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The function g x is continuous hence by the

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The function g ( x ) is continuous, hence by the Intermediate Value Theorem there is a a (0 , 1) such that g ( a ) = 3 / 5. By Rolle’s theorem there is an x ( a, 2) such that g 0 ( x ) = 0. Then f 0 ( x ) = 1 / 5. 5
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Problem 6. 30 points. Consider the sequence of functions f n ( x ) = x - x n defined on [0 , 1]. 1. Show that f n ( x ) converges pointwise to a function f ( x ) on [0 , 1], and find a formula for f ( x ). 2. Does f n ( x ) converge to f ( x ) uniformly on [0 , 1]? Justify. 3. Does f n ( x ) converge to f ( x ) uniformly on [0 , a ] for a < 1? Justify. Solution 1. We have f ( x ) = lim f n ( x ) = x, x < 1 0 , x = 1 2. The limit function f ( x ) is not continuous, hence the convergence is not uniform. 3. On [0 , a ], the function f ( x ) - f n ( x ) = x n increases from 0 to a n . Hence lim sup x [0 , 1 / 2] | f n ( x ) - f ( x ) | = lim a n = 0 , so the convergence is uniform. 6
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Problem 7. 30 points. Suppose that f ( x ) is a continuous function on [ a, b ], suppose that f ( x ) 0 for all x [ a, b ], and suppose that Z b a f ( x ) dx = 0 . Prove that f ( x ) = 0 for all x [ a, b ]. (Hint: suppose that f ( c ) 6 = 0 for some c [ a, b ], and draw a picture). Proof. Suppose that f ( c ) > 0 for some c [ a, b ]. Then for any ε > 0 there exists a δ > 0 such that | x - c | < δ, x [ a, b ] implies | f ( x ) - f ( c
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