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Why can f(x)-17 have no inverse? Watch

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    (Original post by _gcx)
    Bit late to the party, but for the sake of aesthetics, you can write this as,

    \csc^2 x\left(x^5+8x^4 - 9x^2 - 2\right)

    From which you can observe the function as being many to one through the periodicity of \sin.
    woah, what's this csc? i feel a bit stupid since im doing uni level and i still don't get it
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    (Original post by will'o'wisp2)
    woah, what's this csc? i feel a bit stupid since im doing uni level and i still don't get it
    it's just \frac 1 {\sin x}, sorry if that was unclear
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    (Original post by _gcx)
    it's just \frac 1 {\sin x}, sorry if that was unclear
    Ah right, so it's a sin graph with the function of the stuff int he brackets then?
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    (Original post by _gcx)
    No, I intended it to be read as \csc^2(x) \times \ldots . My approach is more heuristic so if you're still confused, investigating turning points per the other posts is probably the way to go.
    I personally don't see how you could easily tell that a function in the form f(x)=\csc^2(x)g(x) is necessarily many to one. Could you elaborate?
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    (Original post by I hate maths)
    I personally don't see how you could easily tell that a function in the form f(x)=\csc^2(x)g(x) is necessarily many to one. Could you elaborate?
    Looking back, I have no idea where I was going with that either, I probably wasn't paying much attention. I've added,


    Edit: I have no idea where I was going with this. I should've really explained why the second term is not one-to-one, ie. x^5 + 8x^4 - 9x^2 - 2 is not one-to-one as its derivative is neither strictly positive nor strictly negative, then perhaps using the product rule to conclude that the function \csc^2 x\left(x^5+8x^4 - 9x^2 - 2\right) is not one-to-one.

    Basically, a continuous function f is one-to-one if its derivative is strictly positive or strictly negative.
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    (Original post by _gcx)
    Basically, a continuous function f is one-to-one if its derivative is strictly positive or strictly negative.
    A continuous function need not have a derivative.
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    (Original post by Zacken)
    A continuous function need not have a derivative.
    Indeed, should I say, a continuous, differentiable function.
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    (Original post by _gcx)
    Indeed, should I say, a continuous, differentiable function.
    If a function is differentiable, it is automatically continuous - so there is no need to specify continuity.

    NB: there wasn't anything technically wrong with your statement -- it just read weirdly because of the aforementioned.

    [Also, you have sufficiency but not necessity, for example x^3 is differentiable and bijective, but has 0 derivative at 0]
 
 
 
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