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I don't understand the question, actually. When they say "the minumum value," are they asking for a value of f(x)? So the answer would be 1^1 = 1? or 0.5^0.5 = 0.707? I'm confused.

2007-09-02 07:12:35 · 6 answers · asked by abc123 1 in Science & Mathematics Mathematics

6 answers

If you drew the graph you would see an "upward parabolic shape" with the bottom point at (0.368, 0.692)
Yes, they are asking for the lowest f(x).

2007-09-02 07:36:41 · answer #1 · answered by ? 5 · 0 0

Yes, what they're asking is the possible minimum value for x.
The minimum possible value of f(x) is 0. But they want x > 0.
f'(x)=2x=0
You can take any number, as small as you want that is greater than 0. There is no such fixed number as you can always find a number smaller than that.

2007-09-02 14:36:09 · answer #2 · answered by cidyah 7 · 0 0

To find the minimum you set the first derivative equal to 0.
For f(x)=x^x, f'(x)=[1+ln x]*x^x and for minimum f'(x)=0 This occurs when
1+ln(x)=0
ln x=-1
x=1/e hence the minimum of f(x) is (1/e)^(1/e)

2007-09-02 15:36:26 · answer #3 · answered by marcus101 2 · 0 0

The derivative of x^x = (lnx +1)x^x
Set this to 0 getting lnx+1 = 0 --> x = 1/e
(1/e)^(1/e) = .6922 is the approximate min value

2007-09-02 14:29:07 · answer #4 · answered by ironduke8159 7 · 0 0

The answer is: X>1-.05(0+f)=1/1<3XABC=7gfr54t 6gfyd783l9i'manidiot.

Be sure to remember this because it will make you a lot of money when you can't find a use for it..

2007-09-02 14:21:45 · answer #5 · answered by Anonymous · 0 0

f(x) = e^(xlnx)
f'(x) = x^x(lnx + 1) = 0
x = e^(-1)
The minimum of f(x)
= f(e^(-1))
= (1/e)^(1/e)
= 0.6922

2007-09-02 14:17:18 · answer #6 · answered by sahsjing 7 · 2 0

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