Explanation Formula used: The approximation f ( x ) ≈ F ( x ) is the linear approximation of f ( x ) near x = a . The linearization of f ( x ) at x = a is the function, L ( x ) = f ( a ) + f ′ ( a ) ( x − a ) ⋯ ( 1 ) Calculation: The given function is y = 300 0.0005 x 2 + 2 − 50 , 0 < x < 100 . Write y as a function f ( x ) . f ( x ) = 300 0.0005 x 2 + 2 − 50 Substitute, x = 50 in f ( x ) . f ( 50 ) = 300 0.0005 × 50 2 + 2 − 50 = 92.31 − 50 = 42.31 ≈ 42 Differentiate f ( x ) with respect to the x . f ′ ( x ) = d d x ( 300 0.0005 x 2 + 2 ) − d d x ( 50 ) = 300 d d x ( 1 0.0005 x 2 + 2 ) = 300 ( − 1 ( 0.0005 x 2 + 2 ) 2 ) ( 0.001 x ) = − 0.3 x ( 0.0005 x 2 + 2 ) 2 Hence, f ′ ( x ) = − 0.3 x ( 0.0005 x 2 + 2 ) 2 . Substitute, x = 50 in f ′ ( x )

Basic Technical Mathematics with Calculus (11th Edition)

11th Edition

ISBN: 9780134437736

Author: Allyn J. Washington, Richard Evans

Publisher: PEARSON

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Question

Chapter 24, Problem 77RE

To determine

To linearize: The function y=3000.0005x2+2−50 with a=50 and sketch that graphs of y and L(x).

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Chapter 24, Problem 76RE

Chapter 24, Problem 78RE

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Chapter 24 Solutions

Basic Technical Mathematics with Calculus (11th Edition)

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Ch. 24.1 - Prob. 9E Ch. 24.1 - Prob. 10E Ch. 24.1 - Prob. 11E Ch. 24.1 - Prob. 12E Ch. 24.1 - In Exercises 11–14, find the equations of the...Ch. 24.1 - Prob. 14E Ch. 24.1 - Prob. 15E Ch. 24.1 - Prob. 16E Ch. 24.1 - Prob. 17E Ch. 24.1 - Prob. 18E Ch. 24.1 - Prob. 19E Ch. 24.1 - Prob. 20E Ch. 24.1 - Prob. 21E Ch. 24.1 - Where does the normal line to the parabola y = x —...Ch. 24.1 - Prob. 23E Ch. 24.1 - Prob. 24E Ch. 24.1 - A certain suspension cable with supports on the...Ch. 24.1 - Prob. 26E Ch. 24.1 - Prob. 27E Ch. 24.1 - Prob. 28E Ch. 24.1 - Prob. 29E Ch. 24.1 - Prob. 30E Ch. 24.2 - In Example 1, let x1 = 0.3, and find x2. 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To linearize: The function y = 300 0.0005 x 2 + 2 − 50 with a = 50 and sketch that graphs of y and L ( x ) .

To linearize: The function y=3000.0005x2+2−50 with a=50 and sketch that graphs of y and L(x).

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Chapter 24 Solutions