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    I understand that a strictly increasing/decreasing function always has a positive/negative gradient unless at a stationary point.
    But look at the attached function.

    The function always has a negative gradient, but upon passing the asymptote
    x=3, the value jumps from -(infinity) to +(infinity), so the function jumps from a
    negative value to a positive value. Is the function still strictly decreasing? Or does the jump from a -ve to a +ve mean it has increased, therefore not strictly decreasing?
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    f(x)= (x+1)/(x-3) i think.
    (Original post by Slewis99)
    I understand that a strictly increasing/decreasing function always has a positive/negative gradient unless at a stationary point.
    But look at the attached function.

    The function always has a negative gradient, but upon passing the asymptote
    x=3, the value jumps from -(infinity) to +(infinity), so the function jumps from a
    negative value to a positive value. Is the function still strictly decreasing? Or does the jump from a -ve to a +ve mean it has increased, therefore not strictly decreasing?
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    (Original post by _gcx)
    Note: a function may be (strictly) increasing over one interval, and (strictly) decreasing over another.
    So would I say it's strictly decreasing over (-infinity,3) and strictly decreasing over (3,+infinity)?
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    (Original post by Slewis99)
    So would I say it's strictly decreasing over (-infinity,3) and strictly decreasing over (3,+infinity)?
    Sorry, wasn't paying much attention. (though it's true it's not relevant to your question, didn't see the graph or the equation) It's strictly decreasing "everywhere" as its derivative, -\frac 4 {(x-3)^2} < 0 for all x, though there's obviously the undefined point at x=3.
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    (Original post by Slewis99)
    So would I say it's strictly decreasing over (-infinity,3) and strictly decreasing over (3,+infinity)?
    Definition of strictly decreasing is that \forall x \in D where D is the domain of f, you have f' < 0

    In your example, f(x) = \frac{x+1}{x-3} and D=(-\infty,3)\cup (3, \infty) and f'(x)=(x-3)^{-1}-(x+1)(x-3)^{-2}=\frac{-2}{(x-3)^2} which is clearly always negative when x \in D, so your function is strictly decreasing.

    Answer your question? I basically put most of the things you've noticed here.
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    (Original post by RDKGames)
    Definition of strictly decreasing is that \forall x \in D where D is the domain of f, you have f' < 0
    No it isn't!

    A function is strictly decreasing on an interval I if whenever x > y (with x, y in I) we have f(x) < f(y). There is no requirement that f is differentiable, and even if f is differentiable, it is not necessary for f' to be < 0 everywhere.

    -x^3 is strictly decreasing on R, despite the fact that it's derivative is 0 at x = 0.

    It is in fact possible to define a (pathological) function f that is strictly decreasing on [0,1] and yet f'(x) = 0 almost everywhere.

    Edit: and we don't usually talk about increasing / decreasing functions whose domain is not an interval, for the exact reason that everything gets a bit meaningless when you go from one interval to the next. But if you really want to, the same definition holds, only replace interval with domain.
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    (Original post by DFranklin)
    No it isn't!

    A function is strictly decreasing on an interval I if whenever x > y (with x, y in I) we have f(x) < f(y). There is no requirement that f is differentiable, and even if f is differentiable, it is not necessary for f' to be < 0 everywhere.

    -x^3 is strictly decreasing on R, despite the fact that it's derivative is 0 at x = 0.

    It is in fact possible to define a (pathological) function f that is strictly decreasing on [0,1] and yet f'(x) = 0 almost everywhere.

    Edit: and we don't usually talk about increasing / decreasing functions whose domain is not an interval, for the exact reason that everything gets a bit meaningless when you go from one interval to the next. But if you really want to, the same definition holds, only replace interval with domain.
    Didn't really mean to throw the word definition in there in the first place, so my bad - just stuck to what they're expected to use at A-Level with the derivatives.
 
 
 
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