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If x -> 0+x -> 0+e^(1/x) -> ∞ (1/0+) -> ∞In summary, the local minimum of the function y=x(e^1/x) is at (1,e) but the absolute minimum is not at this point because the function does not have a defined absolute minimum due to its behavior at the extremes of its domain. As x approaches negative infinity, the function approaches 0, while as x approaches 0, the function approaches infinity. This makes it impossible for the function to have an absolute minimum.

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nickeisenberg said:

Hey nickeisenberg and welcome to the forums.

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

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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

A local minimum is a point on a graph where the function reaches its lowest value within a specific interval. It is lower than all other points immediately surrounding it, but not necessarily the lowest point on the entire graph.

A global minimum is the lowest point on the entire graph, while a local minimum is only the lowest point within a specific interval. A global minimum is also a local minimum, but a local minimum is not always a global minimum.

To find the local minimum of a function, we can take the derivative of the function and set it equal to 0. This will give us the critical points, which are potential local minimums. We then use the second derivative test or graphing techniques to determine which critical point is the local minimum.

To determine if (1,e) is a local minimum, we need to look at the points immediately surrounding it. However, at (1,e), the function is not defined since we cannot divide by 0. Therefore, (1,e) cannot be a local minimum for this function.

Yes, a function can have multiple local minimums. This occurs when the graph has multiple intervals where the function reaches its lowest point within that interval. These local minimums can be equal or different values.

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