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Increasing / decreasing functions

  1. Nov 14, 2012 #1
    I juat have a quick question.

    Which of these is correct;

    f(x) is one-to-one [itex]\Rightarrow[/itex] f'(x) is increasing

    f(x) is one-to-one [itex]\Rightarrow[/itex] f'(x) is non-decreasing.

    I think number one but I'm just not 100% sure.

    Thanks
     
  2. jcsd
  3. Nov 14, 2012 #2
    What about f(x) = arctan (x).
     
  4. Nov 14, 2012 #3

    Ray Vickson

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    It the function f1(x) = 1 - exp(-x) 1:1? If f1'(x) increasing? Is the function f2(x) = exp(x) 1:1? is f2'(x) increasing? Is f3(x) = arctan(x) 1:1? Is f3'(x) increasing everywhere? Is it decreasing everywhere?

    RGV
     
  5. Nov 14, 2012 #4
    arctan in 1:1 and it looks as if it's always increasing.

    What I want to know is that if a function is non-decreasing (i.e. a straight line) is it 1:1
     
  6. Nov 14, 2012 #5

    Zondrina

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    Suppose f(x) is a one to one function. So if f(x) = f(y) then x = y.

    RGV gave a good example of a function that shoots down your theory, either one of them since they both mean the same thing.

    Take f(x) = 1 - e-x. f'(x) = e-x.

    So first note that f(x) is 1-1 and increasing. Now, f'(x) on the other hand is actually decreasing.

    So it is not true that : If f(x) is 1-1, then f'(x) is decreasing.

    To answer your other question about "what if f is a straight line" consider f(x) = c for some constant real number c.

    There are infinitely many choices of x that give the same value for f, so the function is not 1-1.
     
  7. Nov 14, 2012 #6
    Ah right thanks Zondrina.

    Yeah apologies for the confusion but I wrote it down incorrectly. I meant always increasing or always decreasing.

    Thanks for the clarification
     
  8. Nov 14, 2012 #7
    I don't mean to belabor the point, but if your function is 1-1 it need not be increasing or decreasing. e.g. Let f(x) = 1/x for negative x, f(x) = x for x greater than or equal to 0.

    However, if f(x) is continuous and 1-1, then it is true that f must either be increasing or decreasing.

    Also, if f is differentiable, and 1-1, then either [itex] f'(x) \geq 0 [/itex] for all x. OR [itex] f'(x)\leq 0 [/itex] for all x.
     
  9. Nov 14, 2012 #8
    Yeah, I was referring to Continuous functions, apologies.

    Regarding your last point, if f'(x) = 0 doesn't this mean that f is not 1-1?
     
  10. Nov 14, 2012 #9
    It depends on the set of values x such that f'(x)=0. The derivative can be zero at isolated points. For example, if f(x)=x^3, then f'(0)=0. You could even come up with examples where f'(x)=0 for an infinite number of x values. For example, if f(x)=x-sin(x), then f'(x)=1-cos(x) which equals 0 at all integer multiples of 2*pi, but is otherwise positive.

    It is even possible for f'(x) to be zero at an infinite number of points in a finite interval. For example, suppose
    [itex] f'(x) = x^2 sin^2(\pi / x) [/itex]
    This is 0 at all points x=1/n, but otherwise positive.

    The technical requirement is that if f' ≥ 0 then f is strictly increasing if and only if the set of points x such that f'(x)=0 has measure 0.
     
    Last edited: Nov 14, 2012
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