This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Barkley, Thane Review 2 Due: Dec 11 2007, 3:00 pm Inst: Louiza Fouli 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find all the critical values of f ( x ) = x (2 x ) 3 . 1. x = 1 2 2. x = 1 3. x = 1 4. x = 2 , 1 2 correct 5. x = 1 2 6. x = 2 , 1 7. x = 2 , 1 8. x = 2 , 1 2 Explanation: As f is a polynomial, it is differentiable everywhere, so the only critical values of f are those values of c where f ( c ) = 0. Now by the Product and Chain Rules, f ( x ) = (2 x ) 3 3 x (2 x ) 2 = (2 x ) 2 (2 x 3 x ) = (2 x ) 2 (2 4 x ) . Consequently, the critical values of f all occur at x = 2 , 1 2 . 002 (part 1 of 1) 10 points Find the absolute minimum value of f on the interval 1 2 , 2 / when f ( x ) = x 2 + 2 x 1 . 1. abs. min. value = 13 4 2. abs. min. value = 5 2 3. abs. min. value = 1 4. none of these 5. abs. min. value = 2 correct 6. abs. min. value = 4 Explanation: The absolute minimum value of f on the interval 1 2 , 2 / occurs at an endpoint x = 1 2 , 2 or at a critical point of f in ( 1 2 , 2 ) , i.e. , when f ( c ) = 0 for some c , 1 2 < c < 2. But f ( x ) = 2 x 2 x 2 = 2 x 3 1 x 2 . Now the only solution of x 3 1 = 0 in 1 2 , 2 / is at c = 1. Since f 1 2 = 13 4 , f (1) = 2 , f (2) = 4 , we thus see that f has abs. min. value = 2 on the interval 1 2 , 2 / . keywords: absolute minimum, critical point 003 (part 1 of 1) 10 points Let f be the function defined by f ( x ) = 5 x 2 / 3 . Consider the following properties: Barkley, Thane Review 2 Due: Dec 11 2007, 3:00 pm Inst: Louiza Fouli 2 A. concave up on ( , 0) (0 , ) B. has local minimum at x = 0 Which does f have? 1. both of them 2. A only correct 3. B only 4. neither of them Explanation: The graph of f is 2 4 2 4 2 4 On the other hand, after differentiation, f ( x ) = 2 3 x 1 / 3 , f 00 ( x ) = 2 9 x 4 / 3 . Consequently, A. TRUE: f 00 ( x ) > , x 6 = 0 B. FALSE: see graph. keywords: concavity, local maximum, True/False, graph 004 (part 1 of 1) 10 points The derivative, f , of f has graph a b c graph of f Use it to locate the critical point(s) x at which f has a local maximum? 1. x = a 2. x = b correct 3. x = c, a 4. x = a, b 5. none of a, b, c 6. x = b, c 7. x = a, b, c 8. x = c Explanation: Since the graph of f ( x ) has no holes, the only critical points of f occur at the x intercepts of the graph of f , i.e. , at x = a, b, and c . Now by the first derivative test, f will have (i) a local maximum at x if f ( x ) changes from positive to negative as x passes through x ; (ii) a local minimum at x if f ( x ) changes from negative to positive as x passes through x ....
View
Full
Document
 Fall '08
 schultz
 Calculus

Click to edit the document details