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Math Help - real functions

  1. #1
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    real functions

    For any real numbers a,b,c & d such that c & d are not both zero, the formula f(x)=(ax+b)/(cx+d) defines a real function.
    a)find the domain & range of f.
    b)Show that f is a one-to-one function if and only if ad-bc not=0
    c)Show that if f is one-to-one, then f' is given by f-1(x)=(dx-b)/(-cx+a) where c & a are both not zero
    d) show that if f is one-to-one and a=-d then f-1=f
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  2. #2
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    Re: real functions

    Please show us what you've tried.
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    Re: real functions

    a) since c & d are not both zeros, then the Domain is the set of all real numbers, and the range is the set of all real numbers as well.

    right?
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    Re: real functions

    Quote Originally Posted by franios View Post
    a) since c & d are not both zeros, then the Domain is the set of all real numbers, and the range is the set of all real numbers as well.

    right?
    No, for any functions which are defined for all the reals, as linear functions like those are, their quotient will be defined for all the reals EXCEPT for where the denominator is 0. What value of x will make the denominator 0?
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  5. #5
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    Re: real functions

    x=0
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  6. #6
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    Re: real functions

    No, if x = 0, then the denominator cx + d = c(0) + d = d =/= 0.
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    Re: real functions

    cx+d=0 solve for x, xnot=(-d)/c
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    Re: real functions

    That's better, so what is the domain of your function?
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  9. #9
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    Re: real functions

    All real numbers except when x=-d/c?
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    Re: real functions

    Correct, now what is the range? It might help to rewrite this as \displaystyle \begin{align*} \frac{a}{c} + \frac{bc - ad}{c(cx + d)} \end{align*}.
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  11. #11
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    Re: real functions

    the range would still be all real numbers, wouldn't it??
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