Rational functions with infinite critical points?

Question

I know trigonometric functions such as 7x-6cos(3x) 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 1

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.

Answer 2

Well, a non-zero constant rational function (for example, $g(x)=\frac{x+1}{2x+2}$) has infinitely many critical points: every point of its domain. Assuming the rational function is non-constant, it has the form $f(x)=\frac{P(x)}{Q(x)}$ where $P$ and $Q$ are polynomials. Its derivative is $f'(x)=\frac{P'(x)Q(x)-P(x)Q'(x)}{Q^2(x)}$. Because it's differentiable everywhere on its domain, the critical points are $f'(x)=0$ $\Rightarrow P'(x)Q(x)-P(x)Q'(x)=0$. The last expression has to be a polynomial. If it's the zero polynomial, then $f'(x)\equiv0$, 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 $f(x)$.