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M408K Ex. 2 Solutions

# M408K Ex. 2 Solutions - Mahon Kevin Exam 2 Due Nov 3 2005...

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Mahon, Kevin – Exam 2 – Due: Nov 3 2005, 11:00 pm – Inst: Edward Odell 1 This print-out should have 18 questions. Multiple-choice 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 f 0 ( x ) when f ( x ) = x - 2 x + 1 · 2 . 1. f 0 ( x ) = - 6( x + 2) ( x - 1) 3 2. f 0 ( x ) = 6( x - 2) ( x + 1) 3 correct 3. f 0 ( x ) = 6( x + 1) ( x - 1) 3 4. f 0 ( x ) = - 4( x - 2) ( x + 1) 3 5. f 0 ( x ) = 4( x - 1) ( x + 1) 3 6. f 0 ( x ) = - 4( x + 1) ( x - 1) 3 Explanation: By the Chain and Quotient Rules, f 0 ( x ) = 2 x - 2 x + 1 · ( x + 1) - ( x - 2) ( x + 1) 2 . Consequently, f 0 ( x ) = 6( x - 2) ( x + 1) 3 . keywords: Stewart5e, derivative, quotient rule, chain rule 002 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = tan x 3 + 7 sec x . 1. f 0 ( x ) = sec 2 x 3 + 7 sec x tan x 2. f 0 ( x ) = 3 sec x + 7 (3 + 7 sec x ) 2 3. f 0 ( x ) = 3 + 7 cos x (3 cos x + 7) 2 correct 4. f 0 ( x ) = 3 cos x + 7 (3 cos x + 7) 2 5. f 0 ( x ) = 3 sin x + 7 cos x (3 cos x + 7) 2 6. f 0 ( x ) = - sec 2 x (3 + 7 sec x tan x ) 2 Explanation: Now, tan x 3 + 7 sec x = sin x cos x 3 + 7 cos x = sin x 3 cos x + 7 . Thus f 0 ( x ) = cos x 3 cos x + 7 + 3 sin 2 x (3 cos x + 7) 2 = cos x (3 cos x + 7) + 3 sin 2 x (3 cos x + 7) 2 . Consequently, f 0 ( x ) = 3 + 7 cos x (3 cos x + 7) 2 , since cos 2 x + sin 2 x = 1. keywords: Stewart5e, 003 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = x sin 4 x + 1 4 cos 4 x. 1. f 0 ( x ) = - 4 x cos 4 x 2. f 0 ( x ) = 4 x cos 4 x + 2 sin 4 x

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Mahon, Kevin – Exam 2 – Due: Nov 3 2005, 11:00 pm – Inst: Edward Odell 2 3. f 0 ( x ) = 4 cos 4 x 4. f 0 ( x ) = 4 x cos 4 x correct 5. f 0 ( x ) = 4 x cos 4 x - 2 sin 4 x Explanation: Since d dx sin x = cos x, d dx cos x = - sin x, it follows that f 0 ( x ) = sin 4 x + 4 x cos 4 x - sin 4 x. Consequently, f 0 ( x ) = 4 x cos 4 x . keywords: Stewart5e, derivative, product rule, chain rule, trig functions 004 (part 1 of 1) 10 points Find the equation of the tangent line to the graph of y 2 - xy - 6 = 0 , at the point P = (1 , 3). 1. 4 y = 3 x + 9 2. 5 y = 3 x + 12 correct 3. 4 y + 3 x = 9 4. 5 y + 3 x = 12 5. y = 3 x Explanation: Differentiating implicitly with respect to x we see that 2 y dy dx - y - x dy dx = 0 , so dy dx = y 2 y - x . At P = (1 , 3), therefore, dy dx fl fl fl P = 3 5 . Thus by the point slope formula, the equation of the tangent line at P is given by y - 3 = 3 5 ( x - 1) . Consequently, 5 y = 3 x + 12 . keywords: Stewart5e, 005 (part 1 of 1) 10 points The figure below shows the graphs of three functions: 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. s : v : a :
Mahon, Kevin – Exam 2 – Due: Nov 3 2005, 11:00 pm – Inst: Edward Odell 3 2. s : v : a : correct 3. s : v : a : 4. s : v : a : 5. s : v : a : 6. s : v : a : Explanation: Experience tells us that the car is (i) moving forwards when its velocity is positive, (ii) moving backwards when its velocity is negative, (iii) speeding up when its velocity is positive and increasing, i.e. , when both velocity and acceleration are positive, (iv) slowing down when its velocity is pos- itive but decreasing, i.e. , when its velocity is positive but its acceleration is negative.

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