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.
Stacey Warren - Expert brainly.com
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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
i think it is the last one
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
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so it is the 3rd one?
because it can't have more ?
well, it can't be any of the other options, since we gave counterexamples
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
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