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Unformatted text preview: wtm369 – Review 1 – Helleloid – (58250) 1 This printout should have 20 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 ′ (4) when f ( x ) = 2 x 3 / 2 − 4 x 1 / 2 + π. 1. f ′ (4) = 4 2. f ′ (4) = 6 3. f ′ (4) = 11 2 4. none of the other answers 5. f ′ (4) = 5 correct 6. f ′ (4) = 9 2 Explanation: Since d dx ( x α ) = αx α − 1 holds for all α negationslash = 0, we see that f ′ ( x ) = 3 x 1 / 2 − 2 x − 1 / 2 . Consequently, f ′ (4) = 5 . 002 10.0 points Find the derivative of f when f ( x ) = x 2 4 − 2 x − 2 . 1. f ′ ( x ) = x 4 − 8 2 x 3 2. f ′ ( x ) = x 4 + 4 x 3 3. f ′ ( x ) = x 4 + 8 2 x 3 correct 4. f ′ ( x ) = x 3 − 8 2 x 2 5. f ′ ( x ) = x 3 + 8 x 2 Explanation: Since d dx x r = r x r − 1 , it follows that f ′ ( x ) = x 2 + 4 x 3 . Consequently, f ′ ( x ) = x 4 + 8 2 x 3 . 003 10.0 points Find the derivative of f when f ( x ) = sin x cos x − 1 . 1. f ′ ( x ) = 1 sin x − 1 2. f ′ ( x ) = 1 cos x + 1 3. f ′ ( x ) = 1 1 − cos x correct 4. f ′ ( x ) = 1 1 − sin x 5. f ′ ( x ) = 1 cos x − 1 6. f ′ ( x ) = 1 1 + sin x 7. f ′ ( x ) = − 1 sin x + 1 8. f ′ ( x ) = − 1 1 + cos x Explanation: wtm369 – Review 1 – Helleloid – (58250) 2 By the quotient rule, f ′ ( x ) = cos x (cos x − 1) + sin 2 x (cos x − 1) 2 = ( cos 2 x + sin 2 x ) − cos x (cos x − 1) 2 . But sin 2 x + cos 2 x = 1, so f ′ ( x ) = 1 − cos x (cos x − 1) 2 = 1 − cos x (1 − cos x ) 2 . Consequently, f ′ ( x ) = 1 1 − cos x . 004 10.0 points Find the derivative of f ( x ) = x 3 (sec x − tan x ) . 1. f ′ ( x ) = x 2 ( x sec x + 3)(tan x − sec x ) 2. f ′ ( x ) = x 2 ( x sec x − 3)(tan x + sec x ) 3. f ′ ( x ) = x 2 (3 − x sec x )(tan x + sec x ) 4. f ′ ( x ) = x 2 ( x sec x − 3)(tan x − sec x ) correct 5. f ′ ( x ) = x 2 (3 − x sec x )(tan x − sec x ) 6. f ′ ( x ) = x 2 (3 + x sec x )(tan x + sec x ) Explanation: By the Product Rule, f ′ ( x ) = 3 x 2 (sec x − tan x ) + x 3 (sec x tan x − sec 2 x ) . But then f ′ ( x ) = 3 x 2 (sec x − tan x ) + x 3 sec x (tan x − sec x ) . Consequently, f ′ ( x ) = x 2 ( x sec x − 3)(tan x − sec x ) . keywords: 005 10.0 points Find the xcoordinates of all the points on the graph of f at which the tangent line is horizontal when f ( x ) = 2 x 3 + 2 x 2 − 2 x + 3 . 1. xcoord = 1 2. xcoords = 1 3 , − 1 correct 3. xcoord = − 1 3 4. xcoords = − 1 3 , 1 5. xcoord = 1 3 6. xcoord = − 1 Explanation: The tangent line will be horizontal at P ( x , f ( x )) when f ′ ( x ) = 0 . Now f ′ ( x ) = 6 x 2 + 4 x − 2 = 2(3 x − 1)( x + 1) ....
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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

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