Explanation Formula used: The increment of the equation is calculated as Δ y = f ( x + Δ x ) − f ( x ) . The differential of the function is calculated as d y = f ′ ( x ) d x . Calculation: The increment of y = 4 x 3 − 12 . Let, the function f ( x ) is: f ( x ) = 4 x 3 − 12 . Substitute, x = 2 in f ( x ) = 4 x 3 − 12 . f ( x ) = 4 x 3 − 12 f ( 2 ) = 4 × 2 3 − 12 = 32 − 12 = 20 The increment is calculated in x is: f ( x + Δ x ) = 4 ( x + Δ x ) 3 − 12 . Substitute, x = 2 and Δ x = 0.1 in f ( x + Δ x ) = 4 ( x + Δ x ) 3 − 12 . f ( x + Δ x ) = 4 ( x + Δ x ) 3 − 12 f ( 2 + 0.1 ) = 4 ( 2.1 ) 3 − 12 = 37.044 − 12 = 25.044 The increment in Δ y is: Substitute, f ( 2 ) = 20 and f ( 2.1 ) = 25.044 in Δ y = f ( x + Δ x ) − f ( x ) . Δ y = f ( 2.1 ) − f ( 2 ) = 25

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ISBN: 9780134437705

Author: Washington

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Differential Equations

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Chapter 24, Problem 37RE

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To evaluate: The difference of increment and differential of y=4x3−12, for, x=2 and Δx=0.1.

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Chapter 24, Problem 36RE

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The difference of increment and differential of y = 4 x 3 − 12 , for, x = 2 and Δ x = 0.1 .

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Chapter 24 Solutions