Need Help With Derivative Problem

G537

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I want to investigate the family of curves given by f(x) = x^4 + x^3 + cx^2. I understand how to solve this when c > 0 and c = 0, but I don’t know how to solve this for c < 0 using the first derivative test. If I assume that c< 0 and thus f(x) = x^4 + x^3 - cx^2, then f’(x) = 4x^3 + 3x^2 - 2cx = x(4x^2 + 3x - 2c). Examining the discriminant of the quadratic in parentheses, when c = -9/32, there are 2 critical points at x = 0 and x = -3/8; when -9/32 < c < 0, there are 3 critical points at x = 0 and x = [-3 +- sqrt(9 + 32c)]/8; and when c < -9/32, there is one critical point at x = 0. But the book solution says that there are 3 critical points whenever c < 0, which contradicts my solutions. Is my reasoning incorrect?
 

Subhotosh Khan

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I want to investigate the family of curves given by f(x) = x^4 + x^3 + cx^2. I understand how to solve this when c > 0 and c = 0, but I don’t know how to solve this for c < 0 using the first derivative test. If I assume that c< 0 and thus f(x) = x^4 + x^3 - cx^2, then f’(x) = 4x^3 + 3x^2 - 2cx = x(4x^2 + 3x - 2c). Examining the discriminant of the quadratic in parentheses, when c = -9/32, there are 2 critical points at x = 0 and x = -3/8; when -9/32 < c < 0, there are 3 critical points at x = 0 and x = [-3 +- sqrt(9 + 32c)]/8; and when c < -9/32, there is one critical point at x = 0. But the book solution says that there are 3 critical points whenever c < 0, which contradicts my solutions. Is my reasoning incorrect?
You say:
If I assume that c< 0 and thus f(x) = x^4 + x^3 - cx^2​

That is "dangerous" writing. I would write it as:

If c<0, let us assume c1 = -c where c1>0. Then the equation becomes:

f(x) = x^4 + x^3 - c1x^2

f'(x) = 4x^3 + 3x^2 - 2c1x = x (4x^2 + 3x - 2c1)

roots of f'(x) are:

x = 0

x1,2 = \(\displaystyle \frac{-3 \pm \sqrt{9 - 4*(8)*(-c_1)}}{8}\) ...........................edited

You have three roots \(\displaystyle \ \ \to \ \ \) x = 0, x1 and x2
 
Last edited:

Dr.Peterson

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I want to investigate the family of curves given by f(x) = x^4 + x^3 + cx^2. I understand how to solve this when c > 0 and c = 0, but I don’t know how to solve this for c < 0 using the first derivative test. If I assume that c< 0 and thus f(x) = x^4 + x^3 - cx^2, then f’(x) = 4x^3 + 3x^2 - 2cx = x(4x^2 + 3x - 2c). Examining the discriminant of the quadratic in parentheses, when c = -9/32, there are 2 critical points at x = 0 and x = -3/8; when -9/32 < c < 0, there are 3 critical points at x = 0 and x = [-3 +- sqrt(9 + 32c)]/8; and when c < -9/32, there is one critical point at x = 0. But the book solution says that there are 3 critical points whenever c < 0, which contradicts my solutions. Is my reasoning incorrect?
I see no reason to start with cases based on the sign of c. I would just look at the discriminant, keeping c exactly as it is in the problem, and find that when c < 9/32, there will be three (real) solutions -- except when two of those three coincide, so you have to check when that occurs (which is when c = 0).

Please show us the entire problem and the entire answer, exactly as given in your book, so we can be sure what they are saying. I suspect that you are trying to reverse-engineer the solution from the book's answer, and assuming that you need to consider cases separately from the start.
 

Jomo

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So c can't be negative. That is if c is negative then it must be -c???

Consider c+5 = 0, so c =-5 which means -c =-5?? -c=-5 means that c=5. So 5+5 =0??
 

