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Solutions Test 2B - Solutions to Test 2 Version B Math 1501...

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Solutions to Test 2, Version B Math 1501, Fall 02. WG * October 17, 2002 Exercise 1. (a) Assume that f ( x ) = 5 - x. Then f (1+ h ) - f (1) h = 4 - h - 4 h . We multiply and divide the right handside of the equality by the conjugate of 4 - h + 4 to obtain that f (1 + h ) - f (1) h = ( 4 - h - 4)( 4 - h + 4) h ( 4 - h + 4) = - h h ( 4 - h + 4) . Simplifying by h, we deduce that f 0 (1) = lim h 0 f (1 + h ) - f (1) h = lim h 0 - 1 4 - h + 4 = - 1 4 . (b) If f ( x ) = x + 1 if x ≤ - 1 ( x + 1) 2 if x > - 1 then f 0 - ( - 1) = ( x + 1) 0 | x = - 1 = 1 , f 0 + ( - 1) = [( x + 1) 2 ] 0 | x = - 1 = 2( x + 1) | x = - 1 = 0 . Since f 0 - ( - 1) 6 = f 0 + ( - 1) we deduce that f 0 ( - 1) does not exist. Exercise 2. Call Δ c the tangent line at c to the graph of x f ( x ) = x 3 - 3 x. Then the equation of Δ c is Δ c : y - f ( c ) = f 0 ( c )( x - c ) , * School of Mathematics, Georgia Institute of technology. 1
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and f 0 ( x ) = 3 x 2 - 3 . Let Δ be given by the equation 5 y - 3 x - 8 = 0 , or equivalently, y = 3 x 5 + 8 5 . Then Δ is perpendicular to Δ c if and only if 3 5 f 0 ( c ) = - 1 i.e. 3 c 2 - 3 = f 0 ( c ) = - 5 3 . We conclude that c 2 = 4 / 9 and so, c = 2 / 3 or c = - 2 / 3 . Exercise 3. Assume that x ( t ) = t + 5 t +2 is the position at each time t 0 of the object. Then its velocity is x 0 ( t ) = 1 - 5 ( t +2) 2 and so, x 0 ( t ) = ( t + 2 - 5)( t + 2 + 5) ( t + 2) 2 . On the set [0 , + ) x 0 changes sign only at time t =
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