Final 4 Solutions

# Final 4 Solutions - Create assignment 57950 Final 1 Dec 03...

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Create assignment, 57950, Final 1, Dec 03 at 4:39 pm 1 This print-out should have 56 questions. Multiple-choice 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, wording-variable. 001 Find the derivative of f when f ( x ) = e - 2 x 1 - 2 x · . 1. df dx = 4 x e - 2 x (1 - 2 x ) 2 correct 2. df dx = 8 x e - 2 x (1 - 2 x ) 2 3. df dx = - 8 x e - 2 x (1 - 2 x ) 2 4. df dx = - 4 x e - 2 x (1 - 2 x ) 2 5. df dx = 4 x e - 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 x e - 2 x (1 - 2 x ) 2 . CalC7b32b 52:04, calculus3, multiple choice, < 1 min, wording-variable. 002 Find f 0 ( x ) when f ( x ) = e 3 x (cos 4 x + 5 sin 4 x ) . 1. f 0 ( x ) = 3 e 3 x (4 cos 4 x + 20 sin 4 x ) 2. f 0 ( x ) = e 3 x (11 cos 4 x + 23 sin 4 x ) 3. f 0 ( x ) = e 3 x (23 cos 4 x - 11 sin 4 x ) 4. f 0 ( x ) = e 3 x (23 cos 4 x + 11 sin 4 x ) cor- rect 5. f 0 ( x ) = e 3 x (20 cos 4 x - 15 sin 4 x ) 6. f 0 ( x ) = 3 e 3 x (20 cos 4 x - 4 sin 4 x ) Explanation: By the Product and Chain rules, f 0 ( 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 0 ( x ) = e 3 x (23 cos 4 x + 11 sin 4 x ) . CalC7b34a 52:04, calculus3, multiple choice, > 1 min, wording-variable. 003 Find the derivative of f when f ( x ) = 3 2 e 2 x (4 x + 1) 1 / 2 . 1. f 0 ( x ) = 12 xe 2 x (4 x + 1) 1 / 2 2. f 0 ( x ) = 6 xe 2 x (4 x + 1) 3 / 2

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Create assignment, 57950, Final 1, Dec 03 at 4:39 pm 2 3. f 0 ( x ) = 12 xe 2 x (4 x + 1) 3 4. f 0 ( x ) = 6 xe 2 x (4 x + 1) 3 5. f 0 ( x ) = 12 xe 2 x (4 x + 1) 3 / 2 correct Explanation: By the product rule f 0 ( 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 0 ( x ) = 12 xe 2 x (4 x + 1) 3 / 2 . CalC7b43a 52:04, calculus3, multiple choice, > 1 min, wording-variable. 004 Find the x -intercept 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. x -intercept = - 3 correct 2. x -intercept = 0 3. x -intercept = - 1 4. x -intercept = - 2 5. x -intercept = - 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 (0). But f 0 ( x ) = e x (3 cos x - 2 sin x ) + e x ( - 3 sin x - 2 cos x ) , in which case f 0 (0) = 1 . By the point-slope 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 x -intercept of this tangent line is given by x -intercept = - 3 . CalC7b61s 52:04, calculus3, multiple choice, > 1 min, wording-variable. 005 Find the absolute maximum of f on the interval [0 , 4] when f ( x ) = 2 + 3(4 - x ) e x - 3 . 1. abs. max = 2 2. abs. max = 2 . 59 3. abs. max = 5 correct 4. abs. max = 7 5. abs. max = 6 Explanation: As f is differentiable everywhere, its abso- lute maximum on the interval [0 , 4] will occur
Create assignment, 57950, Final 1, Dec 03 at 4:39 pm 3 at an end-point of [0 , 4] or at a local maximum of f in (0 , 4). Now f 0 ( x ) = 3(3 - x ) e x - 3 , and f 00 ( x ) = 3(2 - x ) e x - 3 so x = 3 is a critical point at which f has a local maximum. But f (0) = 2 + 12 e - 3 = 2 . 59 , f (3) = 5 , f (4) = 2 .

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