G537

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Thank you all for your helpful responses. I think that your responses get to the crux of my confusion: first, is it necessary to consider cases in this question and, if not, why? It seems to me that negating the constant c would produce different curves with different characteristics, which is why I assumed I needed to consider the 3 cases. Second, if I do want to examine cases where c < 0, how do I do this? I quickly realized that when using values of c < 0 along with the function f(x) = x^4 + x^3 - cx^2, this means that I’m really investigating f(x) = x*^4 + x^3 + cx^2, so writing the function with a negative sign in front of the last term doesn’t make sense. But I was not sure how to proceed in this case.

Regarding the given solution, it simply states that “for c < 0, there is a maximum at x = 0 and minima at x = [-6 +- sqrt(36 - 96c)]/24.” So it appears they took the approach Dr. Peterson recommended above. Dr. Peterson (or others), I’m curious how you determined that it was not necessary to use cases in this problem. I’d like to improve my intuition rather than simply memorizing my way through the material.
 

Dr.Peterson

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Thank you all for your helpful responses. I think that your responses get to the crux of my confusion: first, is it necessary to consider cases in this question and, if not, why? It seems to me that negating the constant c would produce different curves with different characteristics, which is why I assumed I needed to consider the 3 cases. Second, if I do want to examine cases where c < 0, how do I do this? I quickly realized that when using values of c < 0 along with the function f(x) = x^4 + x^3 - cx^2, this means that I’m really investigating f(x) = x*^4 + x^3 + cx^2, so writing the function with a negative sign in front of the last term doesn’t make sense. But I was not sure how to proceed in this case.
You seem to have been thinking somewhat along the lines of ancient people, before negative numbers were invented. When they talked about quadratic equations, for example, they assumed the coefficients were positive (because there was no such thing as a negative number -- how could you have -3 cows?). As a result, they had to consider separately the equations ax^2 + bx + c = 0 (which could have no solutions, so they didn't even consider it), and ax^2 + bx = c, and ax^2 = bx + c, and so on.

There is absolutely no inherent difference between x^4 + x^3 + cx^2 = 0 and x^4 + x^3 - cx^2 = 0. The latter is just an example of the former, in which c has been replaced by -c; the former includes cases like x^4 + x^3 + 2x^2 = 0, where c = 2, and like x^4 + x^3 - 2x^2 = 0, where c = -2.

But it sounds like you now see that replacing c with -c was not necessary even if you want to distinguish c as negative; you just have to say that c<0.

Regarding the given solution, it simply states that “for c < 0, there is a maximum at x = 0 and minima at x = [-6 +- sqrt(36 - 96c)]/24.” So it appears they took the approach Dr. Peterson recommended above. Dr. Peterson (or others), I’m curious how you determined that it was not necessary to use cases in this problem. I’d like to improve my intuition rather than simply memorizing my way through the material.
When I said, "I see no reason to start with cases based on the sign of c", I meant exactly what I said: not that I had determined that it was not necessary to use cases, but that there was nothing in the problem that called for cases (on the surface, at least -- that might come later, but will be handled when it is found appropriate). So there was no reason to even consider doing so. We use cases when we see cases inherent in the problem, such as if there is an absolute value. To solve |x-1| = 3|x+7|, I would consider using cases x<-7, -7<=x<1, and x>=1 because the equation changes its behavior at x=1 and at x=-7. There is nothing in x^4 + x^3 + cx^2 = 0 that obviously changes according to the sign of c.

Can you explain what leads you to say, "It seems to me that negating the constant c would produce different curves with different characteristics, which is why I assumed I needed to consider the 3 cases"? Every value of c leads to a different curve, but there is no qualitative difference in those curves at c=0, or at least none that I notice until I get far enough through the work to see something that happens to be special there. My question is, are you seeing something deep, or just assuming that negative numbers change things? Or did you simply work backward from the answer, seeing that it involves something different at c=0, so that must be where the work of solving it begins?
 

lookagain

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f'(x) = 4x^3 + 3x^2 - 2c1x = x (4x^2 + 3x - 2c1)

roots of f'(x) are:

x = 0

x1,2 = \(\displaystyle \frac{c_1 \pm \sqrt{9 - 4*(8)*(-c_1)}}{8}\)
Instead of \(\displaystyle \ "c_1" \ \) for the first term of your quadratic formula, I have -3, because the quadratic expression is \(\displaystyle \ 4x^2 + 3x - 2c\).
 

