Share
Explore BrainMass

Maximum-Minimum Theorem : Continuity and Bounded Functions

(d) Does the conclusion of the Maximum-Minimum Theorem always hold for a bounded function f : R --> R that is continuous on R? Prove or give a counterexample.
(a) Fix a, b E R, a < b. Prove that if f [a, b] -->R is continuous on [a, b] and f(x)&#8800;0 for all x E [a, b], then 1/f(x) is bounded on [a, b].
(b) Find a, b E R, a < b, and a function f: (a, b) --> R that is continuous on (a, b), such that f(x)&#8800;0 for all x E (a, b), but 1/f(x)is not bounded on (a, b).

Recall:
Max-Min theorem:
Let f: [a, b] -> R (real numbers) be continuous on [a, b]. Then f has an absolute maximum and an absolute minimum on [a, b]

Please see the attached file for the fully formatted problems.

Attachments

Solution Preview

Max-Min theorem
--------------------------------------------------------------------------------
Recall:
Max-Min theorem:
Let f: [a, b] -> R (real numbers) be continuous on [a, b]. Then f has an absolute maximum and an absolute minimum on [a, b]

1 (d) Does the conclusion of the Max-Min Theorem always hold for a *bounded* function f:R ->R that is continuous on R? Prove or give a counterexample.

No, the theorem only holds on *closed* sets, and R is open.
Take for example y = arctan x
which may also be written y = tan^(-1) x
whichever notation you prefer.
This function is definitely ...

Solution Summary

The Maximum-Minimum Theorem is used to prove Continuity with regard to Bounded Functions. The solution is detailed and well presented. The response received a rating of "5" from the student who originally posted the question.

$2.19