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Math Help - Finding the domain of g(f(x))

  1. #1
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    Finding the domain of g(f(x))

    How would I go upon solving these:

    1. For (f/g)(-1/4) when f(x)=X^2-2 and g(x)=4x+1


    2.
    For each function f, construct and simplify the difference quotient [f(x+h)-F(x)]/h.
    F(x)=3-9x^3

    thank you!
    Last edited by pmcoura; February 8th 2009 at 05:58 PM.
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  2. #2
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    clarification

    Could you clarify what is meant by (f/g)(-1/4)?
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  3. #3
    ixo
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    Quote Originally Posted by pmcoura View Post
    How would I go upon solving these:

    1. For (f/g)(-1/4) when f(x)=X^2-2 and g(x)=4x+1


    2.
    For each function f, construct and simplify the difference quotient [f(x+h)-F(x)]/H.
    F(x)=3-9x^3

    thank you!
    Think of (f/g)(-1/4) the same as (\frac{f(-1/4)}{g(-1/4)})

    For #2 i'm not quite sure what its asking, but i'm sure someone will jump in and get it for you.
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  4. #4
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    Quote Originally Posted by ixo View Post
    Think of (f/g)(-1/4) the same as (\frac{f(-1/4)}{g(-1/4)})
    Oh.. so the first problem is just a simple function problem where you plug in the x-values to solve it?
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  5. #5
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    Quote Originally Posted by Catapult14 View Post
    Oh.. so the first problem is just a simple function problem where you plug in the x-values to solve it?
    I'm thinking that's what it's asking but I'm not sure myself. But if (f/g) is multiplied by (-1/4) shouldn't I just multiply the denominator by -4? (the denominator being g)
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  6. #6
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    Quote Originally Posted by ixo View Post
    Think of (f/g)(-1/4) the same as (\frac{f(-1/4)}{g(-1/4)})
    Well, according to ixo, you are not multiplying (f/g) by (-1/4), but are plugging in (-1/4) into the functions f(x)=X^2-2 and g(x)=4x+1:

    (\frac{f(-1/4)}{g(-1/4)}) = (\frac{(-1/4)^2-2}{4(-1/4)+1})

    Have I understood this correctly?
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  7. #7
    ixo
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    If f(x) = 2x and g(x) = 3x and h(x) = 4x
    then f(g(x)) = 2( 3x ) and f(h(x)) = 2( 4x)
    similarly g(f(x)) = 3( 2x ) and g(h(x)) = 3( 4x )
    as well as
    f(x)g(x) = (2x)(3x)

    Whatever is inside the parenthesis is what you place as x for the equation. If f(x) = 2x and they ask you to evaluate f(2) then its just simply 2(2). Similarly if g(x) = 3x and they ask you to solve g(5) its just 3(5). If its (f \cdotg)(3) thats the same as f(3)g(3) or 2(3) x 3(3) , similarly if its (f/g)(9) its the same as f(9)/f(9) or 2(9) / 3(9).

    Hope this helps
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  8. #8
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    I realized the first problem was really simple and stupid for me to ask. It's not really that I had a hard time solving it it's just that the questions on my homework are badly written. But does anyone know what the second problem is trying to ask?
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  9. #9
    ixo
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    Quote Originally Posted by pmcoura View Post
    How would I go upon solving these:

    1. For (f/g)(-1/4) when f(x)=X^2-2 and g(x)=4x+1


    2.
    For each function f, construct and simplify the difference quotient [f(x+h)-F(x)]/h.
    F(x)=3-9x^3

    thank you!
    I think.. it may mean that you simplify for f(x) (lowercase) for (3-9x) : \frac{f(x)+h(x)-3+9x^3}{h}. But i'm not entirely sure. Why wouldn't they use a different variable instead of having f and F.. thats what i'm not understanding.
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