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Math Help - domain and range of a function

  1. #1
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    domain and range of a function

    Let f(x) = (x+3)/(x+1)

    (a) Find the domain and range of f(x)

    (b) Show that f(x) is one-to-one function and find the inverse function f-1(x)


    My attempt for (a) : since the function is undefined when x = -1 thus the domain is IR\{-1}

    But my problem is how i can find the range of the function. is the range of the function is IR except the value when x=-1. BUT how i can find the value of the function at x=-2???

    If I were to find the range of the function by graph, how f(x) graph looks like???

    next if i want to show that f(x) is one-to-one function then i have to find the inverse of it... is there any other way to show that it is one-to-one???

    the way i find the inverse :

    let f(x) = y

    thus have to find x in terms of y

    y = x+3/x+1

    but how can i group the x so that i can found x in terms of y??? since they are separated by quotient....
    Last edited by bobey; March 30th 2010 at 08:15 AM.
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  2. #2
    Ted
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    Quote Originally Posted by bobey View Post
    Let f(x) = (x+3)/(x+1)

    (a) Find the domain and range of f(x)

    (b) Show that f(x) is one-to-one function and find the inverse function f-1(x)


    My attempt for (a) : since the function is undefined when x = -1 thus the domain is IR\{-1}

    But my problem is how i can find the range of the function. is the range of the function is IR except the value when x=-1. BUT how i can find the value of the function at x=-2???

    If I were to find the range of the function by graph, how f(x) graph looks like???

    next if i want to show that f(x) is one-to-one function then i have to find the inverse of it... is there any other way to show that it is one-to-one???

    the way i find the inverse :

    let f(x) = y

    thus have to find x in terms of y

    y = x+3/x+1

    but how can i group the x so that i can found x in terms of y??? since they are separated by quotient....
    Range of f = domain its inverse

    to find the inverse:

    y=\frac{x+3}{x+1}

    yx+y=x+3

    yx-x=3-y


    x=\frac{3-y}{y-1}=-\frac{y-3}{y-1}

    so f^{-1}(x)=-\frac{x-3}{x-1}

    clearly, domain of f inverse is R / {1}

    so the range of f is R / {1}
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