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Math Help - maximum value of a polynomial

  1. #1
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    maximum value of a polynomial

    Let  f(x)=ax^2+bx+c, \ \ a,b,c \in \mathbb{R} \ \ \ f(-1),f(0),f(1) \in [-1,1].
    Prove that  |f(x)|\leq\frac{3}{2} \ \forall \ x \in [-1,1].

    Any ideas?
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  2. #2
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    well im not to sure how you would do this without differentiation,

    however the max values of the function are found at the global extrama, which are a member of [-1,1] so if we find the turning point by differentiation, it should be at x= -b/2a and then if we assume this is in [-1,1] and then show that the corresponding value of f is less than or equal to 3/2, i think this works
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  3. #3
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    @hmmmm

    There's no need for calculus, just write f as a(x+\frac b{2a})^2 - \frac {b^2}{4a^2}+c. But note that this is insufficient, since you are considering the absolute value of f and also it might get bigger at the endpoints (if a is positive, say). I don't know how to do this though.
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  4. #4
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    ok but why does this show that |f(x)| is always less than 3/2, however if you used the turning point is -b/2a and sub this in for x we can show that the inequality holds, and we have our endpoints from the question
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  5. #5
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    Oh, I see. Sorry, I missed the condition on the endpoints. Its not a pretty substitution though. You still have to use the conditions to show that the value at the vertex will be less than 3/4.
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  6. #6
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    yeah but i think that it works, its the only way i could think to do it
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  7. #7
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    Ah, don't thank me, I totally screwed this up ;[
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