5333-Solns-Assignment10-Sp13 - 5333 Analysis Solutions Assignment 10 Section 30 1(a True This is Theorem 30.1(b False Take for example a function that

# 5333-Solns-Assignment10-Sp13 - 5333 Analysis Solutions...

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Unformatted text preview: 5333: Analysis Solutions, Assignment # 10 Section 30 1. (a) True. This is Theorem 30.1. (b) False. Take, for example, a function that is increasing on [ a, b ] but is not continuous. Here’s a specific example: f ( x ) = x, ≤ x < 1 x + 1 , 1 ≤ x ≤ 2 f is increasing on [0 , 2] and f is discontinuous at x = 1. (c) True. This is Theorem 30.4(b). 5. Let f be continuous on [ a, b ]. Suppose that f ( x ) ≥ 0 for all x ∈ [ a, b ] and suppose that there is a point c ∈ [ a, b ] such that f ( c ) > 0. Suppose, first, that c ∈ ( a, b ). Then, since f is continuous, there is a positive number α and interval ( c- δ, c + δ ) such that f ( x ) ≥ α on ( c- δ, c + δ ). (See Exercise 21.13). Now let P be the partition { a, c- δ, c + δ, b } . Then L ( f, P ) ≥ α · 2 δ > 0. Therefore, L ( f ) > 0 from which it follows that integraldisplay b a f ( x ) dx > 0. Next, suppose that c = a . Then there is a positive number δ and a positive number α such that f ( x ) ≥ α for x ∈ [ a, a + δ ). Let P be the partition { a, a + δ, b } . Then, as above, L ( f, P ) ≥ α · δ > 0. Therefore, L ( f ) > 0 from which it follows that integraldisplay b a f ( x ) dx > 0. Finally, if c = b , then there is a positive number δ and a positive number α such that f ( x ) ≥ α for x ∈ ( b- δ, b ]. Again, we conclude that integraldisplay b a f ( x ) dx > 0. 6. Suppose that f ( c ) negationslash = 0 for some c ∈ ( a, b ). Since f is continuous on [ a, b ] there is a positive number α and a neighborhood N ( c ) = ( c- δ, c + δ ) such that f ( x ) ≥ α on N ( c ). 1 Define g on [ a, b ] by g ( x ) = , a ≤ x < c- δ f ( x ) , c- δ ≤ x ≤ c + δ , c + δ < x ≤ b Then, g is integrable on [ a, b ] since it is piecewise continuous, and a b f ( x ) g ( x ) dx = integraldisplay c + δ c- δ f 2 ( x ) dx ≥ 2 δα 2 > 0, a contradiction....
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