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6 September, 11:15

Describe a way of showing that a continuous function on n open interval (a b,) has a minimum value.

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  1. 6 September, 11:37
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    See explanation below

    Step-by-step explanation:

    In fact, the interval has to be closed (just as [a, b], for b>a). Otherwise, this is not neccesarily true. For example, consider the function f (x) = - 1/x defined on the open interval (0.1). f is continuous (quotient of continuous functions) but it does not have a minimum value: it decreases infinitely near zero.

    To show this result on the interval [a, b], the idea is the following:

    We can use a previous theorem. If f is continuous on [a, b], there exists some N>0 such that N≤f (x) (that is, f is bounded below). Now, we take the biggest N such that N≤f (x) for all x∈[a, b] (this is known as the greatest lower bound)

    The number N is the candidate for the minimum value of f. Next, we have to show that there exists some p∈[a, b] such that f (p) = N. To do this, we must use the continuty of f on [a, b]. There are many ways to do it, and usually they require the epsilon-delta definition of continuity.

    This is just a description of the ideas involved, but of course, a rigorous proof would need more technical details, depending on the theorems you are allowed to use.
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