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Math Help - Composite functions

  1. #1
    Senior Member DivideBy0's Avatar
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    Composite functions

    Hi, I'm trying to understand the example the book gives:

    For the functions f(x)=x^2-1, x \in R, and  g(x) = \sqrt{x},x \geq 0, state why g(f(x)) is not defined.

    Range of f \not\subseteq domain of g.
    Therefore, g(f(x)) is not defined.

    How can it not be defined? I've drawn graphs of y = \sqrt{x^2-1} and they look perfectly reasonable. What is going on? Thanks
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  2. #2
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    Quote Originally Posted by DivideBy0 View Post
    Hi, I'm trying to understand the example the book gives:

    For the functions f(x)=x^2-1, x \in R, and  g(x) = \sqrt{x},x \geq 0, state why g(f(x)) is not defined.

    Range of f \not\subseteq domain of g.
    Therefore, g(f(x)) is not defined.

    How can it not be defined? I've drawn graphs of y = \sqrt{x^2-1} and they look perfectly reasonable. What is going on? Thanks


    It is defined on its domain, and its domain is f^{-1} (D_g), where D_g is the domain of g

    ZB
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  3. #3
    Senior Member DivideBy0's Avatar
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    sorry i don't think i'm getting it
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  4. #4
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    For the functions and , state why is not defined.
    Actually, g[f(x)] is defined on the entire domain of g: zero to infinity EXCEPT for values between zero and one.
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  5. #5
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    Quote Originally Posted by DivideBy0 View Post
    Hi, I'm trying to understand the example the book gives:

    For the functions f(x)=x^2-1, x \in R, and  g(x) = \sqrt{x},x \geq 0, state why g(f(x)) is not defined.

    Range of f \not\subseteq domain of g.
    Therefore, g(f(x)) is not defined.

    How can it not be defined? I've drawn graphs of y = \sqrt{x^2-1} and they look perfectly reasonable. What is going on? Thanks
    Say f(x) is defined on domain A. And g(x) is defined on domain B. Suppose we can compose them f(g(x)) in order for this expression to make sense: first we need that x\in B because we evaluate at B first; then we want g(x)\in A because we evaluate at f(x) second. So the rule is for f(g(x)) to be defined we require that the range (output values) of g(x) to be contained in A, the domain f(x).
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  6. #6
    Senior Member DivideBy0's Avatar
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    Thanks everyone,

    So... the composition is 'undefined', even though there are only a few points where it is actually undefined out of infinity. Well I guess |x| <1 does contain infinity points... but that argument seems a bit strained.

    So if f(x) = x^2-1, \ x \in [0, \infty), then g(f(x)) would be defined, right?
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  7. #7
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    So if , then g(f(x)) would be defined, right?
    Not quite. You should be (and probably are) thinking of

    f(x) = x^2-1, f(x)\in [0,\infty)
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  8. #8
    Senior Member DivideBy0's Avatar
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    Yep, that was what I was 'trying' to think of :P

    I have another question, is it always true that ran f(g(x)) = ranf?
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  9. #9
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    I have another question, is it always true that ?
    No.
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  10. #10
    Senior Member DivideBy0's Avatar
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    Ok how about this:

    dom \ f \circ g=ran\ g?

    (My book says it's dom \ f \circ g = dom \ g, but I don't believe this)

    Sine ran\ f \subseteq dom \ g, so i think you can only choose from the ran\ f subset for the domain of the dom \ f \circ g.

    Then ran \ f \circ g can be determined by only graphing the values x \in ran \ g?

    I think I'm finally getting my head around this, thanks so much for all your help.
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  11. #11
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    ?
    no


    (My book says it's , but I don't believe this)

    Sine , so i think you can only choose from the subset for the domain of the .
    Your book is right. A visualisation that might help you is an g-machine, into which you put in a number that is inside dom g, and an f-machine, into which you put a number which is inside dom f. The machine will then print out a number that is inside its range. f o g is the machine made when you attach the input of f to the output of g. It will only work if ran g \subseteq dom f (you wrote this the wrong way round in your post). Otherwise g may print out a number that f can't handle.
    Since ran g \subseteq dom f, the new machine can handle any input in dom g.

    what confused and irritated me when I was learning this is: Why don't we make a machine that works even if ran g \not\subseteq dom f by making a smaller domain so that the machine only receives input that will produce a g(x) that is inside dom f?
    The answer is that we do, but we don't call it defined.

    I think I'm finally getting my head around this, thanks so much for all your help.
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