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

strictly increasing

I think it hinges on the definition of "strictly increasing", which is based on an interval: f(x) is strictly increasing if a < b implies f(a) < f(b).

See: http://planetmath.org/encyclopedia/IncreasingdecreasingmonotoneFunction.html [Broken] & http://www.mathreference.com/ca,inc.html

Last edited by a moderator: May 1, 2017
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

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