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FA06ma.9Hw7solutions

# FA06ma.9Hw7solutions - does then it must pass through zero...

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Math 1a, Section 1 Solutions to Homework 7 1. 4.12.7, 4.12.10, 4.12.14 (p179) See solutions in the book. 2. 4.12.21 (p180) Refer to the diagram below: Note that D 2 = s 2 + 8 2 . So we differentiate both sides with respect to t and obtain 2 D dD dt = 2 s ds dt (1) At a particular time t , we have D = 10, dD dt = - 4 (which means the plane is moving to the left in the diagram) and s = 10 2 - 8 2 = 6. Hence the velocity is ds dt = - 20 3 . 3. 4.12.31 (p181) We were given x 1 / 2 + y 1 / 2 = 5. We differentiate with respect to x . In the given range 0 < x < 5, 1 2 x = y 2(5 - x ) (2)

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Therefore, y = 5 x - 1. Note that y = 0 implies x = 25, which is beyond the given range of x . Hence y = 0 within the given range and by the intermediate value theorem it cannot change sign (because if it
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Unformatted text preview: does, then it must pass through zero). 4. 4.15.2 (p186) Let f ( x ) = x 3-3 x + b . Suppose x 1 , x 2 are zeros of f ( x ) in the interval-1 ≤ x ≤ 1. Then by the Rolle’s theorem, there exists a point c in ( x 1 , x 2 ) such that f ( c ) = 0. But f ( x ) = 3 x 2-3, so f = 0 only when x = ± 1, contradiction. 5. 4.21.4 (p194) Fix y , let f ( x ) = x 2 + y 2 where x, y > 0. Observe that f ( x ) = x 2 + ( S-x ) 2 , hence f ( x ) = 2 x-2( S-x ). Hence f ( x ) = 0 only when x = S-x = y . Finally, note that f 00 ( x ) = 4 > 0, hence f attains its minimum when x = y ....
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