Subhotosh Khan

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Instead of \(\displaystyle \ "c_1" \ \) for the first term of your quadratic formula, I have -3, because the quadratic expression is \(\displaystyle \ 4x^2 + 3x - 2c\).
You are correct ... and I edited my post. Thanks
 

JeffM

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I am not sure whether this answers your question or not. Yes, OF COURSE, different values of c will alter the shape of the curve, but there is no reason a priori to think that the sign of the coefficient of the squared term in a quartic is particularly relevant. It may be relevant, but that must be determined.

A polynomial is defined for all real x. The global graph of any polynomial of even degree > zero, when looked at on a big enough scale, is solely determined by whether the coefficient of x to the defining degree is positive or negative. On a grand scale, the graph of the curve is shaped like a u if that coefficient is positive or an inverted u if that coefficient is negative.

However, if you look at a fourth degree polynomial with a positive leading coefficient at smaller scales, you will find either (a) one global minimum, or (b) two local minima with one local maximum between the minima. The importance of c in this case is in determining which of these two cases obtains and where the local maximum is located if it exists.

\(\displaystyle f(x) = x^4 + x^3 + cx^2 \implies \\

f'(x) = 4x^3 + 3x^2 + 2cx = x(4x^2 + 3x + 2c) \implies \\

f''(x) = 12x^2 + 6x + 2c.\)
Now obviously f'(0) = 0 = f(0). Whether there exist one or more values of x other than zero such that f'(x) = 0 depends on the discriminant of 4x2 + 3x + 2c.

This does lead us initially to three cases, but not based on the sign of c.

Case 1

\(\displaystyle \text {no real solutions} \iff (3)^2 - 4(4)(2c) < 0 \iff c > \dfrac{9}{32} \implies f(x)\\

\text {has no local maximum and a single minimum at } x = 0.\)
Case 2

\(\displaystyle c = \dfrac{9}{32} \implies f'\left ( - \dfrac{3}{8} \right ) = 0 = f'(0).\)

But there cannot be exactly two distinct extrema for a polynomial of degree 4. Why not?

Let's consider f''(0) when c = 9/32.

\(\displaystyle f''(0) = 12 * 0^2 + 6 * 0 + 2 * \dfrac{9}{32} = \dfrac{9}{16} > 0 \implies f(x)\\

\text {has no local maximum and a single minimum at } x = 0.\)
Case 3

We know f'(0) = 0, but when c < 9/32, there is at least one other distinct value of x for which f'(x) = 0. Again, what that value is depends on the discriminant.

Case 3a

\(\displaystyle -3 + \sqrt{9 - 32c} > 0 \implies 9 - 32c > 9 \implies c < 0 \implies \\

f''(0) < 0 \implies f(x) \text { has two local minima,}\\
\text {one on each side of } x = 0 \text { and a local maximum at } x = 0.\)
Do you see why?

In any case, we are interested in the sign of c because of the specifics of this function rather than because of any general rule that the behavior of a quartic is sensitive to the sign of the squared term.

Case 3b

\(\displaystyle c = 0 \implies \dfrac{-3 - \sqrt{9 - 32c}}{2 * 4} = - \dfrac{3}{4} \implies \\

f'' \left (- \dfrac{3}{4} \right ) = 12 * \left (- \dfrac{3}{4} \right )^2 + 6 * \left (- \dfrac{3}{4} \right ) + 2 * 0 =\\

\dfrac{12 * 9}{16} - \dfrac{6 * 3}{4} = \dfrac{108}{16} - \dfrac{72}{16} > 0 \implies f(x)\\

\text {has no local maximum and a single minimum at } x = - \dfrac{3}{4}.\)
So we end up with five cases rather than three.

\(\displaystyle c < 0,\ c = 0,\ 0 < c < \dfrac{9}{32}, \ c = \dfrac{9}{32}, \text { and } c > \dfrac{9}{32}.\)

Case 3c

\(\displaystyle 0 < c < \dfrac{9}{32}.\)

Can you do it?
 

