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increasing/decreasing function help?

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(edited 6 years ago)
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>0x0.5=x>0dy/dx >0 \Rightarrow x^{0.5}=\sqrt{x} > 0, or

b) a>ba3/2>b3/2a > 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\log is an increasing function (what is its derivative?).

2. considering, for a>b>0a > b >0, a3/2b3/2=a3b3a^{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.
Reply 2
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>0x0.5=x>0dy/dx >0 \Rightarrow x^{0.5}=\sqrt{x} > 0, or

b) a>ba3/2>b3/2a > 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\log is an increasing function (what is its derivative?).

2. considering, for a>b>0a > b >0, a3/2b3/2=a3b3a^{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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