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Thread: Algebra 2 Composition of Functions?

  1. #1
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    Algebra 2 Composition of Functions?

    If g(f(x))= 2x+1, f(x)= (1/4)x-1, then what does g(x) equal? The answer is supposed to be 8x+9. Can anyone explain to me clearly step by step how to do this? Thanks!
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  2. #2
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    Re: Algebra 2 Composition of Functions?

    "Fit" the expression of f(x) into g(f(x)):

    $\displaystyle 8f(x) = 8(\frac{1}{4}x - 1) = 2x - 8$

    $\displaystyle 8f(x) + 9 = (2x - 8) + 9 = 2x + 1 = g(f(x))$

    Therefore, $\displaystyle g(x) = 8x + 9$.
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  3. #3
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    Re: Algebra 2 Composition of Functions?

    Hello, woahitzme!

    $\displaystyle \text{If }g(f(x))\:=\: 2x+1,\;\; f(x)\:=\:\tfrac{1}{4}x-1$
    . . $\displaystyle \text{then what does }g(x)\text{ equal?}$
    $\displaystyle \text{The answer is supposed to be: }\,8x+9.$

    We are told that: .$\displaystyle f(x) \:=\:\tfrac{1}{4}x-1$

    Then: .$\displaystyle g\left(\tfrac{1}{4}x-1\right) \:=\:2x+1$

    Assume that $\displaystyle g(x)$ is a linear function: .$\displaystyle g(x) \:=\:ax+b$

    Then: .$\displaystyle g\left(\tfrac{1}{4}x-1\right) \:=\:a\left(\tfrac{1}{4}x-1\right) + b \;=\;2x + 1$

    We have: .$\displaystyle \tfrac{a}{4}x - a + b \:=\:2x+1 \quad\Rightarrow\quad \tfrac{a}{4}x + (b-a) \:=\:2x+1$

    Equate coefficients: .$\displaystyle \begin{array}{ccc}\dfrac{a}{4} &=& 2 \\ b-a &=& 1 \end{array}$

    Hence: .$\displaystyle a \,=\,8,\;b\,=\,9$

    Therefore: .$\displaystyle g(x) \:=\:8x + 9$
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