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 May 19th, 2011, 01:08 AM #1 Newbie   Joined: May 2011 Posts: 28 Thanks: 0 Polynomial Hello i got this problem and i have no idea what they to do: Among polynomials of third degree with coefficients in $\mathbb{R}$ find such $f(x)$, for which is value $max_{x \in \left \langle -1,1 \right \rangle}\:\: f(x)$ minimal. Thanks a lot
 May 19th, 2011, 06:55 AM #2 Senior Member   Joined: Jun 2010 Posts: 618 Thanks: 0 Re: Polynomial token22, Are there further conditions which you are forgetting to state? There are a couple of points about your problem that ought to be reviewed. First, a continuous function over the reals on an open interval does not necessarily attain a maximum value, so I assume you mean to talk about the supremum of the function. Second, by adding negative constants to certain cubic polynomials, we can find ones with arbitrarily small supremum over (-1,1). For example, the class $f_c(x)\=\ x^3-3x+c$ for c ? 0 is such that the supremum over (-1,1) is 2+c, which of course is unbounded from below as c approaches negative infinity. I wonder, therefore, if you mean to take the supremum of the absolute value, |f|, instead? -Ormkärr-
 May 19th, 2011, 11:01 AM #3 Newbie   Joined: May 2011 Posts: 28 Thanks: 0 Re: Polynomial No, there are not other condictions to this problem. Hm and I don't think absolute value helps something more. Looks like this is just not-well put problem, so the only solution for me will be not solve this. I do not know what should be the right solution. And your solution comes to me too simple for "hard one problem". Propably my teacher didn't think of everything at moment when he was making it. Another problem: Determine if there exist parameters $a,b \in \mathbb{R}$ so that polynomial $a{x}^{2}+{a}^{2}x+b$ satisfy the conditions $P(1)=1,P(2)=-1,P(3)=1$ I solved it using matrix that i filled with a, a^2, b (using given x and y) it gave me $a \neq a^2$ so my answer is, no they do not exist. Is it OK, or is there defferent solution?
 May 19th, 2011, 11:39 AM #4 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Polynomial That's perfectly fine. Another way: P(3) = P(1), so 9a+3a²+b = a+a²+b, so 8a+2a² = 0, and then 8+2a = 0 (since a is not zero). Thus a = -4. And then show that no choice for b will work. A third way is to notice that because P is quadratic, its curve is symmetrical, and therefore if P(a)=P(b) then the vertex (centre) must be the average of a and b. P(1) = P(3), so the vertex is at x=2. But the formula for the vertex gives -a²/2a = -a/2, so if that's 2, then a = -4. But in that case, the curve should be a "frown" rather than a "smile", because the coefficient of x² is negative.

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