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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 4, Problema 3.9.14c

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


c. f(x) = 1/√(1 + x)

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Identify the function f(x) = 1/√(1 + x) and recognize that it can be rewritten as (1 + x)^(-1/2).
Use the linear approximation formula (1 + x)ᵏ ≈ 1 + kx, where k is the exponent of the expression. Here, k = -1/2.
Substitute k = -1/2 into the linear approximation formula to get (1 + x)^(-1/2) ≈ 1 - (1/2)x.
This approximation is valid for values of x near zero, providing a simpler expression to estimate f(x) without complex calculations.
Thus, the linear approximation for f(x) = 1/√(1 + x) near x = 0 is approximately 1 - (1/2)x.

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Linear Approximation

Linear approximation is a method used to estimate the value of a function near a given point using the tangent line at that point. For a function f(x), the linear approximation at x = a is given by L(x) = f(a) + f'(a)(x - a). This technique is particularly useful for simplifying complex functions near a specific point, often x = 0.
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The derivative of a function measures how the function's output value changes as its input changes. It is a fundamental concept in calculus, representing the slope of the tangent line to the function at any given point. For the function f(x) = 1/√(1 + x), finding the derivative is crucial for applying linear approximation, as it provides the rate of change needed for the approximation.
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