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Kassandra Ross

Kassandra Ross

Answered question

2022-06-12

I know trigonometric functions such as 7 x 6 cos ( 3 x ) can have an infinite number of critical points, but what about non-zero rational functions? Does it have something to do with how a non-zero polynomial can have a finite number of roots?

Answer & Explanation

Marlee Guerra

Marlee Guerra

Beginner2022-06-13Added 25 answers

The derivative of a rational function is another rational function. Critical points are where the derivative is 0 or undefined. That happens when the numerator or denominator of the derivative is zero. Both of those are polynomials, so they are zero only a finite number of times.
pokoljitef2

pokoljitef2

Beginner2022-06-14Added 9 answers

Well, a non-zero constant rational function (for example, g ( x ) = x + 1 2 x + 2 ) has infinitely many critical points: every point of its domain. Assuming the rational function is non-constant, it has the form f ( x ) = P ( x ) Q ( x ) where P and Q are polynomials. Its derivative is f ( x ) = P ( x ) Q ( x ) P ( x ) Q ( x ) Q 2 ( x ) . Because it's differentiable everywhere on its domain, the critical points are f ( x ) = 0 P ( x ) Q ( x ) P ( x ) Q ( x ) = 0. The last expression has to be a polynomial. If it's the zero polynomial, then f ( x ) 0 so f ( x ) is a constant, a case we already excluded. So it's a non-zero polynomial with a finite number of roots which are the critical points of f ( x ).

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