128ahw1sum10 - MATH 128A, SUMMER 2010, HOMEWORK 1 SOLUTION...

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Unformatted text preview: MATH 128A, SUMMER 2010, HOMEWORK 1 SOLUTION BENJAMIN JOHNSON Homework 1: Due Monday, June 28 1.1; 1b, 4b, 6, 9abcd, 21, 25abc 1.2; 1b, 3c, 5a, 10ab, 15ab section 1.1 1. Show that the following equations have at least one solution in the given intervals. b. ( x- 2) 2- ln x = 0, [1 , 2] and [ e, 4] solution: Let f ( x ) = ( x- 2) 2- ln x . We have f (1) = 1 and f (2) =- ln 2. Zero is between 1 and- ln 2, so by the intermediate value theorem there is a real number c [1 , 2] with f ( c ) = 0. For the interval [ e, 4], we have f ( e ) = ( e- 2) 2- 1 and f (4) = 2 2- ln 4. Since e- 2 < 1, we have f ( e ) < 0 and since ln 4 < 4 we have f (4) > 0. Again the desired result follows from the intermediate value theorem. 4. Find max a x b | f ( x ) | for the following functions and intervals. b. f ( x ) = (4 x- 3) / ( x 2- 2 x ), [0 . 5 , 1] solution: The denominator is non-zero on [0 . 5 , 1], but the numerator changes sign in this interval. Because of the absolute value we need to check the point at which the numerator changes sign as a possible extremum, along with the endpoints of the interval, and any points in the interval where the derivative might be zero. For the derivative, we have f ( x ) = 4( x 2- 2 x )- (4 x- 3)(2 x- 2) ( x 2- 2 x ) 2 = 4 x 2- 8 x- 8 x 2 + 8 x + 6 x- 6 ( x 2- 2 x ) 2 =- 4 x 2 + 6 x- 6 ( x 2- 2 x ) 2 We see f is always negative on [0 . 5 , 1]. The numerator of f is zero at x = 3 / 4 but zero is the absolute minimum for the function because it is always non-negative; so the max must be at an extreme point. We have f (0 . 5) =- 1- . 75 = 4 / 3, and f (1) = | 1 /- 1 | = 1; so the max of f on this interval is f (0 . 5) = 4 / 3. 6. Suppose f C [ a,b ] and f ( x ) exists on ( a,b ). Show that if f ( x ) 6 = 0 for all x ( a,b ), then there can exist at most one number p in [ a,b ] with f ( p ) = 0. solution: We argue by contradiction. Suppose to the contrary that there are two points p,q [ a,b ] with f ( p ) = f ( q ) = 0. Since f is differentiable on ( a,b ), Rolles theorem implies that there exists a real number c in ( a,b ) with f ( c ) = 0. But this contradicts our hypothesis Date : June 28, 2010....
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128ahw1sum10 - MATH 128A, SUMMER 2010, HOMEWORK 1 SOLUTION...

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