PS1Solutions

PS1Solutions - Problem Set 1: ECO310 Prof. Stephen Morris...

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Unformatted text preview: Problem Set 1: ECO310 Prof. Stephen Morris Due on February 20th Question 1: (45 points) For notational simplicity, let us de&ne f 1 ( x ) and f 2 ( x ) as follows: f 1 ( x ) = x ( x & 4) f 2 ( x ) = 10 x (8 & x ) Note that f ( x ) is written as: f ( x ) = & f 1 ( x ) if & 10 x f 2 ( x ) if < x 10 (a) ( 12 points ) A simple calculation gives you f ( x ) for integer values in the interval [-10, 10 ] , which is shown in the following table. x & 10 & 9 & 8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 f 1 ( x ) 140 117 96 77 60 45 32 21 12 5 x 1 2 3 4 5 6 7 8 9 10 f 2 ( x ) 70 120 150 160 150 120 70 12 & 90 & 200 You can write a graph by connecting these points: 1-10-8-6-4-2 2 4 6 8 10-200-100 100 x y (b) ( 12 points ) The derivatives of functions f 1 and f 2 can be solved as: df 1 dx = 2 x & 4 df 2 dx = & 20 x + 80 Note that a critical point is a point whose derivative is . 1 From the above expression, it is obvious that df 1 dx = 0 when x = 2 and df 2 dx = 0 when x = 4 . Note that f 1 is de&ned on [ & 10 ; 0] and thereby f 1 does not have a critical point in its domain. So, we have a unique critical point at x = 4 . To check this point is a local maximum or minimum (or neither of them), we derive the second derivatives. 2 d 2 f 2 dx 2 = & 20 Since d 2 f 2 dx 2 j x =4 < ; f (4) is a local maximum....
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PS1Solutions - Problem Set 1: ECO310 Prof. Stephen Morris...

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