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Increasing/decreasing f(x)

  1. Feb 18, 2005 #1
    Let's take [itex]y=x^2[/itex] as an example. This function decreases if [itex]-\infty < x < 0[/itex] and increases if [itex]0 > x > \infty[/itex]. But what about [itex]x=0[/itex]? Shouldn't it be included in one of the two ranges of [itex]x[/itex]?

    - Kamataat
     
  2. jcsd
  3. Feb 18, 2005 #2
    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.
     
  4. Feb 18, 2005 #3
    So the function [itex]y=x^3[/itex] is not increasing if [itex]-\infty < x < \infty[/itex] (1), but instead is increasing if [itex]-\infty < x < 0[/itex] and if [itex]0 < x < \infty[/itex], because [itex]y'(0)=0[/itex] cuts the range (1) in two pieces?

    - Kamataat
     
  5. Feb 18, 2005 #4

    Doc Al

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    Staff: Mentor

    Why? At x = 0 the slope is zero: y is neither increasing nor decreasing.
     
  6. Feb 18, 2005 #5
    Yes, I know that, but my confusion arises from my textbook saying that [itex]y=x^3[/itex] is a strictly increasing function for all [itex]x \in X[/itex]. How can that be right, if at one x (namely x=0), y'=0 and the function is thus constant?

    - Kamataat
     
  7. Feb 18, 2005 #6

    Doc Al

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    Staff: Mentor

  8. Feb 18, 2005 #7
    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
     
  9. Feb 18, 2005 #8
    nevermind, the 2nd link seems to explain it. thank you!

    - Kamataat
     
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