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Unformatted text preview: ASSIGNMENT 7  SOLUTIONS Problem 1. i) Find the domain of f ( x ) = ln(ln x ). ii) Write the equation of the tangent line to the graph of f ( x ) = ln √ x 2 + x + 2 √ x + 3 at x = 1. Solution. i) The domain is given by the requirement that the inner logarithm be positive, and ln x > 0 for x > 1 precisely. ii) First, rewrite the function: f ( x ) = ln ð x 2 + x + 2 ln( √ x + 3) = 1 2 ln( x 2 + x + 2) ln( √ x + 3) Then f Í ( x ) = 1 2 ( x 2 + x + 2) Í x 2 + x + 2 ( √ x + 3) Í √ x + 3 = 2 x + 1 2( x 2 + x + 2) 1 2 √ x ( √ x + 3) so f Í (1) = 1 4 . This is the slope of the line we are looking for. Since the line also passes through (1 ,f (1)) = (1 , ln 2), it has equation y ( ln 2) = 1 4 ( x 1) or y = 1 4 x ( 1 4 + ln 2). Problem 2: Consider the function f ( x ) = 1 + 2ln x x 2 . a) What is the domain of f ? b) Find the critical numbers and the points of inflection for f . c) Find the absolute extrema of f on the interval [ e,e 2 ]. Solution. a) The domain is (0 , ∞ ). b) We compute the first and the second derivatives using the quotient rule: f Í ( x ) = (2ln x 1) Í x 2 (2ln x 1)( x 2 ) Í x 4 = (2 1 x ) x 2 (2ln x 1)2 x x 4 = 4 1 ln x x 3 f Í ( x ) = 4 (1 ln x ) Í x 3 (1 ln x )( x 3 ) Í x 6 = 4 ( 1 x ) x 3 (1 ln x )(3 x 2 ) x 6 = 4 4 + 3ln x x 4 The only critical number is x = e , obtained by solving f Í ( x ) = 0. On the other hand, from f Í ( x ) = 0 we obtain...
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This note was uploaded on 07/31/2011 for the course MATH 102 taught by Professor Maryamnamazi during the Spring '10 term at University of Victoria.
 Spring '10
 MARYAMNAMAZI
 Calculus, Logic

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