# Exam2_3 5 - Version 208 – Exam 2 – Radin – (58305) 5...

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Unformatted text preview: Version 208 – Exam 2 – Radin – (58305) 5 7 4. 5. 6. x = 1 7. x = −5 , 8. x = 5 7 B and C only C only 8. 5. x = −1 B only 7. 4. x = 5 , 1 All of them 6. 3. x = − None of them Explanation: The graph of f is 5 7 4 Explanation: As f is a polynomial, it is diﬀerentiable everywhere, so the only critical values of f are those values of c where f ′ (c) = 0. Now by the Product and Chain Rules, ′ 6 f (x) = (5 − x) − 6x(5 − x) 2 −4 5 = (5 − x)5 (5 − x − 6x) Consequently, the critical values of f all occur at 5 x = 5, . 7 011 10.0 points f ′ ( x) = − Consider the following properties: A. derivative exists for all x = 0 ; B. has local minimum at x = 0 ; C. concave up on (−∞, 0) ∪ (0, ∞) ; Which does f have? A and C only correct 2. A and B only 3. A only 2 4 2 , 3x1/3 f ′′ (x) = 2 . 9x4/3 Consequently, A. has: ( f ′ (x) = −(2/3)x−1/3 , x = 0); B. not have: (see graph); C. has: ( f ′′ (x) > 0, x = 0). Let f be the function deﬁned by f (x) = 5 − x2/3 . −2 On the other hand, after diﬀerentiation, = (5 − x)5 (5 − 7x) . 1. 5 012 10.0 points Determine the interval(s) in [−π, π ] on which f (x) = 2 sin x − x is decreasing. 1. 2. 3. 4. π 5π −, 66 ππ −, 66 5π , −π , − 6 2π , −π , − 3 π ,π 6 2π ,π 3 ...
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## This note was uploaded on 02/14/2012 for the course MATH 408 K taught by Professor Clark,c.w./hoy,r.r during the Spring '08 term at University of Texas.

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