Calculus problem

Forum for the GRE subject test in mathematics.
phcooh
Posts: 4
Joined: Mon Oct 13, 2008 7:15 pm

Calculus problem

Postby 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?

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm

Postby 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




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