HW5 (solutions)

HW5 (solutions) - myers(nmm698 – HW05 – Hamrick...

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Unformatted text preview: myers (nmm698) – HW05 – Hamrick – (54868) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine the derivative of f when f ( x ) = parenleftbigg 2 3 parenrightbigg 2 / 3 . 1. f ′ ( x ) = parenleftbigg 2 3 parenrightbigg x − 1 / 3 2. f ′ ( x ) = 4 9 x − 1 / 3 3. f ′ ( x ) = 0 correct 4. f ′ ( x ) does not exist 5. f ′ ( x ) = 2 3 parenleftbigg 2 3 parenrightbigg − 1 / 3 Explanation: The derivative of any constant function is zero. Consequently, f ′ ( x ) = 0 . 002 10.0 points Find the value of f ′ ( − 1) when f ( x ) = 3 x 2 − 1 x 3 + 1 . Correct answer: − 3. Explanation: Since d dx x r = r x r − 1 , we see that f ′ ( x ) = 6 x + 3 x 4 . At x = − 1, therefore, f ′ ( − 1) = − 3 . 003 10.0 points Determine the derivative of f when f ( x ) = √ 2 x − 1 √ 2 x . 1. f ′ ( x ) = 2 x + 1 x √ 2 x 2. f ′ ( x ) = 1 2 parenleftbigg 2 x − 1 x √ 2 x parenrightbigg 3. f ′ ( x ) = 2 x − 1 x √ 2 x 4. f ′ ( x ) = 2 x − 1 √ 2 x 5. f ′ ( x ) = 1 2 parenleftbigg 2 x + 1 x √ 2 x parenrightbigg correct 6. f ′ ( x ) = 1 2 parenleftbigg 2 x + 1 √ 2 x parenrightbigg Explanation: Since d dx a x r = r a x r − 1 for all exponents r negationslash = 0 and constants a , we see that f ′ ( x ) = 1 2 radicalbigg 2 x + 1 2 1 x √ 2 x . After simplification this becomes f ′ ( x ) = 1 2 parenleftbigg 2 x + 1 x √ 2 x parenrightbigg . 004 10.0 points Find the derivative of f when f ( x ) = 7 x 1 4 + 4 x − 1 4 − 4 . 1. f ′ ( x ) = 7 x 1 2 − 4 4 x 5 4 correct myers (nmm698) – HW05 – Hamrick – (54868) 2 2. f ′ ( x ) = 7 x 1 4 + 4 4 x 3 4 3. f ′ ( x ) = 7 x 1 2 − 4 3 x 5 4 4. f ′ ( x ) = 7 x 1 4 − 4 4 x 5 4 5. f ′ ( x ) = 7 x 1 2 + 4 4 x 3 4 Explanation: Because of the linearity of differentiation and the property d dx ( x r ) = rx r − 1 , we see that f ′ ( x ) = 1 4 parenleftbigg 7 x 3 4 − 4 x 5 4 parenrightbigg . Thus f ′ ( x ) = 7 x 1 2 − 4 4 x 5 4 . 005 10.0 points Find the derivative of f when f ( x ) = x x − 1 x . 1. f ′ ( x ) = − 2 x ( x 2 − 1) 2 correct 2. f ′ ( x ) = 1 1 + 1 x 2 3. f ′ ( x ) = − 1 1 − 1 x 2 4. f ′ ( x ) = − 2 x x − 1 5. f ′ ( x ) = 2 x x 2 − 1 6. f ′ ( x ) = 2 x ( x 2 − 1) 2 Explanation: It’s best to simplify the function before dif- ferentiating; for then f ( x ) = x 2 x 2 − 1 . Thus by the Quotient Rule, f ′ ( x ) = 2 x ( x 2 − 1) − 2 x ( x 2 ) ( x 2 − 1) 2 Consequently, f ′ ( x ) = − 2 x ( x 2 − 1) 2 . 006 10.0 points Find the derivative of f ( x ) = ( x − 4) | x − 4 | . 1. f ′ ( x ) = 2 | x − 4 | correct 2. f ′ ( x ) = 2 x − 4 3. f ′ ( x ) = x − 4 4. f ′ ( x ) = | 2 x − 4 | 5. f ′ ( x ) = 2( x − 4) 6. f ′ ( x ) = | x − 4 | Explanation: Since | v | = braceleftbigg v, v ≥ 0, − v, v < 0, we see that f ( x ) = braceleftBigg − ( x − 4) 2 , x < 4,...
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This note was uploaded on 03/10/2011 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas.

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HW5 (solutions) - myers(nmm698 – HW05 – Hamrick...

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