HW05-solutions

# HW05-solutions - vasquez(mpv244 – HW05 – Schultz...

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Unformatted text preview: vasquez (mpv244) – HW05 – Schultz – (56445) 1 This print-out should have 20 questions. Multiple-choice 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 ) = 5 3 x 3 / 2 + 8 x 1 / 2 . 1. f ′ (4) = 17 2 2. f ′ (4) = 13 2 3. f ′ (4) = 7 correct 4. f ′ (4) = 8 5. f ′ (4) = 15 2 Explanation: Since d dx x r = rx r − 1 , we see that f ′ ( x ) = 5 2 x 1 / 2 + 4 x − 1 / 2 . At x = 4, therefore, f ′ (4) = 7 . 002 10.0 points Find the x-coordinate of all points on the graph of f ( x ) = x 3 − x 2 − x + 5 at which the tangent line is horizontal. 1. x-coords = − 1 3 , 1 correct 2. x-coords = 1 3 , − 1 3. x-coord = − 1 4. x-coord = 1 5. x-coord = − 1 3 6. x-coord = 1 3 Explanation: The tangent line will be horizontal at P ( x , f ( x )) when f ′ ( x ) = 0 . Now f ′ ( x ) = 3 x 2 − 2 x − 1 = (3 x + 1)( x − 1) . Consequently, x = − 1 3 , 1 . 003 10.0 points Find the derivative of f when f ( x ) = x 5 2 + 3 x − 7 2 − 4 x . 1. f ′ ( x ) = 5 x 7 2 − 21 x − 5 2 + 8 2 x 2 correct 2. f ′ ( x ) = 5 x 7 2 − 21 x − 5 2 − 4 x 2 3. f ′ ( x ) = 3 x 7 2 + 21 x − 7 2 + 8 2 x 2 4. f ′ ( x ) = 3 x 5 2 − 15 x − 7 2 + 4 2 x 2 5. f ′ ( x ) = 5 x 5 2 − 15 x − 5 2 − 8 2 x 2 Explanation: Since d dx x r = rx r − 1 vasquez (mpv244) – HW05 – Schultz – (56445) 2 holds for all real numbers r , we see that f ′ ( x ) = 5 2 x 3 2 − 21 2 x − 9 2 + 4 x 2 . To simplify this expression we bring the right hand side to a common denominator so that f ′ ( x ) = 5 x 7 2 − 21 x − 5 2 + 8 2 x 2 . 004 10.0 points Find the derivative of g ( x ) = parenleftbigg x + 3 x + 1 parenrightbigg (2 x − 7) . 1. g ′ ( x ) = x 2 − 4 x + 20 x + 1 2. g ′ ( x ) = 2 x 2 − 4 x − 20 ( x + 1) 2 3. g ′ ( x ) = x 2 + 4 x − 20 ( x + 1) 2 4. g ′ ( x ) = 2 x 2 − 4 x − 20 x + 1 5. g ′ ( x ) = 2 x 2 + 4 x + 20 x + 1 6. g ′ ( x ) = 2 x 2 + 4 x + 20 ( x + 1) 2 correct Explanation: By the Quotient and Product Rules we see that g ′ ( x ) = 2 braceleftbigg x + 3 x + 1 bracerightbigg + (2 x − 7) braceleftbigg ( x + 1) − ( x + 3) ( x + 1) 2 bracerightbigg = 2 braceleftbigg x + 3 x + 1 bracerightbigg + 2 braceleftbigg 2 x − 7 ( x + 1) 2 bracerightbigg = 2( x + 3)( x + 1) + 2(2 x − 7) ( x + 1) 2 . But 2( x + 3)( x + 1) − 2(2 x − 7) = 2 x 2 + 4 x + 20 . Consequently g ′ ( x ) = 2 x 2 + 4 x + 20 ( x + 1) 2 . 005 10.0 points Find the derivative of f when f ( x ) = 3 x x + 3 x . 1. f ′ ( x ) = − 18 x x + 3 2. f ′ ( x ) = 18 x x 2 + 3 3. f ′ ( x ) = 18 x ( x 2 + 3) 2 correct 4. f ′ ( x ) = 3 1 − 3 x 2 5. f ′ ( x ) = 3 1 + 3 x 2 6. f ′ ( x ) = − 18 x ( x 2 + 3) 2 Explanation: It’s best to simplify the function before dif- ferentiating: f ( x ) = 3 x 2 x 2 + 3 ....
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## This note was uploaded on 04/18/2009 for the course CH 52375 taught by Professor Ruth during the Spring '09 term at University of Texas.

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HW05-solutions - vasquez(mpv244 – HW05 – Schultz...

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