G537

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Thank you once again to everyone for taking your time to help me! This has been extremely helpful. Jeff, your summary was immensely helpful, as it provided very clear reasoning. I also learned a bit about proper typesetting after reading your post, so thank you for that as well.

When I initially worked the problem, I did obtain cases for \(\displaystyle c>\frac{9}{32}, c=\frac{9}{32}\), and \(\displaystyle c=0\). I also built a table of my results and found that I could obtain 3 extrema for \(\displaystyle 0<c<\frac{9}{32}\). So, I was four-fifths of the way there, but I could not figure out how to reason through the case where \(\displaystyle c<0\). After working through your questions, it became much clearer to me how I might approach the problem. Let me address your questions to see if I'm on the right track.

Case 2

\(\displaystyle c = \dfrac{9}{32} \implies f'\left ( - \dfrac{3}{8} \right ) = 0 = f'(0).\)

But there cannot be exactly two distinct extrema for a polynomial of degree 4. Why not?
For a polynomial of degree 4, \(\displaystyle f(x)\rightarrow\infty\) as \(\displaystyle x\rightarrow\pm\infty\) and so there must be an odd number of extrema. If there were an even number, then either \(\displaystyle f(x)\rightarrow\infty\) as \(\displaystyle x\rightarrow\infty\) and \(\displaystyle f(x)\rightarrow-\infty\) as \(\displaystyle x\rightarrow-\infty\) OR \(\displaystyle f(x)\rightarrow-\infty\) as \(\displaystyle x\rightarrow\infty\) and \(\displaystyle f(x)\rightarrow\infty\) as \(\displaystyle x\rightarrow-\infty\), which would mean that the degree of the polynomial is odd.

Case 3a

\(\displaystyle -3 + \sqrt{9 - 32c} > 0 \implies 9 - 32c > 9 \implies c < 0 \implies \\

f''(0) < 0 \implies f(x) \text { has two local minima,}\\
\text {one on each side of } x = 0 \text { and a local maximum at } x = 0.\)
Do you see why?
Case 3c

\(\displaystyle 0 < c < \dfrac{9}{32}.\)

Can you do it?
I approached your final 2 questions in the same way. For the case where \(\displaystyle c<\frac{9}{32}\), I see that \(\displaystyle c=0\) is a "special case" of this inequality in that it is the only value of \(\displaystyle c<\frac{9}{32}\) for which f(x) has only one extremum, and therefore it breaks the inequality \(\displaystyle c<\frac{9}{32}\) into 2 parts: \(\displaystyle 0<c<\frac{9}{32}\) or \(\displaystyle c<0\). Starting with the first inequality, since \(\displaystyle 0<c<\frac{9}{32}\):

\(\displaystyle b^2>b^2-4ac>b^2-\frac{9a}{8}\). Since a = 4 and b = 3, \(\displaystyle 3>\sqrt{9-16c}>\frac{3}{\sqrt{2}}\), then \(\displaystyle 0<\frac{3-\sqrt{9-16c}}{8}<\frac{6-3\sqrt{2}}{16}\) and thus \(\displaystyle 0<x_1<\frac{6-3\sqrt{2}}{16}\) or \(\displaystyle \frac{3}{4}>\frac{3+\sqrt{9-16c}}{8}>\frac{6+3\sqrt{2}}{16}\) and thus \(\displaystyle \frac{3}{4}>x_2>\frac{6+3\sqrt{2}}{16}\).

So, there are 3 extrema when \(\displaystyle 0<c<\frac{9}{32}\): \(\displaystyle x_1, x_2\), and x = 0 - two local minima and one local maximum.

When \(\displaystyle c<0\), \(\displaystyle b^2-4ac>b^2\). Since a = 4 and b = 3, then \(\displaystyle \sqrt{9-16c}>3\). Thus \(\displaystyle \frac{3-\sqrt{9-16c}}{8}<0\), or \(\displaystyle \frac{3+\sqrt{9-16c}}{8}<\frac{3}{4}\). Therefore \(\displaystyle x_3<0\) and \(\displaystyle x_4<\frac{3}{4}\).

