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

  • Thread starter Olle.s
  • Start date
  • Tags
    Function
In summary: But that's pretty much it. I think it might be easiest to graph it on a coordinate plane and see if there's a pattern. But that's really all I can think of off the top of my head. Thanks for trying though!
  • #1
Olle.s
3
0

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
 
Physics news on Phys.org
  • #2
Welcome to PF!

Hi Olle.s! :smile:Welcome to PF! :smile:
Olle.s said:
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! :redface:

Try a fourth degree equation. :wink:
 
  • #3
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
 
  • #4


tiny-tim said:
Hi Olle.s! :smile:Welcome to PF! :smile:


erm … you had five equations and only four variables! :redface:

Try a fourth degree equation. :wink:

I don't think a fourth degree equation will work; the fact that [itex]f''(x)<0 \text{ if } x>-1[/itex] and [itex]f''(x)>0\text{ if } x<-1[/itex] 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
oops!

gabbagabbahey said:
I don't think a fourth degree equation will work; the fact that [itex]f''(x)<0 \text{ if } x>-1[/itex] and [itex]f''(x)>0\text{ if } x<-1[/itex] 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. :redface:

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

so maybe a fifth degree equation will do it?

Thanks, gabbagabbahey! :smile: It's a good job other people check up on me! :wink:
 
  • #6
Olle.s said:
...The function goes through the points -3,1 -1,3 and 1,4...

Are you sure it doesn't pass through (-1,2) instead? :wink:
 
  • #7


tiny-tim said:
...
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 :smile:
 
  • #8
gabbagabbahey said:
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
gabbagabbahey said:
Are you sure it doesn't pass through (-1,2) instead? :wink:

hmm … never thought of sketching it …

yes, (-1,2) does look a lot more likely! :smile:
...it looks almost cubic, but a little fishy at x=-1.

uhh? what's wrong with looking fishy?? o:)
 
  • #9
tiny-tim said:
uhh? what's wrong with looking fishy?? o:)

The accompanying golden hue distracts one from the mathematical tasks at hand of course. :rolleyes: (-grasps at straws in search of witticisms-)
 
  • #10
gabbagabbahey said:
The accompanying golden hue distracts one from the mathematical tasks at hand of course. :rolleyes: (-grasps at straws in search of witticisms-)

:biggrin: i'll take that as a compliment! :biggrin:​
 
  • #11
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!:frown:
 
  • #12
Olle.s said:
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!:frown:

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
Hm... I'm not sure I follow you.
Why is the graph discontinuous?
 
  • #14
Olle.s said:
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?
 

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

1.

What does it mean when f'(-3) = 0 and f'(1) = 0 in a function?

When the derivative of a function is equal to 0 at a specific point, it means that the slope of the tangent line at that point is 0. This can also be interpreted as the function having a constant value at that point, or a horizontal tangent line on the graph.

2.

How do I solve a function homework with f'(-3) = 0 and f'(1) = 0?

To solve a function with these conditions, you will need to find the original function using integration. Since the derivative of a constant function is always 0, you can set up the integral with the given values and solve for the constant. Once you have the original function, you can use it to find the values at any other point.

3.

Can a function have multiple points where f'(-3) = 0 and f'(1) = 0?

Yes, a function can have multiple points where the derivative is equal to 0. This means that there can be multiple points on the graph where the tangent line is horizontal. However, the function can only have one value of the constant at each point.

4.

What is the significance of finding f'(-3) = 0 and f'(1) = 0 in a function?

Finding the values of the derivative at specific points in a function can help us understand the behavior of the function. It can tell us where the function is increasing or decreasing, and where it has maximum or minimum values. It can also help us identify critical points and inflection points on the graph.

5.

Are there any other methods to solve a function homework with f'(-3) = 0 and f'(1) = 0?

Yes, there are other methods to solve a function with these conditions. One method is using the Mean Value Theorem, which states that if a function is continuous on a closed interval and differentiable on the open interval, then there must exist a point within the interval where the derivative is equal to the average rate of change of the function over that interval. Another method is using the Rolle's Theorem, which is a special case of the Mean Value Theorem and states that if a function has a value of 0 at two distinct points, then there must be a point between them where the derivative is also 0.

Similar threads

Differentiable function proof given ##f''(c) = 1##
Replies
1
Views
606
Medium Hard continuity proof Tutorial Q6
Replies
2
Views
615
Is the given function continuous at x= 0? f(x) = {sin(1/x) if x≠0, 1
Replies
4
Views
3K
Proof given ##x < y < z## and a twice differentiable function
Replies
8
Views
787
Show function series involving arctan is not differentiable at x=0
Replies
26
Views
1K
Taylor Series Expansion of f(x) at 0
Replies
7
Views
1K
Which of the following statements are true? (Real Analysis question)
Replies
15
Views
1K
f(x) = 2x+1, proving that it is continuous when p = 1 with 𝛿 and ε
Replies
7
Views
457
How should I use the Jacobi equation to show this?
Replies
4
Views
1K
The integral of a function ##f(x)## from its graph
Replies
15
Views
2K

Hot Threads

Recent Insights

Back
Top