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.

Question

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 · Answer

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

anonymous · Answer

i think it is the last one

anonymous · Answer

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

anonymous · Answer

so it is the 3rd one?

anonymous · Answer

because it can't have more ?

anonymous · Answer

well, it can't be any of the other options, since we gave counterexamples

anonymous · Answer

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

anonymous · Answer

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