Again, there are 3 extrema when \(\displaystyle c<0\): \(\displaystyle x_3, x_4\), and x = 0 - two local minima and one local maximum.

So, with the exception of c = 0, there are always 3 extrema when \(\displaystyle c<\frac{9}{32}\), and this covers the cases when \(\displaystyle c<0\).

If my reasoning is incorrect, please let me know. Thanks again!
 

JeffM

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For a polynomial of degree 4, \(\displaystyle f(x)\rightarrow\infty\) as \(\displaystyle x\rightarrow\pm\infty\) and so there must be an odd number of extrema. If there were an even number, then either \(\displaystyle f(x)\rightarrow\infty\) as \(\displaystyle x\rightarrow\infty\) and \(\displaystyle f(x)\rightarrow-\infty\) as \(\displaystyle x\rightarrow-\infty\) OR \(\displaystyle f(x)\rightarrow-\infty\) as \(\displaystyle x\rightarrow\infty\) and \(\displaystyle f(x)\rightarrow\infty\) as \(\displaystyle x\rightarrow-\infty\), which would mean that the degree of the polynomial is odd.
Your logic is basically fine. The only thing wrong with it is that you assumed that the leading coefficient is positive. The global features of a polynomial are determined by the sign of its leading coefficient and by whether its degree is odd or even.

A polynomial of even degree has a global minimum if the leading coefficient is positive. It may have additional local minima, but if so, there will be a local maximum between each successive pair of minima. A polynomial of even degree has a global maximum if the leading coefficient is negative. It may have additional local maxima, but if so, there will be a local minimum between each pair of maxima. This is geometric reason for why the number of extrema for a polynomial of even degree is always odd.

Polynomial of degree 2n > 0, odd number of extrema. Minimum number of extrema = 1. Maximum number of extrema n - 1. Leading coefficient positive, odd number of minima and any maximum is preceded and succeeded by a minimum. Leading coefficient negative, odd number of maxima and any minimum is preceded and succeeded by a minimum.

You might want to work out the general rules for polynomials of odd degree. You have the basic ideas.

I approached your final 2 questions in the same way. For the case where \(\displaystyle c<\frac{9}{32}\), I see that \(\displaystyle c=0\) is a "special case" of this inequality in that it is the only value of \(\displaystyle c<\frac{9}{32}\) for which f(x) has only one extremum, and therefore it breaks the inequality \(\displaystyle c<\frac{9}{32}\) into 2 parts: \(\displaystyle 0<c<\frac{9}{32}\) or \(\displaystyle c<0\). Starting with the first inequality, since \(\displaystyle 0<c<\frac{9}{32}\):

\(\displaystyle b^2>b^2-4ac>b^2-\frac{9a}{8}\). Since a = 4 and b = 3, \(\displaystyle 3>\sqrt{9-16c}>\frac{3}{\sqrt{2}}\), then \(\displaystyle 0<\frac{3-\sqrt{9-16c}}{8}<\frac{6-3\sqrt{2}}{16}\) and thus \(\displaystyle 0<x_1<\frac{6-3\sqrt{2}}{16}\) or \(\displaystyle \frac{3}{4}>\frac{3+\sqrt{9-16c}}{8}>\frac{6+3\sqrt{2}}{16}\) and thus \(\displaystyle \frac{3}{4}>x_2>\frac{6+3\sqrt{2}}{16}\).

So, there are 3 extrema when \(\displaystyle 0<c<\frac{9}{32}\): \(\displaystyle x_1, x_2\), and x = 0 - two local minima and one local maximum.

When \(\displaystyle c<0\), \(\displaystyle b^2-4ac>b^2\). Since a = 4 and b = 3, then \(\displaystyle \sqrt{9-16c}>3\). Thus \(\displaystyle \frac{3-\sqrt{9-16c}}{8}<0\), or \(\displaystyle \frac{3+\sqrt{9-16c}}{8}<\frac{3}{4}\). Therefore \(\displaystyle x_3<0\) and \(\displaystyle x_4<\frac{3}{4}\).

Again, there are 3 extrema when \(\displaystyle c<0\): \(\displaystyle x_3, x_4\), and x = 0 - two local minima and one local maximum.

