Ask Question
Today, 02:48

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

+4
Answers (1)
  1. Today, 05:25
    0
    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.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Describe a way of showing that a continuous function on n open interval (a b,) has a minimum value. ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers