# Increasing/decreasing f(x)

1. Feb 18, 2005

### Kamataat

Let's take $y=x^2$ as an example. This function decreases if $-\infty < x < 0$ and increases if $0 > x > \infty$. But what about $x=0$? Shouldn't it be included in one of the two ranges of $x$?

- Kamataat

2. Feb 18, 2005

### hypermorphism

Why do you think it should ? This boils down to whether you want zero to be "both positive and negative" or "neither positive nor negative". The latter is more useful.

3. Feb 18, 2005

### Kamataat

So the function $y=x^3$ is not increasing if $-\infty < x < \infty$ (1), but instead is increasing if $-\infty < x < 0$ and if $0 < x < \infty$, because $y'(0)=0$ cuts the range (1) in two pieces?

- Kamataat

4. Feb 18, 2005

### Staff: Mentor

Why? At x = 0 the slope is zero: y is neither increasing nor decreasing.

5. Feb 18, 2005

### Kamataat

Yes, I know that, but my confusion arises from my textbook saying that $y=x^3$ is a strictly increasing function for all $x \in X$. How can that be right, if at one x (namely x=0), y'=0 and the function is thus constant?

- Kamataat

6. Feb 18, 2005

### Staff: Mentor

7. Feb 18, 2005

### Kamataat

yup, that's how it's defined in the book, but it still confuses me. according to the definition of "strictly increasing", i'd say that x^3 is strictly increasing for all x, but then we have the definition that a function is neither decreasing nor incresing if y'=0 (which is true for x^3). so the second def says that x^3 is not increasing for ALL x.

- Kamataat

8. Feb 18, 2005

### Kamataat

nevermind, the 2nd link seems to explain it. thank you!

- Kamataat