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Increasing functions

  1. Nov 24, 2008 #1

    I'm not too sure about this, I thought as a example function e^x, where I can say as x gets larger then so does e^x and then so does its derivative at those values of x.

    But what if it was a function where as x increased the function was getting smaller, and so it would be a decreasing function.. or perhaps it doesn't matter and I should assume a function like that doesn't exist?

    Is there any other fucntion with that property other than e^x?

    Really confused about the second part, taking e^x as an example, then then inverse would be lnx, and its derivative 1/x. Then surely I'm supposed to assume all they are talking about here is e^x, since no other f'(x) can take the form of 1/x other than f(x) = lnx...
  2. jcsd
  3. Nov 24, 2008 #2
    since f(x)=f`(x) and codomain is R+ f`(x) is always > 0 hence increasing.
  4. Nov 24, 2008 #3
    Would that be considered a proof?
  5. Nov 24, 2008 #4


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    What f(x) said is not quite true. f is NOT into R+ and the f ' is NOT always "> 0" but close:
    Your problem tells you that f(x) is never negative and that f'(x)= f(x) for all x. That is, f'(x) is never negative. What does that tell you?
  6. Nov 24, 2008 #5
    Ah I see, the second part though, so far I can say the slope of the inverse function would be given by 1/f'(x).

    Can someone tell me if the question is asking for the derivative of the inverse function, or the inverse of the derivative?


    I see it's asking for the derivative of the inverse function, so what I did was this:

    the slope of the inverse would be 1/f'(x) = dx/dy taking y = f(x)

    then I went on to say that 1/f'(x) = (f -1)' (x) (derivative the inverse of f(x))

    So howdo i show this is 1/x ?
    Last edited: Nov 24, 2008
  7. Nov 25, 2008 #6
    gta hand this in in 30 mins, hoping someone will read this and help :(
  8. Nov 25, 2008 #7


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    You are confusing x and y in y= f(x) and x= f-1(y).
  9. Nov 25, 2008 #8
    Hello sir,
    Pardon me but could you plz explain why f`(x) >= 0 and not f`(x) > 0 since the codomain of f(x) has an interval (0,∞) exclusive while f(x) = f`(x)
  10. Nov 25, 2008 #9


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    You are right, I was wrong. I misread [itex](0, \infty)[/itex] as [itex][0, \infty)[/itex].
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