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anonymous

  • one year ago

Which statement is true? A polynomial of degree n has at least n – 1 turning points. A polynomial of degree n has at least n – 2 turning points. A polynomial of degree n has at most n – 1 turning points. A polynomial of degree n has at most n – 2 turning points.

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  1. anonymous
    • one year ago
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    how many turning points does \(y=x^3\), a third-degree polynomial have? it has \(0\), so which is neither at least \(n-1=2\) nor at least \(n-2=1\). so the first two are obviously wrong

  2. anonymous
    • one year ago
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    i think it is the last one

  3. anonymous
    • one year ago
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    now consider the fourth-degree polynomial \(y=x(x-1)(x-2)(x-3)\). it has three turning points, one in \((0,1)\), one in \((1,2)\), and one in \((2,3)\). this is more than \(4-2=2\) turns, but not more than \(4-1=3\) turns. so we've found counterexamples for all the other options, and the only answer that makes sense is the third -- at most \(n-1\) turning points

  4. anonymous
    • one year ago
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    so it is the 3rd one?

  5. anonymous
    • one year ago
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    because it can't have more ?

  6. anonymous
    • one year ago
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    well, it can't be any of the other options, since we gave counterexamples

  7. anonymous
    • one year ago
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    anyways, if we make \(a,b,c>0\) big enough, we can make \(a-bx+cx^2-x^3\) have up to \(3\) zeros, and thus necessarily at most \(2\) turning points. if we could make it have another turning point, it would have \(4\) zeros -- which violates teh fundamental theorem of algebra, which says that an \(n\)-th degree polynomial has at most \(n\) zeros

  8. anonymous
    • one year ago
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    the intuition for why we can make that have up to \(3\) zeros follows from the same intuition at work in Descartes' rule of signs

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