Forum for the GRE subject test in mathematics.
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phcooh
- Posts: 4
- Joined: Mon Oct 13, 2008 7:15 pm
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by phcooh » Fri Oct 24, 2008 5:48 pm
compare three values: 2^(1/2), 3^(1/3), 6^(16).
I know the function f(x)=x^(1/x) has max at x=e,
and lim f(x) = 0 when x-> 0,
lim f(x) = 1 when x-> OO
but how can we determine the relative value of f(x) when x=2, 3, 6?
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Nameless
- Posts: 128
- Joined: Sun Aug 31, 2008 4:42 pm
Post
by Nameless » Fri Oct 24, 2008 7:27 pm
All those number are greater than 1 so
[sqrt(2)]^6=2^3
[(3)^(1/3)]^6=3^2
[(6)^(1/6)]^6=6
Q.E.D
