practice2 - Granillo Yvette Review 2 Due Dec 9 2005 6:00 pm...

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Granillo, Yvette – Review 2 – Due: Dec 9 2005, 6:00 pm – Inst: Edward Odell 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. Review two covers the material on test 2, namely it covers sections 3.5-3.10, 4.1-4.5 While you can submit answers and get a score as usual this counts 0your fnal average. 001 (part 1 oF 1) 10 points Determine the derivative oF f when f ( x ) = x 2 sin x + 2 x cos x. 1. f 0 ( x ) = - ( x 2 + 2) sin x 2. f 0 ( x ) = - ( x 2 + 2) cos x 3. f 0 ( x ) = - ( x 2 - 2) sin x 4. f 0 ( x ) = ( x 2 + 2) cos x correct 5. f 0 ( x ) = ( x 2 - 2) cos x 6. f 0 ( x ) = ( x 2 + 2) sin x Explanation: By the Product Rule, f 0 ( x ) = 2 x sin x + x 2 cos x + 2 cos x - 2 x sin x. Consequently, f 0 ( x ) = ( x 2 + 2) cos x . keywords: Stewart5e, di±erentiation, trig Function, product rule 002 (part 1 oF 1) 10 points ²ind f 0 ( x ) when f ( x ) = 1 4 x - x 2 . 1. f 0 ( x ) = x - 2 ( x 2 - 4 x ) 3 / 2 2. f 0 ( x ) = 2 - x ( x 2 - 4 x ) 1 / 2 3. f 0 ( x ) = 2 - x (4 x - x 2 ) 3 / 2 4. f 0 ( x ) = x - 2 (4 x - x 2 ) 3 / 2 correct 5. f 0 ( x ) = x - 2 (4 x - x 2 ) 1 / 2 6. f 0 ( x ) = 2 - x ( x 2 - 4 x ) 3 / 2 Explanation: By the Chain Rule, f 0 ( x ) = - 1 2(4 x - x 2 ) 3 / 2 (4 - 2 x ) . Consequently, f 0 ( x ) = x - 2 (4 x - x 2 ) 3 / 2 . keywords: Stewart5e, 003 (part 1 oF 1) 10 points ²ind an equation For the tangent line to the graph oF f at the point P ( π 4 , f ( π 4 )) when f ( x ) = 7 tan x. 1. y = 6 x + 13 1 - π 4 · 2. y = 14 x + 7 1 - π 2 · correct 3. y = 10 x + 10 1 - π 4 · 4. y = 4 x + 15 1 - π 2 ·
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2 5. y = 9 x + 2 1 - π 4 · Explanation: When x = π 4 , then f ( x ) = 7, so P = ( π 4 , 7). Now f 0 ( x ) = 7 sec 2 x. Thus the slope of the tangent line at P is f 0 π 4 · = 7 sec 2 π 4 · = 14 . By the point-slope formula, therefore, an equation for the tangent line at P is given by y - 7 = 14 x - π 4 · , which after simpliFcation becomes y = 14 x + 7 1 - π 2 · . keywords: Stewart5e, tangent line, trig func- tion 004 (part 1 of 1) 10 points ±ind the slope of the tangent line to the graph of x 3 - 2 y 3 - xy = 0 at the point P ( - 1 , - 1). 1. slope = - 4 5 2. slope = 3 2 3. slope = 5 4 4. slope = - 5 4 5. slope = - 2 3 6. slope = 4 5 correct Explanation: Di²erentiating implicitly with respect to x we see that 3 x 2 - 6 y 2 dy dx - y - x dy dx = 0 . Consequently, dy dx = 3 x 2 - y 6 y 2 + x . Hence at P ( - 1 , - 1) slope = dy dx f f f P = 4 5 . keywords: Stewart5e, implicit di²erentiation, tangent line, slope 005 (part 1 of 1) 10 points The Fgure below shows the graphs of three functions of time t : t One is the graph of the position function s of a car, one is its velocity v , and one is its acceleration a . Identify which graph goes with which function. 1.
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practice2 - Granillo Yvette Review 2 Due Dec 9 2005 6:00 pm...

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