HW7 Solutions - ASSIGNMENT 7 SOLUTIONS Problem 1 i Find the...

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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 x 2 + x + 2 - ln( x + 3) = 1 2 ln( x 2 + x + 2) - ln( x + 3) Then f ( x ) = 1 2 ( x 2 + x + 2) x 2 + x + 2 - ( x + 3) x + 3 = 2 x + 1 2( x 2 + x + 2) - 1 2 x ( x + 3) so f (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 + 2 ln 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 ( x ) = (2 ln x - 1) x 2 - (2 ln x - 1)( x 2 ) x 4 = (2 1 x ) x 2 - (2 ln x - 1)2 x x 4 = 4 1 - ln x x 3 f ( x ) = 4 (1 - ln x ) x 3 - (1 - ln x )( x 3 ) x 6 = 4 ( - 1 x ) x 3 - (1 - ln x )(3 x 2 ) x 6 = 4 - 4 + 3 ln x x 4 The only critical number is x = e , obtained by solving f ( x ) = 0. On the other hand, from f ( x ) = 0 we obtain x = e 4 / 3 . Since f ( x ) < 0 for x < e 4 / 3 , and f ( x ) > 0 for x > e 4 / 3 , we see that ( e 4 / 3 , f ( e 4 / 3 )) = ( e 4 / 3 , 5 3 e - 8 /
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