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Unformatted text preview: Version 072 – Exam 2 – Zheng – (57255) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points When the graph of f is 2 4 6 8 10 − 2 − 4 − 6 2 4 6 − 2 − 4 − 6 which of the following is the graph of f ′′ ? 1. 4 8 − 4 4 − 4 2. 4 8 − 4 4 − 4 3. 4 8 − 4 4 − 4 4. 4 8 − 4 4 − 4 correct 5. 4 8 − 4 4 − 4 Explanation: The graph of f has exactly one point at which it changes concavity, so the graph of f ′′ has exactly one xintercept. This rules out the parabola. But the graph of f changes concavity at x = 2, so the graph of f ′′ must be one of the straight lines having x = 2 as xintercept. This rules out two of the lines, leaving just two lines each with the same x intercept but slopes of opposite sign. An inspection of the concavity of the graph of f to the left and right of x = 2 thus shows that Version 072 – Exam 2 – Zheng – (57255) 2 2 4 6 8 10 − 2 − 4 − 6 2 4 6 − 2 − 4 − 6 must be the graph of f ′′ . 002 10.0 points Find the equation of the tangent line to the graph of f ( x ) = (5 x + 11) 1 4 at the point P = (1 , f (1)) on the graph of f . 1. y + 5 32 x = 59 32 2. y = 5 32 x + 60 3. y + 5 x = 60 4. y = 5 32 x + 59 32 correct 5. y = 5 x + 60 6. y = 5 x + 59 32 Explanation: When x = 1 the corresponding value of f is f (1) = 2, so P = (1 , 2). On the other hand, by the Power rule, f ′ ( x ) = 5 4 (5 x + 11) − 3 4 . Consequently, the slope of the tangent line at P is given by f ′ (1) = 5 4(16) 3 4 = 5 32 , so by the point slope formula, the equation of the tangent line at P is y − 2 = 5 32 ( x − 1) . After simplification this becomes y = 5 32 x + 59 32 . 003 10.0 points Determine f ′ ( x ) when f ( x ) = x − 1 √ x 2 + 2 . 1. f ′ ( x ) = 2 − x ( x 2 + 2) 1 / 2 2. f ′ ( x ) = x − 2 ( x 2 + 2) 1 / 2 3. f ′ ( x ) = 2 + x ( x 2 + 2) 3 / 2 correct 4. f ′ ( x ) = 2 + x ( x 2 + 2) 1 / 2 5. f ′ ( x ) = x − 2 ( x 2 + 2) 3 / 2 6. f ′ ( x ) = 2 − x ( x 2 + 2) 3 / 2 Explanation: By the Product and Chain Rules, f ′ ( x ) = 1 ( x 2 + 2) 1 / 2 − 2 x ( x − 1) 2( x 2 + 2) 3 / 2 = ( x 2 + 2) − x ( x − 1) ( x 2 + 2) 3 / 2 . Consequently, f ′ ( x ) = 2 + x ( x 2 + 2) 3 / 2 . Version 072 – Exam 2 – Zheng – (57255) 3 (Note: the Quotient Rule could have been used, but it’s simpler to use the Product Rule.) keywords: derivative, square root, chain rule, product rule 004 10.0 points Find the second derivative of f when f ( x ) = − 2 cos 2 x − 5 cos 2 x . 1. f ′′ ( x ) = 18 sin2 x 2. f ′′ ( x ) = − 9 sin 2 x 3. f ′′ ( x ) = − 18 sin 2 x 4. f ′′ ( x ) = − 18 cos 2 x 5. f ′′ ( x ) = 9 cos 2 x 6. f ′′ ( x ) = 18 cos 2 x correct Explanation: Differentiating once we see that f ′ ( x ) = 4 sin2 x + 10sin x cos x ....
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 Fall '08
 schultz
 Calculus, Derivative, lim, 072 – Exam

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