted to be normal.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Finding the function

I think I see the problem with your function – though I should have spotted it sooner.

Now, you want

and

, and

for all

. Unfortunately, Rolle’s theorem implies that there would need to be a value

with

such that

.

Therefore such a function cannot exist.

Last edited by JaneFairfax (2008-12-23 08:32:04)

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Finding the function

/ common sense

f''(x) < 0 means the gradient is always decreasing, but if it's decreasing it can't go from 0... back to 0

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