Which equation is the inverse of 5y+4 = (x+3)^2 + 1/2?

Question

Which equation is the inverse of 5y+4 = (x+3)^2 + 1/2?

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7 days 2021-10-10T19:18:26+00:00 1 Answer 0

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    2021-10-10T19:20:21+00:00

    Answer:

    f^{-1}(x)=-3(+/-)\sqrt{\frac{10x+7}{2}}

    Step-by-step explanation:

    we have

    5y+4=(x+3)^{2}+\frac{1}{2}

    Exchange x for y and y for x

    5x+4=(y+3)^{2}+\frac{1}{2}

    Isolate the variable y

    5x+4-\frac{1}{2}=(y+3)^{2}

    5x+\frac{7}{2}=(y+3)^{2}

    \frac{10x+7}{2}=(y+3)^{2}

    Take square root both sides

    (y+3)=(+/-)\sqrt{\frac{10x+7}{2}}

    y=-3(+/-)\sqrt{\frac{10x+7}{2}}

    Let

    f^{-1}(x)=y

    f^{-1}(x)=-3(+/-)\sqrt{\frac{10x+7}{2}}

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