This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: wtm369 Homework 15 Helleloid (58250) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find the value of f ( 1) when f ( x ) = 6 tan 1 x 5 sin 1 x . 1. f ( 1) = correct 2. f ( 1) = 5 2 3. f ( 1) = 3 4. f ( 1) = 3 2 5. f ( 1) = 2 Explanation: Since tan 1 ( 1) = 4 , sin 1 ( 1) = 2 , we see that f ( 1) = parenleftBig 5 2 3 2 parenrightBig = . 002 10.0 points Simplify the expression y = sin parenleftbigg tan 1 x 6 parenrightbigg by writing it in algebraic form. 1. y = x x 2 6 2. y = x x 2 + 6 3. y = x 2 + 6 6 4. y = x x 2 + 6 correct 5. y = 6 x 2 + 6 Explanation: The given expression has the form y = sin where tan = x 6 , 2 &lt; &lt; 2 . To determine the value of sin given the value of tan , we can apply Pythagoras theorem to the right triangle 6 x radicalbig x 2 + 6 From this it follows that y = sin = x x 2 + 6 . Alternatively, we can use the trig identity csc 2 = 1 + cot 2 to determine sin . 003 10.0 points Determine if lim x sin 1 parenleftbigg 1 + x 5 + 2 x parenrightbigg exists, and if it does, find its value. 1. limit = 6 correct 2. limit does not exist 3. limit = 2 4. limit = 4 wtm369 Homework 15 Helleloid (58250) 2 5. limit = 3 6. limit = 0 Explanation: Since lim x 1 + x 5 + 2 x = 1 2 , we see that lim x sin 1 parenleftbigg 1 + x 5 + 2 x parenrightbigg exists, and that the limit = sin 1 1 2 = over 6 . 004 10.0 points Determine the derivative of f ( x ) = 4 sin 1 ( x/ 2) . 1. f ( x ) = 8 1 x 2 2. f ( x ) = 2 4 x 2 3. f ( x ) = 2 1 x 2 4. f ( x ) = 8 4 x 2 5. f ( x ) = 4 1 x 2 6. f ( x ) = 4 4 x 2 correct Explanation: Use of d dx sin 1 ( x ) = 1 1 x 2 , together with the Chain Rule shows that f ( x ) = 4 radicalbig 1 ( x/ 2) 2 parenleftBig 1 2 parenrightBig . Consequently, f ( x ) = 4 4 x 2 . 005 10.0 points Find the derivative of f ( x ) = tan 1 ( e 3 x ) . 1. f ( x ) = 1 1 + e 6 x 2. f ( x ) = 3 1 + e 6 x 3. f ( x ) = 3 1 e 6 x 4. f ( x ) = e 3 x 1 e 6 x 5. f ( x ) = 1 1 e 6 x 6. f ( x ) = 3 e 3 x 1 + e 6 x correct 7. f ( x ) = 3 e 3 x 1 e 6 x 8. f ( x ) = e 3 x 1 + e 6 x Explanation: Since d dx tan 1 x = 1 1 + x 2 , d...
View
Full
Document
This note was uploaded on 12/04/2009 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas.
 Fall '08
 schultz
 Calculus

Click to edit the document details