Polynomial graph, stuck

[Otis]

… I think what [Jomo means] is that, given only what [an intermediate-algebra] student has learned … [such a student could envision] the [absolute maximum] on either side of the y-axis …

I don't believe that Jomo was speaking on behalf of other students, but I agree with the main point.

having lots of experience seeing polynomials, I can tell at a glance that [the graph] would have to bend the wrong way to [reach the point (3,0)] …

My intuition told me the absolute maximum has to be in Quad I, for the same reason. I can't quantify that sense; I think it developed from looking at so many polynomial-graph details, like inflection points and symmetry (or near-symmetry).

?

 

[topsquark]

The max can be in quad 1 or quad 2. In Otis' graph the function was increasing to the y intercept. There is no reason why it can't be decreasing to the y-intercept.

But a higher degree polynomial would look different from the sketch given. If the sketch is presumed to be incorrect in the number of zeros it has then, yes, we can have a higher degree polynomial.

-Dan

 

[JeffM]

But a higher degree polynomial would look different from the sketch given. If the sketch is presumed to be incorrect in the number of zeros it has then, yes, we can have a higher degree polynomial.

-Dan

We can have higher degree polynomials. For example

[MATH]f(x) = -\ \dfrac{1}{81}(x + 3)^4(x + 1)(x - 3)[/MATH]
has exactly three roots.

[MATH]f'(x) = \dfrac{-\ (x + 3)}{81} * (6x^2 - 8x -18).[/MATH]
There is a tangent at x = - 3. And a local minimum in quadrant II.

Sorry for all the earlier screw-ups.

 

[Otis]

Why is this not possible for a 4th degree polynomial?

Hi Jomo. I'm not making fun of your graph (I've done similar, in years past), but look at the distortion in scale required, to get that curve all the way over to positive 3. Not only were the units to the right of zero already "crunched together", but that wasn't enough. You then had to move 3 even farther to the left, to meet the graph.

My short answer to your question is, "Because that doesn't jive with the intent of the exercise". But, also, based on what I've seen about polynomial curves, I think it's too much of a "stretch". (That's my gut speaking, but do look at Jeff's work.)

With an inflection point between -1 and 0, I think we'd need another local inflection point -- or a plateau surrounding the absolute maximum -- before the graph could make it to (3,0). Either of those possibilities would require a higher-degree polynomial, and, if either of those were the case, then the given graph would contain yet another mistake.

?

 

[Dr.Peterson]

If I were trying to get the local maximum to be in quadrant 2, I'd be trying to put in a nonlinear, irreducible factor that wouldn't add more (real) zeros, but would add enough wiggle to move the peak over to the left of the axis. I experimented a little, but the best I could do was to have a peak to the left and another to the right.

 

[JeffM]

If I were trying to get the local maximum to be in quadrant 2, I'd be trying to put in a nonlinear, irreducible factor that wouldn't add more (real) zeros, but would add enough wiggle to move the peak over to the left of the axis. I experimented a little, but the best I could do was to have a peak to the left and another to the right.

Given the implied constraints, the local extrema must be on opposite sides of the y-axis.

 

[topsquark]

We can have higher degree polynomials. For example

[MATH]f(x) = -\ \dfrac{1}{81}(x + 3)^4(x + 1)(x - 3)[/MATH]

Oh, okay. I hadn't thought about giving it more factors of (x + 3). Good catch!

-Dan

 

[Dr.Peterson]

Given the implied constraints, the local extrema must be on opposite sides of the y-axis.

I must be missing something. What constraints are you assuming, and how do they imply that conclusion?

 

[JeffM]

I must be missing something. What constraints are you assuming, and how do they imply that conclusion?

OK

Here is the fourth go. I shall try to abbreviate it because in post # 20 I went into painful detail after having messed up some details in previous posts by skipping steps.

Our quantitative constraints are that f(x) has zeroes only at x = - 3, x = - 1, and x = 3 and is tangent to the x-axis at x = - 3. Our qualitative constraints are that f(x) is a polynomial with a positive slope if x < - 3, a negative slope if x > 3, a local minimum at some unspecified x such that - 3 < x < 0, and a second local maximum to the right of the local minimum. That is actually a lot of constraints. Satisfying them will be any polynomial of the following form:

[MATH]f(x) = -\ a(x + 3)^{(2n)}(x + 1)(x - 3), \text { where } a = \dfrac{1}{3^{(2n)}} \text { and } n \in \mathbb Z^+..[/MATH]
Any such polynomial will have the correct zeroes and correct y-intercept. To look at the other constraints we need the first derivative.

