Solving Function Homework w/ f'(-3)=0 & f'(1)=0

[Olle.s]

Homework Statement

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?

Homework Equations

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

 

[tiny-tim]

Welcome to PF!

Hi Olle.s! Welcome to PF!

I take it this is a third degree equation?
f(x) = ax^3 + bx^2 + cx + d
…
0 =27a– 6b+ c
0 =3a+ 2b+ c
…
1 =-27a +9b− 3c+ d
3 =-a+ b− c+ d
4 =a+ b+ c+ d
…
What did I do wrong? ?

erm … you had five equations and only four variables!

Try a fourth degree equation.

 

[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:
http://wims.unice.fr/wims/en_home.html

 

[gabbagabbahey]

Hi Olle.s! Welcome to PF!


erm … you had five equations and only four variables!

Try a fourth degree equation.

I don't think a fourth degree equation will work; the fact that f&#039;&#039;(x)&lt;0 \text{ if } x&gt;-1 and f&#039;&#039;(x)&gt;0\text{ if } x&lt;-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.

 

[tiny-tim]

oops!

I don't think a fourth degree equation will work; the fact that f&#039;&#039;(x)&lt;0 \text{ if } x&gt;-1 and f&#039;&#039;(x)&gt;0\text{ if } x&lt;-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.

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!

 

[gabbagabbahey]

...The function goes through the points -3,1 -1,3 and 1,4...

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

 

[gabbagabbahey]

...
so maybe a fifth degree equation will do it?...

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

 

[tiny-tim]

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

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

hmm … never thought of sketching it …

yes, (-1,2) does look a lot more likely!

...it looks almost cubic, but a little fishy at x=-1.

uhh? what's wrong with looking fishy??

 

[gabbagabbahey]

uhh? what's wrong with looking fishy??

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

 

[tiny-tim]

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

i'll take that as a compliment! ​

 

[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!

 

[gabbagabbahey]

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!

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?

 

[Olle.s]

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

 

[gabbagabbahey]

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

I haven't said that it is.

Is there any reason to assume otherwise though?