FA06ma.9H6solutions

# FA06ma.9H6solutions - Math 1a Section 1 Solutions to...

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Math 1a, Section 1 Solutions to Homework 6 1. 4.6.4, 4.6.6, 4.6.9 (p167) 1(a) f 0 ( x ) = 4 x 3 + cos x 1(b) f 0 ( x ) = - 1 ( x +1) 2 1(c) f 0 ( x ) = - - 1 2+cos x ( - sin x ) 2. 4.6.18, 4.6.20, 4.6.23 (p167-168) 2(a) f 0 ( x ) = 1 2 x 2(b) f 0 ( x ) = 1 / 2 x - 1 / 2 + 1 / 3 x - 2 / 3 + 1 / 4 x - 3 / 4 2(c) f 0 ( x ) = 1 1+ x - x (1+ x ) 2 1 2 x 3. 4.6.37 (p168) g 0 ( x ) = (2 ax + b ) sin x +( ax 2 + bx + c ) cos x +(2 dx + e ) cos x - ( dx 2 + ex + f ) sin x We compare coeﬃcients x sin x 2 a + e = 0 sin x b - f = 0 x cos x b + 2 d = 0 cos x c + e = 0 x 2 cos x a = 0 x 2 sin x - d = 1 So the solution is: a = 0 , b = 2 , c = 0 , d = - 1 , e = 0 , f = 2. 4. 4.9.9, 4.9.11 (p173) 4(a) For x < c , f 0 ( x ) = 2 x . For x > c , f 0 ( x ) = a . In order for f 0 ( c ) to exist, we must have f being continuous at c and the derivative satisﬁes the following relation lim x c - f 0 ( x ) = lim x c + f 0 ( x ) Hence, ac + b = c 2 and 2 c = a , this means b = - c 2 .

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4(b) For x < c , f 0 ( x ) = cos x . For x > c
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## This note was uploaded on 12/15/2009 for the course MA 1a taught by Professor Borodin,a during the Spring '08 term at Caltech.

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FA06ma.9H6solutions - Math 1a Section 1 Solutions to...

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