Explanation Given: A function satisfying the hypothesis of Mean Value Theorem on [ 0 , 4 ] for which there are exactly three values c ∈ ( 0 , 4 ) satisfying the conclusion of the theorem Calculation: If function ‘f’ is continuous on [ a , b ] and differentiable on (a, b), then there exists at least one value c ∈ ( a , b ) such that, f ' ( c ) = f ( b ) − f ( a ) b − a Consider the function, f ( x ) = x 2 − 2 It is a polynomial and therefore, is continuous and differentiable; in particular it is continuous on [0, 2] and differentiable on (0, 2). That is why; the Mean Value Theorem applies to this function on the interval [0, 2]. The slope of the line from (0, f (0)) to (2, f (2)) is f ( 2 ) − f ( 0 ) 2 − 0 = ( 2 2 − 2 ) − ( 0 2 − 2 ) 2 = 4 − 2 + 2 2 = 2 By the Mean Value Theorem, there must exist at least on

5th Edition

ISBN: 9781429241861

Author: Laura Taalman, Peter Kohn

Publisher: W. H. Freeman

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Chapter 3.1, Problem 16C

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The graph for the function satisfying the given description

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Chapter 3.1, Problem 15C

Chapter 3.1, Problem 17C

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Question 2 Let F be a solenoidal vector field, suppose V × F = (-8xy + 12z², −9x² + 4y² + 9z², 6y²), and let (P,Q,R) = V²F(.725, —.283, 1.73). Then the value of sin(2P) + sin(3Q) + sin(4R) is -2.024 1.391 0.186 -0.994 -2.053 -0.647 -0.588 -1.851 1 pts

1 pts Let F and G be vector fields such that ▼ × F(0, 0, 0) = (0.76, -9.78, 3.29), G(0, 0, 0) = (−3.99, 6.15, 2.94), and G is irrotational. Then sin(5V (F × G)) at (0, 0, 0) is Question 1 -0.246 0.072 -0.934 0.478 -0.914 -0.855 0.710 0.262 .

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Chapter 3 Solutions

Calculus

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