asst2-f10-solns

asst2-f10-solns - MATH 137 Solutions to Assignment #2, Part...

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MATH 137 Fall 2010 Solutions to Assignment #2, Part II 1. (a) When a = b = c = 1 and d = 3, f ( x ) = x + 1 x + 3 . Then, y = x + 1 x + 3 y ( x + 3) = ( x + 1) xy + 3 y = ( x + 1) xy - x = 1 - 3 y x ( y - 1) = 1 - 3 y x = 1 - 3 y y - 1 Therefore, the inverse function is f - 1 ( x ) = 1 - 3 x x - 1 . (b) Repeating the process from part (a) for the general function, y = ax + b cx + d y ( cx + d ) = ( ax + b ) cxy + dy = ( ax + b ) cxy - ax = b - dy x ( cy - a ) = b - dy x = b - dy cy - a Therefore in general, the inverse function is f - 1 ( x ) = b - dx cx - a . Alternatively, we could have completed part (b) first, then used the general inverse function above to answer part (a) by simply substituting a = b = c = 1 and d = 3. (c) If f ( x 1 ) = f ( x 2 ), then ax 1 + b cx 1 + d = ax 2 + b cx 2 + d ( ax 1 + b )( cx 2 + d ) = ( ax 2 + b )( cx 1 + d ) acx 1 x 2 + adx 1 + bcx 2 + bd = acx 1 x 2 + adx 2 + bcx 1 + bd adx 1 + bcx 2 = adx 2 + bcx 1 adx 1 - adx 2 + bcx 2 - bcx 1 = 0 ad ( x 1 - x 2 ) - bc ( x 1 - x 2 ) = 0 ( ad - bc )( x 1 - x 2 ) = 0 Thus if ad = bc , then f ( x 1 ) = f ( x 2 ) regardless of what x 1 and x 2 equal and hence f ( x ) would not be one-to-one. For f ( x ) to be one-to-one, it is necessary that x 1 = x 2 when f ( x 1 ) = f ( x 2 ), and therefore we must restrict a , b , c , and d so that ad 6 = bc .
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2. (a) In order for x to be in the domain of f , we must have x 2 + 1 0, which is true for all x , and we also must have that x + x 2 + 1 > 0 in order for ln to be defined. Notice that x 2 + 1 > x 2 0 for all x R . Therefore, x 2 + 1 > x 2 = | x | , since the square root function is strictly increasing on its domain. So we have that
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asst2-f10-solns - MATH 137 Solutions to Assignment #2, Part...

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