So, with the exception of c = 0, there are always 3 extrema when \(\displaystyle c<\frac{9}{32}\), and this covers the cases when \(\displaystyle c<0\).
There is nothing AT ALL wrong with your reasoning in terms of logic. It just is much more elaborate than it needs to be.

Except in the special case where c = 0, c < 9/32 entails that we have extrema at three different values of x. We know the middle one will be a maximum and the two end ones will be minima. We also know that there will be extrema at

\(\displaystyle x = 0, \ x = \dfrac{-3 - \sqrt{9 - 32c}}{8}, \text { and } x = \dfrac{-3 + \sqrt{9 - 32c}}{8}.\)

The second location is clearly the farthest to the left and is just as clearly negative. The third location could be negative, zero, or positive. The zero possibility is a special case already addressed. If c is not 0, it is either less than 0 or greater than zero.

\(\displaystyle c < 0 \implies 9 - 32c > 9 \implies \dfrac{- 3 + \sqrt{9 - 32c}}{8} > \dfrac{dfrac{-3 + 3}{8} = 0.\)

Thus, if c negative, zero is in the middle and locates a maximum.

\(\displaystyle 0 < c < \dfrac{9}{32} \implies - 9 < - 32c < 0 \implies 0 < 9 - 32c < 0 \implies 0 < \ sqrt{9 - 32c} < 3 \implies \\

\dfrac{-3 + \sqrt{9 - 32c}}{8} < \dfrac{-3 + 3}{8} = 0.\)
Thus, 0 is the rightmost location of the extrema and is a minimum.

Good work. If you have more questions, please ask.
 

G537

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Polynomial of degree 2n > 0, odd number of extrema. Minimum number of extrema = 1. Maximum number of extrema n - 1. Leading coefficient positive, odd number of minima and any maximum is preceded and succeeded by a minimum. Leading coefficient negative, odd number of maxima and any minimum is preceded and succeeded by a minimum.
I assume you meant that, for a polynomial of degree 2n > 0 and a positive leading coefficient, there will be an even number of minima, with a maximum preceded and succeeded by a minimum. Likewise for a polynomial with a negative leading coefficient, where a minimum (if it exists) will be preceded and succeeded by a maximum, and therefore there are an even number of maxima.

You might want to work out the general rules for polynomials of odd degree. You have the basic ideas.
For a polynomial of odd degree, there must be either no extrema, or an equal number of maxima and minima, so the number of local extrema is an even number. Consequently, there are a minimum of zero extrema and a maximum of \(\displaystyle n-1\) extrema.

Thank you again for working through this problem with me. It was a very useful exercise for me to work through your questions and then repeat the the entire process on my own with the second derivative, again taking the discriminant by cases. I also worked through the first and second derivative tests using the method you illustrated at the end of your previous post, and I can see that this is much easier.

I'm sure I'll have more math questions in the future. I'm glad I discovered this online community.
 

JeffM

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I assume you meant that, for a polynomial of degree 2n > 0 and a positive leading coefficient, there will be an even number of minima, with a maximum preceded and succeeded by a minimum. Likewise for a polynomial with a negative leading coefficient, where a minimum (if it exists) will be preceded and succeeded by a maximum, and therefore there are an even number of maxima.
Consider a quadratic with a positive coefficient, for example f(x) = x2.

There is one minimum at x = 0. That is an odd number of minima. So I meant what I said.

Here is general rule about functions that are everywhere differentiable. If a function has multiple extrema, a minimum is never next to a minimum, nor is a maximum next to a maximum

You don't need to memorize what's below because you can derive them, but I have found them useful to keep in mind.

The derivative of a polynomial of degree > 0 is also a polynomial.

The maximum number of extrema of a polynomial of degree n is n - 1.

A polynomial of odd degree may have no extrema.

A polynomial of even degree has at least one extremum.

A polynomial of odd degree has as many minima as maxima.

A polynomial of even degree and a positive leading coefficient has one more minimum than it has maxima.

A polynomial of odd degree and a negative leading coefficient has one more maximum than it has minima.
 
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