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sole12 - ANALYSIS EXERCISE 12SOLUTIONS 1 Since the function...

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ANALYSIS EXERCISE 12–SOLUTIONS 1. Since the function f is continuous on [ a, b ] there are { x 1 , x 2 } ⊂ [ a, b ] such that f ( x 1 ) = m, f ( x 2 ) = M . If m = M then f is a constant function, and the result is vacuously true (as ( m,M ) = .) Otherwise set x 0 = min { x 1 , x 2 } , x 00 = max { x 1 , x 2 } and consider f on the interval [ x 0 , x 00 ]. The statement now follows from the intermediate-value theorem. 2. lim h 0 f ( a + h ) + g ( a + h ) - f ( a ) - g ( a ) h = lim h 0 ± f ( a + h ) - f ( a ) h + g ( a + h ) - g ( a ) h ² = f 0 ( a ) + g 0 ( a ) . 3. Let n = 1. Then f ( x ) = x , and f 0 ( x ) = 1. Suppose that the statement is true for n = k , i.e. f k ( x ) = x k is differentiable and f 0 k ( a ) = ka k - 1 . Let n = k + 1. Then f k +1 = x k +1 = x · f k ( x ) is differentiable as a product of two differentiable functions and f 0 k +1 ( a ) = 1 · f k ( a ) + a · f 0 k ( a ) = a k + a · ka k - 1 = ( k + 1) a k . 4. Assume that
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