sol1 - ECE 5620 Spring 2011 Homework #1: Solutions 1. For x...

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ECE 5620 Spring 2011 Homework #1: Solutions 1. For x < 1 , let f ( x ) = e - x + x - 1 . Now f (0) = 0 , and by differentiating, we see that f is nondecreasing on (0 , 1] and nonincreasing on ( -∞ , 0] . It follows that f is nonnegative 1 - x e - x Since both sides are nonnegative, we can then raise both sides to the power 1 /y . 2. (a) By the second derivative test, log( x ) , log(1+ x ) , and x +5 are concave, while x 2 , x + 5 , and x log( x ) are convex. (b) The sum f + g is concave: θ ( f ( x ) + g ( x )) + (1 - θ )( f ( y ) + g ( y )) f ( θx + (1 - θ ) y ) + g ( θx + (1 - θ ) y ) . The product, however, is not necessary concave. Let f ( x ) = g ( x ) = x then f and g are both concave but the product, x 2 , is not. The maxi- mum, too, is not necessarily concave. Let f ( x ) = -| x - 1 | and g ( x ) = -| x + 1 | . Then max( f (1) ,g (1)) + max( f ( - 1) ,g ( - 1)) = 0 but max( f (0) ,g (0)) = - 1 . On the other hand,
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sol1 - ECE 5620 Spring 2011 Homework #1: Solutions 1. For x...

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