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Macroeconomics Exam Review 145

Macroeconomics Exam Review 145 - Solutions for Foundations...

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3.133 If 𝑓 is convex 𝑓 ( 𝛼 x 1 + (1 𝛼 ) x 2 ) 𝛼𝑓 ( x 1 ) + (1 𝛼 ) 𝑓 ( x 2 ) Since 𝑔 is increasing 𝑔 ( 𝑓 ( 𝛼 x 1 + (1 𝛼 ) x 2 ) ) 𝑔 ( 𝛼𝑓 ( x 1 ) + (1 𝛼 ) 𝑓 ( x 2 ) ) 𝛼𝑔 ( 𝑓 ( x 1 ) ) + (1 𝛼 ) 𝑔 ( 𝑓 ( x 2 ) ) since 𝑔 is also convex. The concave case is proved similarly. 3.134 Let 𝐹 = log 𝑓 . If 𝐹 is convex, 𝑓 ( x ) = 𝑒 𝐹 ( x ) is an increasing convex function of a convex function and is therefore convex (Exercise 3.133). 3.135 If 𝑓 is positive and concave, then log 𝑓 is concave (Exercise 3.51). Therefore log 1 𝑓 = log 1 log 𝑓 = log 𝑓 is convex. By the previous exercise (Exercise 3.134), this implies that 1 /𝑓 is convex. If 𝑓 is negative and convex, then 𝑓 is positive and concave, 1 / 𝑓 is convex, and therefore 1 /𝑓 is concave. 3.136 Consider the identity 𝑔 ( 𝑓 ( 𝑥 1 𝑥 2 ) ) + 𝑔 ( 𝑓 ( 𝑥 1 𝑥 2 ) ) 𝑔 ( 𝑓 ( 𝑥 1 ) ) 𝑔 ( 𝑓 ( 𝑥 2 ) ) = 𝑔 ( 𝑓 ( 𝑥 1 𝑥 2 ) ) + 𝑔 ( 𝑓 ( 𝑥 1 𝑥 2 ) ) 𝑔 ( 𝑓 ( 𝑥 1 ) ) 𝑔 ( 𝑓 ( 𝑥 1 𝑥 2
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