Is x=0 Included in the Increasing or Decreasing Range of y=x^2?

[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

 

[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.

 

[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

 

[Doc Al]

But what about x=0? Shouldn't it be included in one of the two ranges of x?

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

 

[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

 

[Doc Al]

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 & http://www.mathreference.com/ca,inc.html

 

[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

 

[Kamataat]

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

- Kamataat