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Math Help - Composite Functions

  1. #1
    Senior Member sakonpure6's Avatar
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    Composite Functions

    For each of the following, 1- Determine the definite equation for f[g(x)] and g[f(x)]. 2- Determine the domain and range of f[g(x)] and g[f(x)]

    e) f(x) = 10^x , g(x)= log x (log x base 10)
    This is what I have for
    f[g(x)] = 10 ^ ( log x) = x
    g[f(x)] = log 10^x = x.

    f[g(x)] = {x E R , x > 0 } because in g(x) x can not be negative.......................... g[f(x)] = { x E R, x > 0 } because in g(x) x can not be negative
    f[g(x)] ={y E R, y > 0}................................................ ....................................g[f(x)] = {y E R, y > 0}


    However my text book says
    f[g(x)] = {x E R , x > 0 } ......................g[f(x)] = { x E R }
    f[g(x)] = { y E R} ...................... ... g[f(x)] ={y E R}


    Who is right? Thank you in advance.
    Last edited by sakonpure6; December 15th 2013 at 01:23 PM.
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  2. #2
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    Re: Composite Functions

    Quote Originally Posted by sakonpure6 View Post
    This is what I have for
    e-1)f[g(x)] = 10 ^ ( log x) = x ..... e-2)g[f(x)] = log 10^x = x.

    e-1) Domain {x E R , x > 0 } because in g(x) x can not be negative e-2) Domain { x E R, x > 0 } because in g(x) x can not be negative
    e-1) Range { y E R, y > 0} e-2) Range {y E R, y > 0}


    However my text book says
    e-1) Domain {x E R , x > 0 } e-2) Domain { x E R }
    e-1) Range { y E R} e-2) Range {y E R}


    Who is right? Thank you in advance.
    I'm sorry this is all very confusing the way you have it written. Could you repost this along the lines of

    f(g(x)): Domain = {}, Range = {}

    g(f(x)): Domain = {}, Range = {}

    where you would fill in the braces.
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  3. #3
    Senior Member sakonpure6's Avatar
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    Re: Composite Functions

    is this better (I edited post)
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  4. #4
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    Re: Composite Functions

    ok I think I see the confusion.

    It's the subtle difference between the domain of

    f(x) vs. f(g(x)) ; and

    g(x) vs. g(f(x))

    you don't like that g(f(x)) has domain all of R because g(x) has domain {x:x > 0}

    but you don't pass x to g here.. you pass f(x). You pass x to f, and f has domain of all of R.

    the second thing you don't agree with is that the range of g(f(x)) is all R and not just {y:y>0}

    let u=f(x)=10x

    well log(u) is negative for {u<1}. {u<1} is certainly in the range of f(x). It occurs for {x<0}

    so the range of g(f(x)) is in fact all of R
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