Solutions Test 2A

# Solutions Test 2A - Solutions to Test 2 Version A Math 1501...

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Math 1501, Fall 02. WG * October 5, 2002 Exercise 1. (a) Assume that f ( x )= 6 - x. Then f (2+ h ) - f (2) h = 4 - h - 4 h . We multiply and divide the right handside of the equality by the conjugate of 4 - h + 4toobta in that f (2 + h ) - f (2) h = ( 4 - h - 4)( 4 - h + 4) h ( 4 - h + 4) = - h h ( 4 - h + 4) . Simplifying by h, we deduce that f 0 (2) = lim h 0 f (2 + h ) - f (2) 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 - x 2 . 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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Solutions Test 2A - Solutions to Test 2 Version A Math 1501...

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