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Unformatted text preview: Homework 12 Solutions May 1, 2009 Exercise 1, page 669 Determine the stationary (extreme) points for the following functions: (a) f ( x ) = x 3 + x . (b) f ( x ) = x 4 + x 2 . (c) f ( x ) = 4 x 4 x 2 + 5. Solution : We find the extreme (stationary) points as the solutions of the equation setting the derivative of the function to zero, i.e., as solutions of the equation f ( x ) = 0. (a) For f ( x ) = x 3 + x , the derivative is f ( x ) = 3 x 2 + 1, and the equation is 3 x 2 + 1 = 0 ⇐⇒ x 2 = 1 3 . This equation has only imaginary (complex) number solutions. Hence, this function does not have stationary points in the set of real numbers (the domain of the function). (b) For function f ( x ) = x 4 + x 2 , we have f ( x ) = 4 x 3 + 2 x , and the equation is 4 x 3 + 2 x = 0 ⇐⇒ 2 x (2 x 2 + 2) = 0 ⇐⇒ x = 0 or 2 x 2 + 2 = 0 . The only real solution is x = 0, hence, this is the only stationary point of f . (c) For f ( x ) = 4 x 4 x 2 + 5, we have f ( x ) = 16 x 3 2 x and the equation is 16 x 3 2 x = 0 ⇐⇒ 2 x (8 x 3 1) = 0 ⇐⇒ x = 0 or 8 x 3 = 1 ⇐⇒ x = 0 or x = 1 2 . Thus, the stationary (extreme) points are x = 0 and x = 1 2 . Exercise In the preceding problem, determine whether each of the given functions is convex, concave, or neither convex nor concave. Solution : To determine whether the function is convex, concave or neither of the two, we can look at the second derivative f 00 ( x ), and use the following criteria: 1 • If f 00 ( x ) ≥ 0 for all x , then f is convex. • If f 00 ( x ) ≤ 0 for all x , then f is concave. • If f 00 ( x ) changes the sign, then f is neither convex nor concave. (a) For f ( x ) = x 3 + x , the first and second derivatives are f ( x ) = 3 x 2 + 1 and f 00 ( x ) = 6 x . The scond derivative is negative for x < 0 and nonnegative for x ≥ 0. Thus, this function is neither convex nor concave. (b) For function f ( x ) = x 4 + x 2 , we have f ( x ) = 4 x 3 + 2 x and f 00 ( x ) = 12 x 2 + 2. The second derivative is positive for all x , and therefore f is convex. (c) For f ( x ) = 4 x 4 x 2 + 5, we have f ( x ) = 16 x 3 2 x and f 00 ( x ) = 48 x 2 2. The second derivative takes both negative and positive values. For example, f 00 (0) = 2 while f 00 (1) = 46. Thus, the given function is neither convex nor concave. Exercise 2, page 669 Determine the stationary points of the following functions: (a) f ( x 1 , x 2 ) = x 3 1 + x 3 2 3 x 1 x 2 . (b) f ( x 1 , x 2 , x 3 ) = 2 x 2 1 + x 2 2 + x 2 3 + 6( x 1 + x 2 + x 3 ) + 2 x 1 x 2 x 3 . Solution : We find the extreme (stationary) points as the solutions of the equation setting the gradient of the function to zero, i.e., as solutions of the equation ∇ f ( x ) = 0....
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This note was uploaded on 08/31/2010 for the course IESE GE 330 taught by Professor Nedich during the Spring '09 term at University of Illinois at Urbana–Champaign.
 Spring '09
 Nedich

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