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    (Original post by katJ1873)
    ok so I have a supposedly simple function but I can't seem to understand it;

    y=x^(1.5)
    dy/dx=1.5x^0.5

    show y is an increasing function?

    ..so all values of x are meant to give a positive y values meaning it's increasing but I don't think it does? and how do I show this?

    thanks
    What level is this? A level, university? The answer will depend on what you already know.

    Anyway, you need to show either that, for +ve x:

    a) dy/dx >0 \Rightarrow x^{0.5}=\sqrt{x} > 0, or

    b) a > b \Rightarrow a^{3/2} > b^{3/2}

    But the first is essentially trivial since the +ve square root is +ve by definition, and the second can be done by, say:

    1. using the fact that \log is an increasing function (what is its derivative?).

    2. considering, for a > b >0, a^{3/2}-b^{3/2}=\sqrt{a^3}-\sqrt{b^3}, rationalising the numerator using the difference-of-two-squares, and proceeding from there to show that the expression is +ve.
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    (Original post by atsruser)
    What level is this? A level, university? The answer will depend on what you already know.

    Anyway, you need to show either that, for +ve x:

    a) dy/dx >0 \Rightarrow x^{0.5}=\sqrt{x} > 0, or

    b) a > b \Rightarrow a^{3/2} > b^{3/2}

    But the first is essentially trivial since the +ve square root is +ve by definition, and the second can be done by, say:

    1. using the fact that \log is an increasing function (what is its derivative?).

    2. considering, for a > b >0, a^{3/2}-b^{3/2}=\sqrt{a^3}-\sqrt{b^3}, rationalising the numerator using the difference-of-two-squares, and proceeding from there to show that the expression is +ve.
    it is a level - I think I understand now you said that, thanks
 
 
 
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