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31 July, 17:59

How do I show that this function is one-to-one algebraically?

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  1. 31 July, 20:24
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    For a one-to-one function, f (x) = f (y).

    So, lets find f (x) and f (y).

    We know,

    f (x) = (x - 2) ³ + 8

    Now,

    f (y) = (y - 2) ³ + 8

    Now, we said earlier, for a function to be one-to-one, f (x) = f (y).

    Therefore,

    f (x) = f (y)

    (x - 2) ³ + 8 = (y - 2) ³ + 8

    (x - 2) ³ = (y - 2) ³

    x - 2 = y - 2

    x = y

    Since we got x = y, we know for every x, there is one and only one y. Therefore, the function is one-to-one function.
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