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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 109

Find the linearization of ƒ(x) = √(1 + x) + sin x - 0.5 at x = 0.

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Start by identifying the function ƒ(x) = √(1 + x) + sin x - 0.5. We need to find its linearization at x = 0.
Recall that the linearization of a function at a point x = a is given by L(x) = ƒ(a) + ƒ'(a)(x - a). Here, a = 0.
Calculate ƒ(0) by substituting x = 0 into the function: ƒ(0) = √(1 + 0) + sin(0) - 0.5.
Find the derivative ƒ'(x). Use the derivative rules: the derivative of √(1 + x) is 1/(2√(1 + x)), and the derivative of sin x is cos x.
Evaluate ƒ'(0) by substituting x = 0 into the derivative: ƒ'(0) = 1/(2√(1 + 0)) + cos(0).

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Linearization

Linearization is the process of approximating a function near a specific point using its tangent line. This involves calculating the function's value and its derivative at that point. The linear approximation can be expressed as L(x) = f(a) + f'(a)(x - a), where 'a' is the point of tangency.
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Derivative

The derivative of a function at a point measures the rate at which the function's value changes as its input changes. It is defined as the limit of the average rate of change of the function as the interval approaches zero. In the context of linearization, the derivative at the point of interest provides the slope of the tangent line.
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Function Evaluation

Function evaluation involves substituting a specific value into a function to determine its output. For linearization, it is essential to evaluate the function and its derivative at the point of interest, which in this case is x = 0. This step is crucial for constructing the linear approximation.
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Evaluating Composed Functions