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Thread: 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?

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

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

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

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

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

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


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

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

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

    Take logs: .$\displaystyle \ln(x - 4)\:=\:\ln(2^{y+1})\quad\Rightarrow\quad \ln(x - 4)\:=\y + 1)\ln(2) $

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

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


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

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

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

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

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

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