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Math Help - Inverse Equation

  1. #1
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    Exclamation Inverse Equation

    How would I find out the inverse for these equations?

    F(x) = 2√(x+3)
    G(x) = 2^(x+1)+4
    Y= 2(x-5)^3
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  2. #2
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    Hello, takkun0486!

    Do you know the procedure for finding an inverse function?

    [1] Replace the function with y.
    [2] "Switch" the x's and y's.
    [3] Solve for y.

    How would I find out the inverses for these functions?

    (a)\;\;F(x) \:= \:2\sqrt{x+3}
    (b)\;\;G(x) \:= \:2^{x+1} + 4
    (c)\;\;Y\:= \:2(x-5)^3
    (a) We have: [1]\;\;y\;=\;2\sqrt{x+3}

    "Switch": [2]\;\;x\;=\;2\sqrt{y + 3}

    Solve for y: . [3]\;\;\frac{x}{2}\;=\;\sqrt{y + 3}

    Square: . \left(\frac{x}{2}\right)^2\;=\;(\sqrt{y + 3})^2\quad\Rightarrow\quad \frac{x^2}{4}\;=\;y + 3

    Then: . \frac{x^2}{4} - 3 \;= \;y

    Therefore: . f^{-1}(x)\;=\;\frac{x^2}{4} - 3


    (b) We have: . [1]\;y \:= \:2^{x+1} + 4

    Switch: . [2]\;\;x\;=\;2^{y+1} + 4

    Solve for y: . [3]\;\;x - 4\:=\:2^{y+1}

    Take logs: . y + 1)\ln(2) " alt="\ln(x - 4)\:=\:\ln(2^{y+1})\quad\Rightarrow\quad \ln(x - 4)\:=\y + 1)\ln(2) " />

    Then: . \frac{\ln(x - 4)}{\ln(2)}\:=\:y + 1\quad\Rightarrow\quad\frac{\ln(x-4)}{\ln(2)} - 1\:=\:y

    Therefore: . G^{-1}(x)\;=\;\frac{\ln(x-4)}{\ln(2)} - 1


    (c) We have: . y \:=\:2(x - 5)^3

    Switch: . x \;= \;2(y - 5)^3

    Solve for y: . \frac{x}{2}\;=\;(y - 5)^3

    Take cube roots: . \sqrt[3]{\frac{x}{2}}\;=\;y - 5

    Then: . \sqrt[3]{\frac{x}{2}} + 5 \;= \;y

    Therefore: . Y^{-1}\;=\;\sqrt[3]{\frac{x}{2}} + 5 . . . or: Y^{-1} \;= \;\frac{\sqrt[3]{4x}}{2} + 5
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