HW01-solutions - kim(tk5895 – HW01 – Henry –(54974 1...

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Unformatted text preview: kim (tk5895) – HW01 – Henry – (54974) 1 This print-out should have 22 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 lim x →− 4 6 x + 4 parenleftbigg 4 x 2 + 8 − 1 6 parenrightbigg . 1. limit does not exist 2. limit = 1 3 correct 3. limit = 1 2 4. limit = 1 4 5. limit = 1 6 Explanation: After the second term in the product is brought to a common denominator it becomes 24 − x 2 − 8 6( x 2 + 8) = 16 − x 2 6( x 2 + 8) . Thus the given expression can be written as 6(16 − x 2 ) 6( x + 4)( x 2 + 8) = 6(4 − x ) 6( x 2 + 8) so long as x negationslash = − 4. Consequently, lim x →− 4 6 x + 4 parenleftbigg 4 x 2 + 8 − 1 6 parenrightbigg = lim x →− 4 6(4 − x ) 6( x 2 + 8) . By properties of limits, therefore, limit = 1 3 . 002 10.0 points Find the derivative of f when f ( x ) = 1 − 2 cos x sin x . 1. f ′ ( x ) = − 2 + cos x sin 2 x 2. f ′ ( x ) = 2 + sin x cos 2 x 3. f ′ ( x ) = 1 − 2 cos x sin 2 x 4. f ′ ( x ) = sin x − 2 cos 2 x 5. f ′ ( x ) = 2 sin x + 1 cos 2 x 6. f ′ ( x ) = 2 sin x − 1 cos 2 x 7. f ′ ( x ) = − 1 + 2 cos x sin 2 x 8. f ′ ( x ) = 2 − cos x sin 2 x correct Explanation: By the quotient rule, f ′ ( x ) = 2 sin 2 x − cos x (1 − 2 cos x ) sin 2 x = 2(sin 2 x + cos 2 x ) − cos x sin 2 x . But cos 2 x + sin 2 x = 1. Consequently, f ′ ( x ) = 2 − cos x sin 2 x . 003 10.0 points Find the derivative of f when f ( x ) = 3 x cos4 x − 5 sin 4 x . 1. f ′ ( x ) = − 20cos 4 x + 12 x sin4 x kim (tk5895) – HW01 – Henry – (54974) 2 2. f ′ ( x ) = − 12 x sin4 x − 17 cos4 x correct 3. f ′ ( x ) = 12 x sin 4 x − 17 cos4 x 4. f ′ ( x ) = 20 cos 4 x − 17 x sin4 x 5. f ′ ( x ) = − 12 x sin4 x − 20cos 4 x Explanation: Using formulas for the derivatives of sine and cosine together with the Product and Chain Rules, we see that f ′ ( x ) = 3 cos 4 x − 12 x sin4 x − 20cos 4 x = − 12 x sin 4 x − 17 cos 4 x . 004 10.0 points Find f ′ ( x ) when f ( x ) = 1 √ x 2 − 6 x . 1. f ′ ( x ) = x − 3 (6 x − x 2 ) 3 / 2 2. f ′ ( x ) = 3 − x ( x 2 − 6 x ) 3 / 2 correct 3. f ′ ( x ) = x − 3 ( x 2 − 6 x ) 3 / 2 4. f ′ ( x ) = 3 − x (6 x − x 2 ) 3 / 2 5. f ′ ( x ) = x − 3 (6 x − x 2 ) 1 / 2 6. f ′ ( x ) = 3 − x ( x 2 − 6 x ) 1 / 2 Explanation: By the Chain Rule, f ′ ( x ) = − 1 2( x 2 − 6 x ) 3 / 2 (2 x − 6) . Consequently, f ′ ( x ) = 3 − x ( x 2 − 6 x ) 3 / 2 . 005 10.0 points Determine f ′ ( x ) when f ( x ) = 3 sec 2 x − 2 tan 2 x . 1. f ′ ( x ) = 10 sec 2 x tan x 2. f ′ ( x ) = − 2 sec 2 x tan x 3. f ′ ( x ) = 2 sec 2 x tan x correct 4. f ′ ( x ) = 2 tan 2 sec x 5. f ′ ( x ) = − 2 tan 2 sec x 6. f ′ ( x ) = 10 tan 2 sec x Explanation: Since d dx sec x = sec x tan x, d dx tan x = sec 2 x, the Chain Rule ensures that f ′ ( x ) = 6 sec 2...
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This note was uploaded on 04/09/2011 for the course M 408 L taught by Professor Cepparo during the Fall '08 term at University of Texas.

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HW01-solutions - kim(tk5895 – HW01 – Henry –(54974 1...

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