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Math Help - Composite f, g, and/or h

  1. #1
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    Composite f, g, and/or h

    For questions 1 and 2 below, let f(x) = x^2, g(x) = 3x and
    h(x) = (sqrt{x}) + 1. Express each function as a composite of f, g and/or h.

    (1) P(x) = 3(sqrt{x}) + 3

    (2) Q(x) = sqrt({sqrt{x} + 1}) + 1

    NOTE:

    For question 2, the square root of x lies within another square root.

    We have a square root within a square root.

    Clear?
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  2. #2
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    Quote Originally Posted by symmetry View Post
    For questions 1 and 2 below, let f(x) = x^2, g(x) = 3x and
    h(x) = (sqrt{x}) + 1. Express each function as a composite of f, g and/or h.

    (1) P(x) = 3(sqrt{x}) + 3

    (2) Q(x) = sqrt({sqrt{x} + 1}) + 1

    NOTE:

    For question 2, the square root of x lies within another square root.

    We have a square root within a square root.

    Clear?
    Hello,

    to (1):

    P(x)=g(h(x)) and then expand the bracket.

    to (2):

    Q(x)= h(h(x))

    Substitute a complete term instead of the x of the "outer" function.

    EB
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  3. #3
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    ok

    Sorry...but I don't follow what you're saying.

    Can you show me by doing one of the questions?

    Thanks!
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  4. #4
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    Quote Originally Posted by symmetry View Post
    For questions 1 and 2 below, let f(x) = x^2, g(x) = 3x and
    h(x) = (sqrt{x}) + 1. Express each function as a composite of f, g and/or h.

    (1) P(x) = 3(sqrt{x}) + 3

    (2) Q(x) = sqrt({sqrt{x} + 1}) + 1
    Second attempt:

    to (1)

    P(x)=3 \sqrt{x} +3=3 \underbrace{(\sqrt{x} +1)}_{\text{this is h}}=\underbrace{3 \cdot h(x)}_{\text{this is g with h(x) instead of x}}. Therefore:

    P(x)=g(h(x))

    to (2)

    Q(x)=\underbrace{\sqrt{\underbrace{(\sqrt{x} +1)}_{\text{this is h}}}+1}_{\text{this is h with h instead of x}}. Therefore:

    Q(x)=h(h(x))

    EB
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  5. #5
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    ok

    Thanks. I now know what you are talking about.
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