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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 3, Problema 3.9.14a

Use the linear approximation (1 + x)ᵏ ≈ 1 + kx to find an approximation for the function f(x) for values of x near zero.


a. f(x) = (1 − x)⁶

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Identify the function f(x) = (1 - x)⁶ and recognize that it is in the form of (1 + x)ᵏ with k = 6 and x replaced by -x.
Apply the linear approximation formula (1 + x)ᵏ ≈ 1 + kx to the function. Here, substitute x with -x and k with 6.
The linear approximation becomes: (1 - x)⁶ ≈ 1 + 6(-x).
Simplify the expression: 1 + 6(-x) becomes 1 - 6x.
Thus, the linear approximation for f(x) = (1 - x)⁶ near x = 0 is f(x) ≈ 1 - 6x.

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