Linear Algebra - Find a Polynomial

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Homework Help Overview

The problem involves finding polynomials of the form a + bx + cx² that pass through specific points and satisfy a derivative condition. The points given are (1,1) and (3,3), and the derivative at x=2 must equal 1.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss setting up equations based on the polynomial form and the conditions provided. There are attempts to derive a matrix representation for solving the system of equations. Some participants express confusion about the role of the constant term after differentiation and how to properly set up the equations.

Discussion Status

There is active engagement with participants questioning the correctness of their expressions and the setup of their equations. Some guidance has been offered regarding rewriting equations, but no consensus has been reached on the correct approach or solution method.

Contextual Notes

Participants note the challenge of maintaining the constant term after differentiation and the implications of the conditions on the polynomial form. There are indications of potential misunderstandings regarding the relationship between the equations derived from the conditions and the original polynomial form.

dkotschessaa
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Homework Statement

Find all polynomials of the form a + bx + cx^2 that:

Goes through the points (1,1) and (3,3)

and such that f'(2) = 1

Homework Equations



a + bx + cx^2
f'(x) = x+2cx
f'(2) = 2 + 4c

polynomial through (1,1) = a + b1 + c1 = 1
polynomial through (3,3) = a + b3+ c3^2 = 3

The Attempt at a Solution



I have the general idea that this should result in a series of equations that I need to do gauss Jordan on. Similar problems like this resulted in 3 similar equations and were quite simple.

My problem here is that, since I have taken the derivative of f(x) I have lost my constant a. So I'm not sure what my matrix should look like. I've tried:

1 1 1 1
1 3 9 3

But for the one with the derivative, my f"(2) isn't in the same form - it's 2+4c = 1

so I"m not sure whether to use

1 2 4 1
0 2 4 1

or something more general. If I use something more general though like:

a 2 4 1

I can't get a pivot in my first column...

'elp!

-Dave K
 
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You can rewrite 2+4c = 1 to 4c = -1.
That's of the form ... a + ... b + ... c = ...
 
That helps, but I'm still not sure whether my a's and b's for that equation should be 0s or 1's. I can try both, I suppose, but clearly my understanding of the problem is muddled or I wouldn't have to guess...
 
Is 1 \cdot a + 1 \cdot b + 4 \cdot c = -1 the same as 4c = -1?
Or is 0 \cdot a + 0 \cdot b + 4 \cdot c = -1 the same as 4c = -1?
 
Thanks for that clarification. I'm still getting some weird stuff (fractions I shouldn't be getting) but there must be another problem, since I'm pretty sure I'm not supposed to get those and that I should wind up with a free variable instead.

I'll check back later if I'm still working on it. Thanks again!
-Dave K
 
1 1 1 1
1 3 9 3
0 2 4 1
do you know why?

Since x is such a polynomial it is easier to let
f(x)=a + bx + cx^2
g(x)=f(x)-x
f(1)=1
f(3)=3
f'(2)=1
imply
g(1)=0
g(3)=0
g'(2)=0
the matrix is
1 1 1 0
1 3 9 0
0 2 4 0
 
dkotschessaa said:

Homework Statement

Find all polynomials of the form a + bx + cx^2 that:

Goes through the points (1,1) and (3,3)

and such that f'(2) = 1

Homework Equations



a + bx + cx^2
f'(x) = x+2cx
f'(2) = 2 + 4c

polynomial through (1,1) = a + b1 + c1 = 1
polynomial through (3,3) = a + b3+ c3^2 = 3

The Attempt at a Solution



I have the general idea that this should result in a series of equations that I need to do gauss Jordan on. Similar problems like this resulted in 3 similar equations and were quite simple.

My problem here is that, since I have taken the derivative of f(x) I have lost my constant a. So I'm not sure what my matrix should look like. I've tried:

1 1 1 1
1 3 9 3

But for the one with the derivative, my f"(2) isn't in the same form - it's 2+4c = 1

so I"m not sure whether to use

1 2 4 1
0 2 4 1

or something more general. If I use something more general though like:

a 2 4 1

I can't get a pivot in my first column...

'elp!

-Dave K

Your expression for f '(2) is incorrect.
 
Last edited:
Good catch Ray! It took three people to see that :)
 
Yes, indeed. That was a problem too. But your question "Is 1⋅a+1⋅b+4⋅c=−1 the same as 4c=−1?" was also instructive.

Thanks to the whole team. :)
 

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