hw12_sol - Homework 12 Solutions May 1 2009 Exercise 1 page...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

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.

Page1 / 10

hw12_sol - Homework 12 Solutions May 1 2009 Exercise 1 page...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online