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problemset5s

# problemset5s - MAT 137Y 2008-09 Winter Session Solutions to...

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MAT 137Y 2008-09 Winter Session, Solutions to Problem Set 5 1 (a) Suppose f ( x ) = g ( x + c ) . Using the definition of derivative, g 0 ( x + c ) = lim h 0 g ([ x + h ]+ c ) - g ( x + c ) h = lim h 0 f ( x + h ) - f ( x ) h = f 0 ( x ) . (b) Suppose f ( x ) = g ( cx ) for some constant c . Then f 0 ( x ) = lim h 0 f ( x + h ) - f ( x ) h = lim h 0 g ( c [ x + h ]) - g ( cx ) h = lim h 0 g ( cx + ch ) - g ( cx ) h = lim t 0 g ( cx + t ) - g ( cx ) t / c = c lim t 0 g ( cx + t ) - g ( cx ) t = c · g 0 ( cx ) , where we’ve used the substitution t = ch . (c) If f is periodic with period a , then f ( x ) = f ( x + a ) . By part (a), since f is differentiable it follows that f 0 ( x ) = f 0 ( x + a ) , so f 0 is periodic with period a . (d) Let f ( x ) = 1 + x 1 - x . Then f 0 ( x ) = lim h 0 1 +( x + h ) 1 - ( x + h ) - 1 + x 1 - x h = lim h 0 ( 1 + x + h )( 1 - x ) - ( 1 + x )( 1 - x - h ) h ( 1 - x - h )( 1 - x ) = lim h 0 1 - x + x - x 2 + h - hx - ( 1 - x - h + x - x 2 - xh ) h ( 1 - x - h )( 1 - x ) = lim h 0 2 h h ( 1 - x - h )( 1 - x ) = lim h 0 2 ( 1 - x - h )( 1 - x ) = 2 ( 1 - x ) 2 . (e) Let f ( x ) = 1 + x . Then f 0 ( 3 ) = lim h 0 f ( 3 + h ) - f ( 3 ) h = lim h 0 4 + h - 2 h = lim h 0 ( 4 + h - 2 )( 4 + h + 2 ) h ( 4 + h + 2 ) = lim h 0 ( 4 + h ) - 4 h ( 4 + h + 2 ) = lim h 0 h h ( 4 + h + 2 ) = lim h 0 1 ( 4 + h + 2 ) = 1 4 . Alternatively, f 0 ( 3 ) = lim x 3 f ( x ) - f ( 3 ) x - 3 = lim x 3 1 + x - 2 x - 3 = lim x 3 ( 1 + x - 2 )( 1 + x + 2 ) ( x - 3 )( 1 + x + 2 ) = lim x 3 ( 1 + x ) - 4 ( x - 3 )( 1 + x + 2 ) = lim x 3 x - 3 ( x - 3 )( 1 + x + 2 ) = lim x 3 1 1 + x + 2 = 1 4 . 2 (i) (SHE 3.2 54) Suppose f ( x ) = Ax 3 + Bx 2 + Cx + D , so f 0 ( x ) = 3 Ax 2 + 2 Bx + C . The curve intersects the point ( 1 , 0 ) , so 0 = A + B + C + D . Furthermore, the slope of the tangent line at ( 1 , 0 ) is 3, so f 0 ( 1 ) = 3, hence 3 = 3 A + 2 B + C . The curve also intersects ( 2 , 9 ) where the slope of the tangent line is 18. Therefore, 9 = 8 A + 4 B + 2 C + D and 18 = 12 A + 4 B + C . We thus have four equations and four unknowns; solving the system gives us A = 3, B = - 6, C = 6, and D = - 3.

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problemset5s - MAT 137Y 2008-09 Winter Session Solutions to...

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