ex2sp02s

Solution let f x = x ln x 1 note that for any x ≥ 0

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Unformatted text preview: Solution: Let f ( x ) = x- ln( x + 1). Note that for any x ≥ 0, f ( x ) = 1- 1 / ( x + 1) ≥ 0. Thus f is increasing on the interval [0 , ∞ ). Thus if x ≥ 0, then f ( x ) ≥ f (0). Noting that f (0) = 0, this is exactly the inequality we had to prove. 2 5. (15 pts) Suppose [ a, b ] is a closed, bounded, nondegenerate interval. Is the following statement true? For any continuous function f : [ a, b ] → R , the function | f | , defined by | f | ( x ) = | f ( x ) | , is integrable on [ a, b ] . Briefly justify your answer. Solution: Yes, since f is continuous on [ a, b ], so is | f | , hence, by theorem 5.10, | f | is integrable on [ a, b ]. 2 6. (20 pts) Suppose f : [ a, b ] → R is continuous and increasing. Prove that sup f ( E ) = f (sup E ) for every nonempty set E ⊆ [ a, b ]. Solution: Let E ⊆ [ a, b ]. Thus E is bounded, so s = sup E is finite and s ∈ [ a, b ]. Because f is increasing, for any x ∈ [ a, b ], we have f ( a ) ≤ f ( x ) ≤ f ( b ). In particular f ( E ) is bounded, so sup f ( E ) is finite. Again because f is increasing and s ≥ x, ∀ x ∈ E , we have f ( s ) ≥ f ( x ) ∀ x ∈ E . Thus, f ( s ) is an upper bound for f ( E ), so we get f (sup E ) ≥ sup f ( E ). To obtain the other inequality, let x n be a sequence of elements from E such that x n → sup E (such a sequence exists, by the approximation property of the supremum). Then sup f ( E ) ≥ f ( x n ), for any n ∈ N (because x n ∈ E ). Taking the limit on n in this inequality and using the fact that f is continuous, we get sup f ( E ) ≥ f ( s ) = f (sup E ). 2...
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Solution Let f x = x ln x 1 Note that for any x ≥ 0 f x =...

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