# What do we find infimum of exp(-x) without graph?

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#1
I know it is 0 but how to find it without graph?
0
2 years ago
#2
(Original post by Infinity...)
I know it is 0 but how to find it without graph?
You could show its a decreasing monotonic function (-ive gradient), so the minimum value must be when ...
0
#3
(Original post by mqb2766)
You could show its a decreasing monotonic function (-ive gradient), so the minimum value must be when ...
yeah it wil be its limit point which is 0. May i know how
to find supremum without using graph?
0
2 years ago
#4
Just use the definition, you know exp(-x) > 0. If e > 0 is a lower bound then take x = ln(2/e) so that exp(-x) = e/2. So e > 0 can't be a lower bound. Hence...
0
#5
(Original post by Zacken)
Just use the definition, you know exp(-x) > 0. If e > 0 is a lower bound then take x = ln(2/e) so that exp(-x) = e/2. So e > 0 can't be a lower bound. Hence...
hence 0 is lower bound. And how do we find supremum of exp(-x)?
0
2 years ago
#6
(Original post by Infinity...)
hence 0 is lower bound. And how do we find supremum of exp(-x)?
No. Hence 0 is the infimum...

There is no supremum of exp(-x) over R. Do you mean some other set?
0
2 years ago
#7
(Original post by Infinity...)
yeah it wil be its limit point which is 0. May i know how
to find supremum without using graph?
Opposite direction for x? Again use the fact its monotonic.
0
#8
(Original post by mqb2766)
Opposite direction for x? Again use the fact its monotonic.
So is it not bounded above because it is monotonic?
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#9
(Original post by Zacken)
No. Hence 0 is the infimum...

There is no supremum of exp(-x) over R. Do you mean some other set?
Yes, there.is no upper bound of exp(-x). Is there some way to prove this or it is just the nature of this function?
0
2 years ago
#10
(Original post by Infinity...)
So is it not bounded above because it is monotonic?
No, it's not bounded above because it's not bounded above. Not because it's monotonic.
1
2 years ago
#11
(Original post by Infinity...)
Yes, there.is no upper bound of exp(-x). Is there some way to prove this or it is just the nature of this function?
Why can't you just go back to the definition? If A > 0 is an upper bound of exp(-x) then just pick x = - ln(2A) so that exp(-x) = 2A > A. So there is no upper bound for exp(-x).
1
2 years ago
#12
(Original post by Infinity...)
So is it not bounded above because it is monotonic?
Yes (assuming x -> -inf).
For a bit more thorough analysis you need to show the gradient doesn't go to zero as well (which happens when x-> inf) as otherwise it would converge to some value
But if you have a monotonic function, the min max values must occur at the ends?
1
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