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## Main Question or Discussion Point

I found that the local minimum of y=x(e^1/x) is (1,e) but can someone tell me why the absolute minimum is also not (1,e) and can someone tell me why this function does not have an absolute minimum? Thanks

- Thread starter nickeisenberg
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I found that the local minimum of y=x(e^1/x) is (1,e) but can someone tell me why the absolute minimum is also not (1,e) and can someone tell me why this function does not have an absolute minimum? Thanks

- #2

chiro

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Hey nickeisenberg and welcome to the forums.I found that the local minimum of y=x(e^1/x) is (1,e) but can someone tell me why the absolute minimum is also not (1,e) and can someone tell me why this function does not have an absolute minimum? Thanks

Just to clarify, what is the domain of your function?

- #3

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all reals number except 0

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y= x (e^1/x)

Think about the two components.

If x -> -∞

x -> -∞

e^(1/x) -> 1 (1/-∞) ~ 0, e^0 = 1

Think about the two components.

If x -> -∞

x -> -∞

e^(1/x) -> 1 (1/-∞) ~ 0, e^0 = 1

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