Problem 11 - University Calculus: Early Transcendentals

Question

Find a linearization at a suitably chosen integer near \(a\) at which the given function and its derivative are easy to evaluate. $$f(x)=\sqrt[3]{x}, \quad a=8.5$$

Step-by-Step Solution

Answer

The linearization at 8 is \(L(x) = 2 + \frac{1}{12}(x - 8)\).

Step 1: Choose a Suitable Integer Near a

To find the linearization of the function at a point near \(a = 8.5\), we should choose an integer where the function \(f(x) = \sqrt[3]{x}\) and its derivative are easy to evaluate. The nearest integer to 8.5 that fits this criterion is 8.

Step 2: Define the Linearization Formula

The general formula for the linearization of a function \(f(x)\) at a point \(a\) is given by:\[ L(x) = f(a) + f'(a)(x-a) \]

Step 3: Evaluate the Function at the Chosen Point

Evaluate \(f(8)\) for the function \(f(x) = \sqrt[3]{x}\):\[ f(8) = \sqrt[3]{8} = 2 \]

Step 4: Differentiate the Function

Find the derivative of \(f(x) = \sqrt[3]{x}\) using the power rule. Let \(f(x) = x^{1/3}\), then the derivative \(f'(x)\) is:\[ f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}} \]

Step 5: Evaluate the Derivative at the Chosen Point

Evaluate \(f'(8)\):\[ f'(8) = \frac{1}{3\sqrt[3]{8^2}} = \frac{1}{3\times 4} = \frac{1}{12} \]

Step 6: Construct the Linearization

Substitute the values into the linearization equation:\[ L(x) = 2 + \frac{1}{12}(x - 8) \]

Key Concepts

Cubic Root Function

When we talk about cubic root functions, we're referring to a type of function that assigns to each number its cube root. Specifically, for a function like \( f(x) = \sqrt[3]{x} \), it transforms the input \( x \) into the number that, when multiplied by itself three times, gives \( x \) back. This is simpler than it sounds:

• If \( x = 8 \), then \( f(x) = \sqrt[3]{8} = 2 \), since \( 2 \times 2 \times 2 = 8 \).
• The cubic root function is an example of a power function where the exponent is \( \frac{1}{3} \). This notation means that for any \( x \), we find its cubic root by raising it to the power of \( \frac{1}{3} \).

Cubic root functions have unique properties. They are defined for all real numbers, unlike square root functions, which only handle non-negative numbers. Additionally, cubic root functions grow more slowly than linear functions but never flatten out entirely. This makes them excellent for exploring small neighborhood changes, which is key in linearization.

Derivative Evaluation

Evaluating the derivative of a function is fundamental in calculus. It gives the rate at which the function's value changes with respect to changes in input. When we talk about derivatives in the context of the cubic root function, we are looking to understand how its slope behaves.
To evaluate the derivative of \( f(x) = \sqrt[3]{x} \) at a point, we use rules of differentiation like the power rule. Studying the derivative at specific points, like \( x = 8 \), informs us about the tangent line’s slope at that point.

• The derivative \( f'(x) = \frac{1}{3}x^{-2/3} \) shows how 'steep' the cubic root graph is for any value of \( x \).
• At \( x = 8 \), \( f'(8) = \frac{1}{12} \) tells us the exact slope of the function at this point, helping us construct a linear approximation or tangent line.

This tangent line provides an approximation of the function in the neighborhood of \( x = 8 \), denoting small changes effectively.

Power Rule Differentiation

Differentiating using the power rule is crucial when dealing with functions expressed as power expressions, like \( f(x) = x^{n} \), where \( n \) is any real number. The power rule states:

• If \( f(x) = x^{n} \), then the derivative \( f'(x) = nx^{n-1} \).

Let's apply this to the cubic root function \( f(x) = x^{1/3} \):

1. To find its derivative \( f'(x) \), use the power rule: \( f'(x) = \frac{1}{3}x^{-2/3} \).
2. Notice that each application of the power rule involves multiplying by the original power, \( n \), and reducing the power by one.

This rule is efficient and universally applicable to a variety of functions encountered in calculus.
While this differentiation process might seem formulaic, it equips us with a clear framework to tackle complex functions, providing a foundation to understand more intricate calculus problems. The importance of this rule cannot be understated, as it simplifies the task of finding derivatives and applying them in various contexts like linear approximation.