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: Create assignment, 57950, Final 1, Dec 03 at 4:39 pm 1 This printout should have 56 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. CalC7b30c 52:04, calculus3, multiple choice, > 1 min, wordingvariable. 001 Find the derivative of f when f ( x ) = ‡ e 2 x 1 2 x · . 1. df dx = 4 xe 2 x (1 2 x ) 2 correct 2. df dx = 8 xe 2 x (1 2 x ) 2 3. df dx = 8 xe 2 x (1 2 x ) 2 4. df dx = 4 xe 2 x (1 2 x ) 2 5. df dx = 4 xe 2 x (1 2 x ) Explanation: By the product rule df dx = n d dx ‡ 1 1 2 x e 2 x ·o = n 2 e 2 x (1 2 x ) 2 2 e 2 x 1 2 x o = 2 e 2 x (1 2 x ) n 1 1 2 x 1 o . Thus df dx = 4 xe 2 x (1 2 x ) 2 . CalC7b32b 52:04, calculus3, multiple choice, < 1 min, wordingvariable. 002 Find f ( x ) when f ( x ) = e 3 x (cos 4 x + 5 sin 4 x ) . 1. f ( x ) = 3 e 3 x (4 cos 4 x + 20 sin 4 x ) 2. f ( x ) = e 3 x (11 cos 4 x + 23 sin 4 x ) 3. f ( x ) = e 3 x (23 cos 4 x 11 sin 4 x ) 4. f ( x ) = e 3 x (23 cos 4 x + 11 sin 4 x ) cor rect 5. f ( x ) = e 3 x (20 cos 4 x 15 sin 4 x ) 6. f ( x ) = 3 e 3 x (20 cos 4 x 4 sin 4 x ) Explanation: By the Product and Chain rules, f ( x ) = 3 e 3 x (cos 4 x + 5 sin 4 x ) + e 3 x (20 cos 4 x 4 sin 4 x ) . Consequently, f ( x ) = e 3 x (23 cos 4 x + 11 sin 4 x ) . CalC7b34a 52:04, calculus3, multiple choice, > 1 min, wordingvariable. 003 Find the derivative of f when f ( x ) = 3 2 e 2 x (4 x + 1) 1 / 2 . 1. f ( x ) = 12 xe 2 x (4 x + 1) 1 / 2 2. f ( x ) = 6 xe 2 x (4 x + 1) 3 / 2 Create assignment, 57950, Final 1, Dec 03 at 4:39 pm 2 3. f ( x ) = 12 xe 2 x (4 x + 1) 3 4. f ( x ) = 6 xe 2 x (4 x + 1) 3 5. f ( x ) = 12 xe 2 x (4 x + 1) 3 / 2 correct Explanation: By the product rule f ( x ) = 3 2 n d dx 1 (4 x + 1) 1 / 2 × e 2 x o = 3 2 n 2 e 2 x (4 x + 1) 3 / 2 + 2 e 2 x (4 x + 1) 1 / 2 o = 3 e 2 x (4 x + 1) 1 / 2 n 1 1 4 x + 1 o . Thus f ( x ) = 12 xe 2 x (4 x + 1) 3 / 2 . CalC7b43a 52:04, calculus3, multiple choice, > 1 min, wordingvariable. 004 Find the xintercept of the tangent line to the graph of f at the point P (0 ,f (0)) when f ( x ) = e x (3 cos x 2 sin x ) . 1. xintercept = 3 correct 2. xintercept = 0 3. xintercept = 1 4. xintercept = 2 5. xintercept = 4 Explanation: Since f (0) = 3, the point P = (0 , 3). Now the slope of the tangent line at P is the value of f (0). But f ( x ) = e x (3 cos x 2 sin x ) + e x ( 3 sin x 2 cos x ) , in which case f (0) = 1 . By the pointslope formula, therefore, an equation for the tangent line to the graph of f at P is given by y 3 = ( x 0) , i.e., y = x + 3 . Consequently, the xintercept of this tangent line is given by xintercept = 3 ....
View
Full
Document
This note was uploaded on 02/28/2010 for the course M 56200 taught by Professor Radin during the Fall '09 term at University of Texas.
 Fall '09
 RAdin

Click to edit the document details