# hw3so - Create assignment 57321 Homework 3 Feb 07 at 7:41...

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Unformatted text preview: Create assignment, 57321, Homework 3, Feb 07 at 7:41 pm 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. CalC3e01a 49:03, calculus3, multiple choice, > 1 min, wording-variable. 001 The following functions are continuous for all x : F 1 ( x ) = sin | x | , F 2 ( x ) = | x 2 | x , x 6 = 0 , x = 0 , F 3 ( x ) = cos | x | . Which are differentiable for all x ? 1. F 2 and F 3 only correct 2. F 3 only 3. all of them 4. F 2 only 5. F 1 and F 3 only 6. none of them 7. F 1 only 8. F 1 and F 2 only Explanation: The absolute value function has the prop- erties | a 2 | = a 2 = | a | 2 , |- a | = | a | . On the other hand cos x = cos(- x ) , sin(- x ) =- sin x. Consequently, A. False. (Not differentiable at x = 0.) B. True. ( = x .) C. True. ( = cos x .) keywords: True/False, continuous function, differentiable function, continuous v. differ- entiable CalC3e09a 49:03, calculus3, multiple choice, > 1 min, wording-variable. 002 Find the derivative of f when f ( x ) = cos x + 1 sin x . 1. f ( x ) = 1 cos x- 1 correct 2. f ( x ) =- 1 sin x + 1 3. f ( x ) = 1 sin x- 1 4. f ( x ) = 1 cos x + 1 5. f ( x ) = 1 1- cos x 6. f ( x ) = 1 1 + sin x 7. f ( x ) = 1 1- sin x 8. f ( x ) =- 1 1 + cos x Explanation: By the quotient rule, f ( x ) =- sin 2 x- cos x (cos x + 1) sin 2 x =- cos x- ( sin 2 x + cos 2 x ) sin 2 x . Create assignment, 57321, Homework 3, Feb 07 at 7:41 pm 2 But sin 2 x + cos 2 x = 1, so f ( x ) =- cos x + 1 1- cos 2 x = cos x + 1 (cos x- 1)(cos x + 1) . Consequently, f ( x ) = 1 cos x- 1 . keywords: differentiation, trig function, quo- tient rule CalC3e36d 49:03, calculus3, multiple choice, < 1 min, wording-variable. 003 Find the value of lim x → 1 sin2 x ‡ 3 2- 4 x + 3 3 x + 2 · . 1. limit = 1 8 correct 2. limit = 1 2 3. limit = 1 4 4. limit = 3 8 5. limit doesn’t exist Explanation: By bringing the term in parentheses to a common denominator, we see that 3 2- 4 x + 3 3 x + 2 = 1 2 µ x 3 x + 2 ¶ . Thus 1 sin2 x ‡ 3 2- 4 x + 3 3 x + 2 · = 1 4 ‡ 2 x sin2 x · 1 3 x + 2 . But lim x → 2 x sin2 x = 1 , lim x → (3 x + 2) = 2 . Consequently, by properties of limits, lim x → 1 sin2 x ‡ 3 2- 4 x + 3 3 x + 2 · = 1 8 . keywords: limit, trigonometric function, ra- tional function CalC3e40d 49:03, calculus3, multiple choice, > 1 min, wording-variable. 004 Find the value of lim x → sec 2 3 x- 1 5 x 2 if the limit exists. 1. limit = 9 5 correct 2. limit = 11 5 3. limit = 8 5 4. limit = 2 5. limit doesn’t exist Explanation: Since lim x → sin 2 ax x 2 = µ lim x → sin ax x ¶ 2 = a 2 , we see that lim x → sec 2 3 x- 1 5 x 2 = 1- cos 2 3 x 5 x 2 cos 2 3 x = lim x → sin 2 3 x x 2 ‡ 1 5cos 2 3 x · = 9 5 . Create assignment, 57321, Homework 3, Feb 07 at 7:41 pm 3 keywords: limit, trig function CalC3e48a 49:03, calculus3, multiple choice, > 1 min, wording-variable....
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hw3so - Create assignment 57321 Homework 3 Feb 07 at 7:41...

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