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Unformatted text preview: Version 055 – Exam 2 – seckin – (57195) 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 correct 2. 4 8 − 4 4 − 4 3. 4 8 − 4 4 − 4 4. 4 8 − 4 4 − 4 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 055 – Exam 2 – seckin – (57195) 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 ) = (6 x + 10) 1 4 at the point P = (1 , f (1)) on the graph of f . 1. y = 6 x + 29 16 2. y = 6 x + 59 3. y + 3 16 x = 29 16 4. y = 3 16 x + 29 16 correct 5. y + 6 x = 59 6. y = 3 16 x + 59 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 ) = 3 2 (6 x + 10) − 3 4 . Consequently, the slope of the tangent line at P is given by f ′ (1) = 3 2(16) 3 4 = 3 16 , so by the point slope formula, the equation of the tangent line at P is y − 2 = 3 16 ( x − 1) . After simplification this becomes y = 3 16 x + 29 16 . 003 10.0 points Determine f ′ ( x ) when f ( x ) = x − 2 √ x 2 − 1 . 1. f ′ ( x ) = 2 x − 1 ( x 2 − 1) 1 / 2 2. f ′ ( x ) = 1 − 2 x ( x 2 − 1) 3 / 2 3. f ′ ( x ) = 2 x − 1 ( x 2 − 1) 3 / 2 correct 4. f ′ ( x ) = 1 − 2 x ( x 2 − 1) 1 / 2 5. f ′ ( x ) = 1 + 2 x ( x 2 − 1) 1 / 2 6. f ′ ( x ) = 1 + 2 x ( x 2 − 1) 3 / 2 Explanation: By the Product and Chain Rules, f ′ ( x ) = 1 ( x 2 − 1) 1 / 2 − 2 x ( x − 2) 2( x 2 − 1) 3 / 2 = ( x 2 − 1) − x ( x − 2) ( x 2 − 1) 3 / 2 . Consequently, f ′ ( x ) = 2 x − 1 ( x 2 − 1) 3 / 2 . Version 055 – Exam 2 – seckin – (57195) 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 ) = − 4 cos 2 x − 3 cos 2 x . 1. f ′′ ( x ) = − 22 sin 2 x 2. f ′′ ( x ) = 22 cos 2 x correct 3. f ′′ ( x ) = − 11 sin 2 x 4. f ′′ ( x ) = 22 sin2 x 5. f ′′ ( x ) = 11 cos 2 x 6. f ′′ ( x ) = − 22 cos 2 x Explanation: Differentiating once we see that f ′ ( x ) = 8 sin2 x + 6 sin x cos x ....
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This note was uploaded on 09/12/2010 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas.
 Fall '08
 schultz
 Calculus

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