This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Version 103 – Exam 2 – gualdani – (57180) 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 correct 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 = 3, so the graph of f ′′ must be one of the straight lines having x = 3 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 = 3 thus shows that Version 103 – Exam 2 – gualdani – (57180) 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 + 2) 1 3 at the point P = (1 , f (1)) on the graph of f . 1. y + 6 x = 19 2. y = 1 2 x + 19 3. y + 1 2 x = 3 2 4. y = 6 x + 19 5. y = 6 x + 3 2 6. y = 1 2 x + 3 2 correct 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 ) = 2 (6 x + 2) − 2 3 . Consequently, the slope of the tangent line at P is given by f ′ (1) = 2 8 2 3 = 1 2 , so by the point slope formula, the equation of the tangent line at P is y − 2 = 1 2 ( x − 1) . After simplification this becomes y = 1 2 x + 3 2 . 003 10.0 points Determine f ′ ( x ) when f ( x ) = x − 2 √ x 2 + 1 . 1. f ′ ( x ) = 2 x − 1 ( x 2 + 1) 3 / 2 2. f ′ ( x ) = 1 + 2 x ( x 2 + 1) 1 / 2 3. f ′ ( x ) = 1 + 2 x ( x 2 + 1) 3 / 2 correct 4. f ′ ( x ) = 1 − 2 x ( x 2 + 1) 3 / 2 5. f ′ ( x ) = 1 − 2 x ( x 2 + 1) 1 / 2 6. f ′ ( x ) = 2 x − 1 ( x 2 + 1) 1 / 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 ) = 1 + 2 x ( x 2 + 1) 3 / 2 . Version 103 – Exam 2 – gualdani – (57180) 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 ) = − 3 cos 2 x + 2 cos 2 x . 1. f ′′ ( x ) = 8 sin 2 x 2. f ′′ ( x ) = 8 cos 2 x correct 3. f ′′ ( x ) = 4 cos 2 x 4. f ′′ ( x ) = − 8 sin 2 x 5. f ′′ ( x ) = − 8 cos 2 x 6. f ′′ ( x ) = − 4 sin 2 x Explanation: Differentiating once we see that f ′ ( x ) = 6 sin2 x − 4 sin x cos x ....
View
Full
Document
This note was uploaded on 05/12/2010 for the course M 408k taught by Professor Schultz during the Spring '08 term at University of Texas.
 Spring '08
 schultz

Click to edit the document details