FINAL REVIEW 2

# FINAL REVIEW 2 - ngac(xtn62 – NewFinalReview02 –...

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Unformatted text preview: ngac (xtn62) – NewFinalReview02 – Gilbert – (58365) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine lim x →∞ 5 x 2 − 3 x + 2 7 + 6 x − 2 x 2 . 1. limit = 0 2. none of the other answers 3. limit = − 5 2 correct 4. limit = 5 4 5. limit = ∞ Explanation: Dividing the numerator and denominator by x 2 we see that 5 x 2 − 3 x + 2 7 + 6 x − 2 x 2 = 5 − 3 x + 2 x 2 7 x 2 + 6 x − 2 . On the other hand, lim x →∞ 1 x = lim x →∞ 1 x 2 = 0 . By Properties of limits, therefore, the limit = − 5 2 . 002 10.0 points Find the derivative of f when f ( x ) = 3 tan2 x cos 3 2 x . 1. f ′ ( x ) = 3 cos 2 x (3 sin 2 2 x − 1) 2. f ′ ( x ) = 6 cos 2 x (1 − 3 sin 2 2 x ) correct 3. f ′ ( x ) = 6 cos2 x (1 − 3 cos 2 2 x ) 4. f ′ ( x ) = 3 cos2 x (1 − 3 cos 2 2 x ) 5. f ′ ( x ) = 6 cos2 x (1 + 3 sin 2 2 x ) Explanation: Using the fact that d dx tan x = 1 cos 2 x , d dx cos x = − sin x, together with the Chain rule, we obtain f ′ ( x ) = 6 cos 2 2 x cos 3 2 x − 18tan2 x cos 2 2 x sin2 x. Consequently, f ′ ( x ) = 6 cos2 x (1 − 3 sin 2 2 x ) . Notice that the problem slightly simpler if we observe that tan2 x = sin2 x cos 2 x , so that f ( x ) = 3 sin2 x cos 2 2 x, and then differentiate this function using the known derivatives of sin2 x and cos 2 x . 003 10.0 points Find f ′ ( x ) when f ( x ) = parenleftBig x 4 x 2 + 1 parenrightBig 5 . 1. f ′ ( x ) = x 4 (1 − 4 x ) (4 x 2 + 1) 5 2. f ′ ( x ) = x 4 (1 − 4 x 2 ) (4 x 2 + 1) 6 3. f ′ ( x ) = 5 x 4 (1 − 4 x ) (4 x 2 + 1) 5 ngac (xtn62) – NewFinalReview02 – Gilbert – (58365) 2 4. f ′ ( x ) = 5(1 − 4 x 2 ) (4 x 2 + 1) 6 5. f ′ ( x ) = 5(1 − 4 x 2 ) (4 x 2 + 1) 5 6. f ′ ( x ) = 5 x 4 (1 − 4 x 2 ) (4 x 2 + 1) 6 correct Explanation: By the Power rule, f ′ ( x ) = 5 parenleftBig x 4 x 2 + 1 parenrightBig 4 × d dx parenleftBig x 4 x 2 + 1 parenrightBig . But, by the Quotient rule, d dx parenleftBig x 4 x 2 + 1 parenrightBig = (4 x 2 + 1) − 8 x 2 (4 x 2 + 1) 2 . Consequently, f ′ ( x ) = 5 x 4 (1 − 4 x 2 ) (4 x 2 + 1) 6 . 004 10.0 points Find the slope of the tangent line to the graph of x 3 + 2 y 3 − xy = 0 at the point P (1 , − 1). 1. slope = − 2 3 2. slope = − 5 4 3. slope = 3 2 4. slope = − 4 5 correct 5. slope = 4 5 6. slope = 5 4 Explanation: Differentiating implicitly with respect to x we see that 3 x 2 + 6 y 2 dy dx − y − x dy dx = 0 . Consequently, dy dx = − 3 x 2 − y 6 y 2 − x . Hence at P (1 , − 1) slope = dy dx vextendsingle vextendsingle vextendsingle P = − 4 5 . 005 10.0 points Determine dy/dx when 5 cos x sin y = 4 ....
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FINAL REVIEW 2 - ngac(xtn62 – NewFinalReview02 –...

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