problemset6s

# problemset6s - MAT 137Y 2008-09 Winter Session Solutions to...

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MAT 137Y 2008-09 Winter Session, Solutions to Problem Set 6 1 (SHE 4.1) 12. We have f ( x ) = x 2 / 3 - 1. (a) f 0 ( x ) = 2 3 x - 1 / 3 has no zeroes in ( - 1 , 1 ) . (b) f 0 ( 0 ) clearly does not exist, so f is not differentiable at 0, and as such we can not apply Rolle’s Theorem since we require differentiability for all x ( - 1 , 1 ) . Hence this does not contradict Rolle’s Theorem. 24. Consider the function f ( x ) = 6 x 5 + 13 x + 1, which is continuous and differentiable for all x since it is a polynomial. Since f ( - 1 ) = - 18 < 0 < 1 = f ( 0 ) , by the Intermediate Value Theorem there exists c ( - 1 , 0 ) such that f ( c ) = 0, so f ( x ) has at least one root. Now suppose c 1 and c 2 are distinct real roots (where c 1 < c 2 ). Then f ( c 1 ) = f ( c 2 ) , so by Rolle’s Theorem there exists γ ( c 1 , c 2 ) such that f 0 ( γ ) = 0. But f 0 ( x ) = 30 x 4 + 13 > 0 for all x R , so this contradicts the assumption that there exist distinct real roots. Hence f ( x ) has exactly one real root. 26. For part (a), suppose that f has two zeroes x 1 , x 2 ( a , b ) . f is differentiable on ( x 1 , x 2 ) and continuous on [ x 1 , x 2 ] . By Rolle’s Theorem, f 0 has a zero in ( x 1 , x 2 ) which contradicts the hy- pothesis. For part (b), if f had three zeroes in ( a , b ) , then by Rolle’s Theorem, f 0 would have at least two zeroes in ( a , b ) and f 00 would have at least one zero in ( a , b ) which contradicts the hypothesis. 38. We prove for all a , b R that | cos b - cos a | ≤ | b - a | . If a = b the proof is obvious. Otherwise, assume without loss of generality that a < b . As f ( x ) = cos x is continuous and differentiable for all x , then by the Mean Value Theorem, there exists c ( a , b ) such that f 0 ( c ) = f ( b ) - f ( a ) b - a = ⇒ - sin c = cos b - cos a b - a = ⇒ |- sin c | = ± ± ± ± cos b - cos a b - a ± ± ± ± . But |- sin c | ≤ 1 for all c , so ± ± ± ± cos b - cos a b - a ± ± ± ± 1 = ⇒ | cos b - cos a | ≤ | b - a | . 4 (SHE 4.2) 30. Since d dx ( - 1 4 x - 4 - 25 4 x 4 / 5 ) = f 0 ( x ) , it follows that f ( x ) = - 1 4 x - 4 - 25 4 x 4 / 5 + C . But f ( 1 ) = 0, so 0 = - 1 4 - 25 2 + C , so C = 13 2 . Thus, f ( x ) = - 1 4 x - 4 - 25 4 x 4 / 5 + 13 2 , x > 0. 56. (a) Suppose f 0 ( x ) > g 0 ( x ) for all x ( 0 , c ) and f ( 0 ) = g ( 0 ) . Now consider the function F ( x ) = f ( x ) - g ( x ) . Then F 0 ( x ) = f 0 ( x ) - g 0 ( x ) > 0 for all x ( 0 , c ) , so F is increasing on ( 0 , c ) . Since F ( 0 ) = f ( 0 ) - g ( 0 ) = 0, it follows that F ( x ) > 0 on ( 0 , c ) . Therefore f ( x ) - g ( x ) > 0 or f ( x ) > g ( x ) on ( 0 , c ) . (b) Let

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problemset6s - MAT 137Y 2008-09 Winter Session Solutions to...

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