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# cal1 - choi(kc24547 – HW01 – BERG –(56525 1 This...

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Unformatted text preview: choi (kc24547) – HW01 – BERG – (56525) 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 Determine if lim x → 7 √ x + 2 − 3 x − 7 exists, and if it does, find its value. 1. limit = 1 3 2. limit = 1 6 correct 3. limit = 6 4. limit = 5 5. limit does not exist 6. limit = 1 7 Explanation: Since ( √ x + 2 − 3)( √ x + 2 + 3) = ( x + 2) − 9 = x − 7 , we see by rationalizing the numerator that √ x + 2 − 3 x − 7 = x − 7 ( x − 7)( √ x + 2 + 3) = 1 √ x + 2 + 3 provided x negationslash = 7. On the other hand, lim x → 7 √ x + 2 + 3 = 6 . Consequently, by properties of limits, lim x → 7 √ x + 2 − 3 x − 7 exists and has limit = 1 6 . 002 10.0 points Find the value of lim x →− 3 7 x + 3 parenleftbigg 3 x 2 + 6 − 1 5 parenrightbigg . 1. limit = 14 25 correct 2. limit = 7 25 3. limit does not exist 4. limit = 7 15 5. limit = 14 15 Explanation: After the second term in the product is brought to a common denominator it becomes 15 − x 2 − 6 5( x 2 + 6) = 9 − x 2 5( x 2 + 6) . Thus the given expression can be written as 7(9 − x 2 ) 5( x + 3)( x 2 + 6) = 7(3 − x ) 5( x 2 + 6) so long as x negationslash = − 3. Consequently, lim x →− 3 7 x + 3 parenleftbigg 3 x 2 + 6 − 1 5 parenrightbigg = lim x →− 3 7(3 − x ) 5( x 2 + 6) . By properties of limits, therefore, limit = 14 25 . 003 10.0 points choi (kc24547) – HW01 – BERG – (56525) 2 Find the derivative of f when f ( x ) = 2 − sin x cos x . 1. f ′ ( x ) = 2 + sin x cos 2 x 2. f ′ ( x ) = − 2 + cos x sin 2 x 3. f ′ ( x ) = 2 sin x − 1 cos 2 x correct 4. f ′ ( x ) = sin x − 2 cos 2 x 5. f ′ ( x ) = 1 − 2 cos x sin 2 x 6. f ′ ( x ) = 2 − cos x sin 2 x 7. f ′ ( x ) = 2 sin x + 1 cos 2 x 8. f ′ ( x ) = − 1 + 2 cos x sin 2 x Explanation: By the quotient rule, f ′ ( x ) = − cos 2 x + sin x (2 − sin x ) cos 2 x = − sin 2 x − cos 2 x + 2 sin x cos 2 x . But cos 2 x + sin 2 x = 1 . Consequently, f ′ ( x ) = 2 sin x − 1 cos 2 x . 004 10.0 points Find the derivative of f when f ( x ) = 4 x cos 2 x − 3 sin2 x . 1. f ′ ( x ) = 6 cos 2 x − 2 x sin2 x 2. f ′ ( x ) = − 8 x sin2 x − 2 cos2 x correct 3. f ′ ( x ) = 8 x sin2 x − 2 cos2 x 4. f ′ ( x ) = − 6 cos 2 x + 8 x sin2 x 5. f ′ ( x ) = − 8 x sin2 x − 6 cos2 x Explanation: Using formulas for the derivatives of sine and cosine together with the Product and Chain Rules, we see that f ′ ( x ) = 4 cos 2 x − 8 x sin2 x − 6 cos 2 x = − 8 x sin2 x − 2 cos2 x ....
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