2010 Calculus Test 2

2010 Calculus Test 2 - 1. Math1151 Calculus T1 2010. Test 2...

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1. Math1151 Calculus T1 2010. Test 2 Version 2a. (1) Determine all a, b R such that the function f ( x ) = ± ax + b if x 6 1 tan ( πx 4 ) if 1 < x < 2 is diﬀerentiable at x = 1. Solution. For f to be diﬀerentiable at x = 1, it must satisfy lim x 1 + f ( x ) = lim x 1 - f ( x ) lim h 0 + f ( x + h ) - f ( x ) h = lim h 0 - f ( x + h ) - f ( x ) h . The ﬁrst equation gives a + b = tan( π/ 4) = 1. The second equation gives lim h 0 + f ( x + h ) - f ( x ) h = d d x tan ² πx 4 ³ ´ ´ ´ ´ x =1 = π 4 sec 2 ² π 4 ³ = a. Hence a = π/ 2 and b = 1 - π/ 2. (2) Determine how many real numbers satisfy the equation p ( x ) = x 3 - 6 x 2 + 1 = 0. Solution. First we compute p 0 ( x ) = 3 x ( x - 4) so p has stationary points at 0 and 4. Since p is continuous on R , we can apply the intermediate value theorem on the closed intervals [ - 1 , 0], [0 , 4] and [4 , 6] and conclude, by computing the values of p at the endpoints x - 1 0 4 6 p ( x ) - 6 1 - 31 1 that p has at least three real roots. By the fundamental theorem of algebra, the poly-

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2010 Calculus Test 2 - 1. Math1151 Calculus T1 2010. Test 2...

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