Polynomial function, only 1 solution

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?).

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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

Your "family" of solutions: $\;f_a(x) = ax^4-6ax^2$
meets all the criteria, as stated. Read below, I would simply add $a < 0$.
(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 $x = 0$.)
(+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