# Monotonic function

## Homework Statement

$$y=x^3$$

## The Attempt at a Solution

I know that function is increasing when $$f'(x) > 0$$ but in $$x=0$$ there is $$f'(x) = 0$$, so is function increasing there or not? From definition I know that its increasing there, but how can I connect this with theorem that function is increasing when $$f'(x) > 0$$?

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vela
Staff Emeritus
Homework Helper
What precisely is your definition of an increasing function?

$$\forall_{x_1,x_2\in X} \left( x_1 < x_2 \Rightarrow f(x_1)<f(x_2) \right)$$

If a function is non-decreasing (weakly increasing), and $$f(x_1)=f(x_2)$$, $$x_1\neq x_2$$, then f is constant on $$[x_1, x_2]$$, so a single zero of a derivative cannot spoil injectivity.

aha, so if derivative is positive and 0 only for countable set of points the function will be increasing?

Yes. It can probably be strenghtened a little bit, but that should be enough for all practical purpose.

HallsofIvy
$$\forall_{x_1,x_2\in X} \left( x_1 < x_2 \Rightarrow f(x_1)<f(x_2) \right)$$
With that definition, it makes no sense to talk about a function being "increasing" at a point. It is easy to prove that $y= x^3$ is increasing on any interval.