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In calculus, a derivative provides us with the rate at which a function is changing at any given point. For a function like \( f(x) = \sqrt[3]{x} \), knowing the derivative allows us to understand its behavior and how it responds to small changes in \( x \). When we find the derivative of the function, we're essentially determining how the function 'moves' or 'slopes' as \( x \) changes. This is crucial for linearization because it helps us approximate the function using a line at a point.To find the derivative of \( \sqrt[3]{x} \), we first rewrite it in terms of exponents so that \( f(x) = x^{1/3} \). This transformation makes it easier to apply calculus rules like the power rule, which we'll explain next. Derivatives can seem tough at first, but they are a powerful tool that makes complex curves easier to work with.

The power rule is a simple and effective formula for differentiating functions of the form \( x^n \), where \( n \) is any real number. The rule states that the derivative of \( x^n \) is \( nx^{n-1} \). This rule significantly simplifies the process of finding derivatives, making it straightforward to handle functions like \( f(x) = x^{1/3} \).For our exercise, applying the power rule gives \( f'(x) = \frac{1}{3}x^{-2/3} \). Essentially, you bring the exponent \( 1/3 \) down in front and decrease the power by one. The resulting derivative, \( \frac{1}{3}x^{-2/3} \), provides the rate of change of \( \sqrt[3]{x} \) at any x-value, which is a key part of forming the linear approximation. The power rule is a fundamental and widely used technique in calculus.

Evaluating a function means calculating the value of the function at a specific point. When we're asked to find a linearization of \( f(x) = \sqrt[3]{x} \) at \( a = 8.5 \), we first need to evaluate the function at a manageable integer nearby, which is \( x = 8 \).Function evaluation is straightforward: you simply plug the given x-value into the function. For instance, \( f(8) = \sqrt[3]{8} = 2 \). Similarly, evaluating the derivative function at this point gives us \( f'(8) = \frac{1}{12} \). These evaluations are critical because they provide the necessary components for constructing the linear approximation: the value of the function and its slope at the point of interest.

The cubic root is a special type of root function where you're finding a number that, when multiplied by itself three times, equals the original number. In mathematical terms, if \( y^3 = x \), then \( y = \sqrt[3]{x} \).In our exercise, the function \( f(x) = \sqrt[3]{x} \) involves cubic roots. A notable example from the problem is evaluating the cubic root at \( x = 8 \), which results in \( \sqrt[3]{8} = 2 \). Cubic roots can initially appear challenging, but they are just an extension of the more familiar square roots. Understanding cubic roots is essential for working with functions like \( \sqrt[3]{x} \), especially when you need to linearize or differentiate them. As such, knowing cube roots helps simplify many mathematical expressions and functions.