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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.
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how many turning points does \(y=x^3\), a thirddegree polynomial have? it has \(0\), so which is neither at least \(n1=2\) nor at least \(n2=1\). so the first two are obviously wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think it is the last one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now consider the fourthdegree polynomial \(y=x(x1)(x2)(x3)\). it has three turning points, one in \((0,1)\), one in \((1,2)\), and one in \((2,3)\). this is more than \(42=2\) turns, but not more than \(41=3\) turns. so we've found counterexamples for all the other options, and the only answer that makes sense is the third  at most \(n1\) turning points

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it is the 3rd one?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because it can't have more ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well, it can't be any of the other options, since we gave counterexamples

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0anyways, if we make \(a,b,c>0\) big enough, we can make \(abx+cx^2x^3\) 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 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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