HW4Sol - S and c [0 , 1] we have that (1) f ( cx + (1-c ) y...

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Fall 2009 IEOR 160 October 1, 2009 Page 1 of 1 HOMEWORK 4 SOLUTIONS Chapter 12.3 1. f ′′ ( x ) = 6 x 0, (for x 0) so f ( x ) is convex on S . 3. f ′′ ( x ) = 2 x 3 > 0 (for x > 0). Thus f ( x ) is convex on S . 5. f ′′ ( x ) = - x 2 < 0, so f ( x ) is a concave function on S . 7. f ( x 1 , x 2 ) is the sum of convex functions and is therefore a convex function. (See problem 12) 9. H = - 2 0 . 5 0 0 . 5 - 2 0 0 0 - 4 1st Order PM’s are -2, -2, -4 which are all negative. 2nd Order PM’s are all positive: det b - 2 0 . 5 0 . 5 - 2 B = 3 . 75 det b - 2 0 0 - 4 B = 8 det b - 2 0 0 - 4 B = 8 Expanding by Row 3 we ±nd 3rd Order PM = - 4(( - 2)( - 2) - ( . 5)( . 5)) = - 15 < 0. Thus f ( x 1 , x 2 , x 3 ) is concave on R 3 . 12. Since f and g are covex we know that for all x, y
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Unformatted text preview: S and c [0 , 1] we have that (1) f ( cx + (1-c ) y ) cf ( x ) + (1-c ) f ( y ) (2) g ( cx + (1-c ) y ) cg ( x ) + (1-c ) g ( y ) Adding (1) and (2) yields h ( cx + (1-c ) y ) ch ( x ) + (1-c ) h ( y ) which shows that h is also a convex function. 13. Since f is convex we know that for 0 k 1 (1) f ( kx + (1-k ) y ) kf ( x ) + (1-k ) f ( y ). Multiplying both sides of (1) by c 0 shows that g is also a convex function. Similarly, multiplyng both sides of (1) by c 0 shows that g is a concave function....
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