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I am given the following task:

The graph of the polynomial function $f(x)$ is symmetric to the y-axis. It has exactly one local minimum on the x-axis and an inflection point at $x = 1$. Find the the only possible function $f$.

My thoughts: Because $f$ is symmetric and it only has one minimum on the x-axis, this has to be at $x = 0$. Therefore: $f(0) = 0$, $f'(0) = 0$. Because of the inflection point, $f''(1) = 0$.

Now suppose $\deg(f) = 4$ (could also be $6, 8, \ldots$). Then $f(x) = ax^4+bx^2 +c$. From the above, I get $c = 0$ and $b = -6a$.

So I have multiple functions $f_a(x) = ax^4-6ax^2$, although the task stated that there is only one solution.

I can't find my mistake (or is the homework wrong?).

Thanks in advance.

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Your condition "It has eactly one local minimum on the x-axis" is confusing. Does it mean "It has exactly one local minimum (which happens to be on the x-axis)" or "Of all the local minima, exactly one is on the x-axis"? – Rick Decker Feb 18 '13 at 13:39
I assume it means the latter, although I can't tell exactly because it was an addition to the text in my book (my teacher just told us to "modify" the question. This is in german, but I will try to translate: It originally said "it has one local minimum on the x-axis" ("Er hat einen Tiefpunkt auf der x-Achse"), and she said "add the word exactly here". – pascalhein Feb 18 '13 at 13:47
It should be clear that if $f(x)$ is a solution then $af(x)$ is also a solution for $a>0$.... – N. S. Feb 18 '13 at 18:01
up vote 3 down vote accepted

Your "family" of solutions: $\;f_a(x) = ax^4-6ax^2$

meets all the criteria, as stated. Read below, I would simply add $a < 0$.

Perhaps this is the "only" (family of) function(s) your teacher intended for you to find, else, there is missing information, or sufficient ambiguity in the question to justify your result.

(Note, there are two other possible local minimum: $x = \sqrt 3$, $x = -\sqrt 3$, may or may not be local minimums/maximums depending on the value of $a$: for $a > 0$, they will be local minimum. For $a < 0,$, they will be local maximums. And we must have that they are local maximums, since we must have $a < 0$ if there is to be a local minimum at $x = 0$.)

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(+1): We have to have $a<0$ in order to get a local minimum at $x=0$, so the other two local extrema are nothing to worry about. – Cameron Buie Feb 18 '13 at 17:27
@Cameron thanks ;-) And indeed, you are correct, we must restrict $a$ to $a < 0$ – amWhy Feb 18 '13 at 17:33
+1 i am jealousy of you cause you did it sooner than me. – Babak S. Feb 18 '13 at 18:13

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