Creating a new curve sketching problem

Hello folks

I am a math teacher and my usual style of making curve sketching questions for my students is to beg, borrow, or steal from other sources. I decided recently to create my own problems in curve sketching of polynomials, only to run into the same problem over and over no matter what coefficients I chose (and the problem only existed if I chose coefficients that factored properly).

I am being a bit fussy in my choice of polynomials for f(x). The polynomials for f(x) and its first and second derivatives must have rational zeroes, no irrational or complex roots. f(x) itself is a quartic.

What follows is a methodology I thought would work, and since this is a small school I don't have easy access to colleagues who know more math than me, I have no idea as to why my idea appears to lead to lackluster results:

I know that if I have, say $\displaystyle f(x) = x^4$, then its subsequent derivatives exhibit the behaviour of factorials done partially, as in: $\displaystyle f'(x) = 4x^3$; and $\displaystyle f''(x) = 3 \times 4x^2 = 12x^2$; and finally: $\displaystyle f'''(x) = 2 \times 3 \times 4x = (4!)x = 24x$. This gave me the idea to work backwards from $\displaystyle f''(x)$ to $\displaystyle f(x)$ to get a factorable second derivative. I would use synthetic division on an unknown constant term to get the completed polynomials for $\displaystyle f'(x)$ and $\displaystyle f(x)$ as I worked backward, that I knew would factor using techniques familiar to my Grade 12 students.

The result is a polynomial f(x) with x-intercepts, max, min, and inflection points that can be found by either simple factoring or synthetic division, with therefore enough info to plot a reasonable graph. This sounds like an airtight idea, except that all I have gotten so far is an f(x) with a triple root and a single root, as in: $\displaystyle f(x) = x^4 - x^3 - 3x^2 - 5x - 2 = (x-2)(x+1)^3$ Another one I found resulted in a triple root as well: $\displaystyle f(x) = x^4 +x^3 - 9x^2 + 11x - 4 = (x-1)^3(x+4)$.

Is this the best I can hope for with my conjecturing? Is there a way of getting a greater variety of roots with modifications to this method or by using an altogether different method? Remember: it must factor completely, no complex or irrational roots.

Phred

Re: Creating a new curve sketching problem

Quote:

Originally Posted by

**Phred** Hello folks

I am a math teacher and my usual style of making curve sketching questions for my students is to beg, borrow, or steal from other sources. I decided recently to create my own problems in curve sketching of polynomials, only to run into the same problem over and over no matter what coefficients I chose (and the problem only existed if I chose coefficients that factored properly).

I am being a bit fussy in my choice of polynomials for f(x). The polynomials for f(x) and its first and second derivatives must have rational zeroes, no irrational or complex roots. f(x) itself is a quartic.

What follows is a methodology I thought would work, and since this is a small school I don't have easy access to colleagues who know more math than me, I have no idea as to why my idea appears to lead to lackluster results:

I know that if I have, say $\displaystyle f(x) = x^4$, then its subsequent derivatives exhibit the behaviour of factorials done partially, as in: $\displaystyle f'(x) = 4x^3$; and $\displaystyle f''(x) = 3 \times 4x^2 = 12x^2$; and finally: $\displaystyle f'''(x) = 2 \times 3 \times 4x = (4!)x = 24x$. This gave me the idea to work backwards from $\displaystyle f''(x)$ to $\displaystyle f(x)$ to get a factorable second derivative. I would use synthetic division on an unknown constant term to get the completed polynomials for $\displaystyle f'(x)$ and $\displaystyle f(x)$ as I worked backward, that I knew would factor using techniques familiar to my Grade 12 students.

The result is a polynomial f(x) with x-intercepts, max, min, and inflection points that can be found by either simple factoring or synthetic division, with therefore enough info to plot a reasonable graph. This sounds like an airtight idea, except that all I have gotten so far is an f(x) with a triple root and a single root, as in: $\displaystyle f(x) = x^4 - x^3 - 3x^2 - 5x - 2 = (x-2)(x+1)^3$ Another one I found resulted in a triple root as well: $\displaystyle f(x) = x^4 +x^3 - 9x^2 + 11x - 4 = (x-1)^3(x+4)$.

Is this the best I can hope for with my conjecturing? Is there a way of getting a greater variety of roots with modifications to this method or by using an altogether different method? Remember: it must factor completely, no complex or irrational roots.

Phred

The derivative of a function of the form $\displaystyle f(x) = A(x - a)^2(x - b)^2$ will have rational roots. The double derivative won't in general, but the quadratic formula can be used to get the roots.

If you have to have all derivatives with rational roots, I suggest you read this: http://www.newcastle.edu.au/school-o...macdougall.pdf

Of related interest: http://journals.cambridge.org/downlo...90e75cf703ee45

Re: Creating a new curve sketching problem

Thanks a lot for your great help. I found the Newcastle article interesting, but I can't seem to be able to download the file from journals.cambridge.org (your second link). The server returned "File Not Found".

Re: Creating a new curve sketching problem

Quote:

Originally Posted by

**Phred** Thanks a lot for your great help. I found the Newcastle article interesting, but I can't seem to be able to download the file from journals.cambridge.org (your second link). The server returned "File Not Found".

Try finding it using a Google search: quartic function rational roots derivative - Google Search

Re: Creating a new curve sketching problem

The first paper you referred me to, by Galvin et al., seems to start on the wrong foot. The equation at the start of the paper is: $\displaystyle f(x) = x (x - pr) (x - qs)$, which, when expanded is: $\displaystyle f(x) = x^3 - qsx^2 - prx^2 + pqrsx$. I couldn't factor the derivative of this, which turns out to be: $\displaystyle f'(x) = 3x^2 - 2qsx - 2prx + pqrs$. Looking at Equation(1) is what is claimed to be the derived polynomial: $\displaystyle f'(x) = (3x - ps)(x - qr)$. But when you expand this, you don't get the derivative. The middle two terms are different: $\displaystyle f'(x) = 3x^2 - 3qrx - psx + pqrs$.

Do you think equation (1) is correct? If so, I can apply $\displaystyle f(x) = \int f'(x)$. It would seem that no constant need be added.

Re: Creating a new curve sketching problem

Quote:

Originally Posted by

**Phred** The first paper you referred me to, by Galvin et al., seems to start on the wrong foot. The equation at the start of the paper is: $\displaystyle f(x) = x (x - pr) (x - qs)$, which, when expanded is: $\displaystyle f(x) = x^3 - qsx^2 - prx^2 + pqrsx$. I couldn't factor the derivative of this, which turns out to be: $\displaystyle f'(x) = 3x^2 - 2qsx - 2prx + pqrs$. Looking at Equation(1) is what is claimed to be the derived polynomial: $\displaystyle f'(x) = (3x - ps)(x - qr)$. But when you expand this, you don't get the derivative. The middle two terms are different: $\displaystyle f'(x) = 3x^2 - 3qrx - psx + pqrs$.

Do you think equation (1) is correct? If so, I can apply $\displaystyle f(x) = \int f'(x)$. It would seem that no constant need be added.

You have not used the fact that p, q, r and s are in arithmetic progression.

Using this fact, it's simple to show that 2qs + 2pr = 3qr + ps

(which in turn shows why p, q, r and s were chosen to be in aritmetic progression ....)

Re: Creating a new curve sketching problem