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Bounded on a b prove that f is uniformly continuous

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bounded on ( a, b ). Prove that f is uniformly continuous on ( a, b ). Solution: By assumption M > 0 such that | f ( x ) | ≤ M , for any x ( a, b ). Let > 0 and take δ = /M . Let y, z ( a, b ) such that | z - y | < δ . By the mean value theorem applied to f on the interval between y and z , there exists c such that f ( z ) - f ( y ) = f ( c )( z - y ). Thus | f ( z ) - f ( y ) | = | f ( c ) || z - y | ≤ M | z - y | < Mδ = . Since this is true for any y, z such that | z - y | < δ , it follows that f is uniformly continuous on ( a, b ). (b) (10 pts) Give an example to show that if the hypothesis f bounded on ( a, b ) is omitted, then the statement of (a) is no longer true. Solution: Let f : (0 , 1) R , defined by f ( x ) = ln x . f is differentiable on (0 , 1), and its derivative f ( x ) = 1 /x is clearly not bounded on (0 , 1). The function f is not uniformly continuous because it cannot be extended by continuity at 0 (the limit of f as x 0 + is -∞ ). 4. (15 pts) Show that ln( x + 1) x , for all x 0. 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.
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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.
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