[MATH]f'(x) = -\ a(x + 3)^{(2n-1)} \ {2(n + 1)x^2 - 4(n - 1)x - 6(n + 1)}[/MATH].

This gives the desired point of tangency at x = - 3.

Moreove, the determinant of the quadratic in the derivative is positive because n is a positive integer.

If you work through the quadratic formula to find where the other local extrema are, you find that the local minimum is necessarily at a negative value of x and that the other local maximum is necessarily at a positive value of x. The details can be found in post 20, but the constraints do require that result. Of course, this problem was posted in algebra, and these issues are far beyond what the student can use. But it is a very nice calculus problem.

I think otis provided the answer expected. I got interested in the questions of uniqueness (no) and whether the x values of the local minimum and the other local maximum necessarily have opposite signs (yes).

 

[Dr.Peterson]

I don't have time right now to think carefully through this; I'll just point out (again) that a polynomial with only the specified x-intercepts (real zeros) might also have some non-real zeros (which can be introduced by multiplying by an irreducible quadratic factor or two). And that might possibly change things. Rather than adding zeros, which we don't want to do, that would introduce what I've been calling "wiggles", which is exactly what we need to accomplish the goal.

Now, I don't say your conclusion is wrong; I have a feeling that the kind of wiggles we can add may not be what we'd need. I just don't have a proof yet.

 

[JeffM]

I don't have time right now to think carefully through this; I'll just point out (again) that a polynomial with only the specified x-intercepts (real zeros) might also have some non-real zeros (which can be introduced by multiplying by an irreducible quadratic factor or two). And that might possibly change things. Rather than adding zeros, which we don't want to do, that would introduce what I've been calling "wiggles", which is exactly what we need to accomplish the goal.

Now, I don't say your conclusion is wrong; I have a feeling that the kind of wiggles we can add may not be what we'd need. I just don't have a proof yet.

I get your point. I shall think about it.

 

[JeffM]

I don't have time right now to think carefully through this; I'll just point out (again) that a polynomial with only the specified x-intercepts (real zeros) might also have some non-real zeros (which can be introduced by multiplying by an irreducible quadratic factor or two). And that might possibly change things. Rather than adding zeros, which we don't want to do, that would introduce what I've been calling "wiggles", which is exactly what we need to accomplish the goal.

Now, I don't say your conclusion is wrong; I have a feeling that the kind of wiggles we can add may not be what we'd need. I just don't have a proof yet.

Been thinking about this, without yet resorting to paper and pencil. It is what I call coffee thinking.

You are thinking about f(x) = g(x) * h(x), where g(x) is a polynomial of the form already analyzed and h(x) is a product of quadratics without real roots. This would not affect the location of the real zeroes. Obviously, the value of h(0) affects the value of the leading coefficient of f(x), but this is not a meaningful constraint. Now let's consider the first derivative.

f'(x) = g'(x) * h(x) + g(x) * h'(x).

Because g(- 3) = 0 = g'(- 3), f'(- 3) = 0. This too imposes no constraints on h(x).

However, we know that g'(x) is zero at only two other values of x, and we know that f'(x) must be zero and f''(x) not zero at an even number of values of x. This does impose meaningful constraints on h(x).

[MATH]f'(x) = 0 \implies g'(x) * h(x) = - \ g(x) * h'(x).[/MATH]
If the first derivatives of h(x) and g(x) are not zero, then

[MATH]\dfrac{g'(x)}{g(x)} = - \ \dfrac{h'(x)}{h(x)}.[/MATH]
Economists work with fractions like those all the time and call them elasticities. There may be a lot of information about their mathematical properties lying around.

We also know where g(x) and g'(x) are zero, and from that perhaps we figure out some constraints on h(x). Now whether the resulting constraints are permissible depends in part on a graph that we have not seen. It also depends on what jomo's question really is. If it is whether the slope of f(x) is necessarily positive at x = 0, that may give one answer. If it is whether the number of local extrema in (- 3, 0) is the same as in (0, 3), that may give another answer. Just drinking coffee, I do not see how to proceed. Maybe I shall have to work.