Question

Find the range of the function \(f(x) = x^\frac{1}{x}\)

Answer 1

i think
[-infinity,infinity)

Answer 2

Domain: (-∞,1) U (1,∞)
Range: (-∞,0) U (0,∞)

Answer 3

y= x^(1/x)
log y = x/logx
y'/y = (1/logx) - 1
y' = y( (1/logx) -1)
since y>0,
1/logx - 1 =0 (for finding the extremity)
this yield x=e
hence maxima or minima occurs at x=e or at (e,e^1/e)

Answer 4

It can be easily seen that it'd be a point of maxima since
1) it is the only extremity of the graph
2) 2<e and 4>e ; (2^1/2) = (4^1/4)
Hence the graph has the same heights at both x=2 and 4. It must have come up till e^1/e and then again fallen down.

Answer 5

Hmmm............ I am not sure if you have the second line of your first comment right. I think it should be
y= x^(1/x)
log y = (1/x) log x
log y = (log x) / x

Answer 6

y= x^(1/x)
logy = (logx)/x
y' / y = (1-logx) / x^2
y' = y [ (1-logx) / x^2 ]
Set y' = 0, it is known that y > 0,
y' =0
y [ (1-logx) / x^2 ] =0
(1-logx) / x^2 =0
x = e
y = e^(1/e) when x=e.
Then applies your argument here :D
Thanks for the idea! I forgot we can use calculus here!!

Answer 7

oops :P
i miscalculated the derivative :P

Answer 8

I guess the first problem was you made a mistake when you took log on both sides?!

Answer 9

Right. Sorry about that. That was silly of me

Answer 10

It's alright! Human errs :) Thanks a lot again for your kind help! :D

Answer 11

Glad if I could help ^^