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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
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Tutti i libri di testoHass 15th EditionCh. 3 - DerivativesProblema 3.9.11
Capitolo 3, Problema 3.9.11

Linearization for Approximation


In Exercises 7–12, find a linearization at a suitably chosen integer near a at which the given function and its derivative are easy to evaluate.


f(x) = ∛x, a = 8.5

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Identify the function f(x) = ∛x and the point a = 8.5 where you want to approximate the function using linearization.
Choose a nearby integer to a = 8.5 where the function and its derivative are easy to evaluate. In this case, choose a = 8 because ∛8 is easy to compute.
Find the derivative of the function f(x) = ∛x. The derivative f'(x) can be found using the power rule for derivatives. Rewrite ∛x as x^(1/3) and differentiate: f'(x) = (1/3)x^(-2/3).
Evaluate the function and its derivative at the chosen integer a = 8. Calculate f(8) = ∛8 = 2 and f'(8) = (1/3)(8)^(-2/3).
Use the linearization formula L(x) = f(a) + f'(a)(x - a) to approximate the function near x = 8.5. Substitute the values found: L(x) = 2 + f'(8)(x - 8).

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Linearization

Linearization is a method used to approximate a function near a given point using its tangent line. It involves finding the linear function that best approximates the original function at a specific point, typically where the function and its derivative are easy to evaluate. This technique simplifies complex functions into linear ones, making calculations more manageable.
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Linearization

Derivative

The derivative of a function represents the rate at which the function's value changes with respect to changes in its input. It is a fundamental concept in calculus used to find the slope of the tangent line at any point on the function. For linearization, the derivative is crucial as it provides the slope of the tangent line, which is used to construct the linear approximation.
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Derivatives

Choosing a Suitable Point

Choosing a suitable point near the given value is essential for effective linearization. This point should be an integer where the function and its derivative are easy to compute, ensuring the linear approximation is accurate and calculations are simplified. In this context, selecting a point like x = 8 for f(x) = ∛x allows for straightforward evaluation, as the cube root and its derivative are simple to calculate.
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Critical Points
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