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Math Help - functions and limits

  1. #1
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    functions and limits

    I have trouble with this questions

    f:[3,infinity)-->R f(x)=x^2 -6x +10

    The answer for this (in mcqs) is
    f^-1:[1,infi)-->[3,infi), f^-1(y)= 3+ (y-1)^-.5

    I do know how to get the f inverse but i do not know how to get the domains and ranges.
    Can you explain what is meant from [1,infi)-->[3,infi) ?

    2nd question

    f(x)={x x is rational
    -x x is irrational

    g(x)={sinx x is rational
    x^3 x is irrational

    Can you comment on the existance of limx->0f(x) and lim->0g(x)

    Thank you

    PS. sorry about the written format. I don't know how to use maths sybols(is there a link which tells how)
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Dili View Post
    2nd question

    f(x)={x x is rational
    -x x is irrational

    g(x)={sinx x is rational
    x^3 x is irrational

    Can you comment on the existance of limx->0f(x) and lim->0g(x)

    Thank you

    PS. sorry about the written format. I don't know how to use maths sybols(is there a link which tells how)
    (A)

    -x <= f(x) <= x

    hence as x -> 0 f(x) is trapped between two continuous function both of
    which go to 0, so the limit exists and is 0.

    (B)

    for |x|<1

    -x <= g(x) <= x,

    so again the limit of g(x) as x->0 is 0.

    RonL
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  3. #3
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    Can anyone post an answer to the first question?
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  4. #4
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    Quote Originally Posted by Dili View Post
    f:[3,infinity)-->R f(x)=x^2 -6x +10

    The answer for this (in mcqs) is
    f^-1:[1,infi)-->[3,infi), f^-1(y)= 3+ (y-1)^-.5

    I do know how to get the f inverse but i do not know how to get the domains and ranges.
    Can you explain what is meant from [1,infi)-->[3,infi) ?
    You have a function f with a domain  [3, \infty) and a given range which is all real numbers.

    I don't know what "mcqs" is supposed to be, but it looks like you are trying to find the inverse function.

    I'm going to first change the form of f a bit. This is not strictly necessary, but will simplify finding the inverse function.
    f(x) = x^2 - 6x + 10

    f(x) = (x^2 - 6x) + 10

    f(x) = (x^2 - 6x + 9) - 9 + 10

    f(x) = (x - 3)^2 + 1

    Now to find the inverse function. I'm going to let y = f(x) represent the function. Then x = f(y) represents the inverse function. So I'm going to solve x = f(y) for y:
    x = (y - 3)^2 + 1

    (y - 3)^2 = x - 1

    y - 3 = \sqrt{x - 1}

    y = 3 + \sqrt{x - 1}

    So the function f^{-1} is given by f^{-1}(y) =  3 + \sqrt{x - 1}. (I changed the variable back to y to match with your problem. It's a "dummy" variable and can be anything we wish.)

    How about domain and range? Well, the domain of f is  [3, \infty) so this is the range of f^{-1}. The allowed domain of f^{-1} is  [1, \infty). This domain gives the exact range as we expect (it must give a range at least as large as the expected range, else we need to modify the domain such that it will.) So the function f^{-1} may be expressed as:
    f^{-1}: [1, \infty) \to [3, \infty) \mapsto f^{-1}(y) = 3 + \sqrt{x - 1}

    -Dan
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