increasing polynomial graph with two points having slope zero

Question

We recently learned about the slope of a polynomial graph, and that an odd-degree graph can be strictly increasing despite having a point with $dy/dx=0$. Example: $x \mapsto x^3.$

i was wondering if there is graph possible with two points having $dy/dx=0$, while remaining injective (one-one).

Answer 1

Your example works because $0$ is a double root of the derivative of $x^3$. Take $f'(x) = x^2(x-1)^2 = x^4 -2x^3+ x^2$. Notice that $f'(x)$ is always positive (because it's a square.) And note that $f'(0) = f'(1) =0.$

So now just undifferentiate to get $f(x) = x^5/5 - x^4/2 +x^3/3.$

Answer 2

Of course!

Why would you think otherwise?

Answer 3

Take any postive polynomial with two roots and find the antiderivative of it the antiderivative will be increasing since its derivative is positive and will have two points having slope $0$ .