Problem 9 - Thomas Calculus

Question

You want linearizations that will replace the functions in Exercises \(5-10\) over intervals that include the given points \(x_{0} .\) To make your subsequent work as simple as possible, you want to center each linearization not at \(x_{0}\) but at a nearby integer \(x=a\) at which the given function and its derivative are easy to evaluate. What linearization do you use in each case? $$ f(x)=\sqrt[3]{x}, \quad x_{0}=8.5 $$

Step-by-Step Solution

Answer

Linearization: \( L(x) = \frac{1}{12}x + \frac{4}{3} \)

Step 1: Find Suitable Integer Center

We need to choose a nearby integer to \( x_0 = 8.5 \) for the linearization. The closest integer is \( a = 8 \). It is a good choice because evaluating the function and its derivative at \( x = 8 \) is simple.

Step 2: Evaluate the Function at Chosen Center

For the function \( f(x) = \sqrt[3]{x} \), calculate \( f(a) \) where \( a = 8 \). Therefore, \( f(8) = \sqrt[3]{8} = 2 \).

Step 3: Determine the Derivative

Find the derivative of \( f(x) = \sqrt[3]{x} \). The derivative is \( f'(x) = \frac{1}{3}x^{-\frac{2}{3}} \).

Step 4: Evaluate the Derivative at Chosen Center

Evaluate \( f'(x) \) at \( x = 8 \). Thus, \( f'(8) = \frac{1}{3} \, 8^{-\frac{2}{3}} = \frac{1}{3} \, \frac{1}{4} = \frac{1}{12} \).

Step 5: Write the Linearization

The linearization of the function at \( x = a \) is given by \( L(x) = f(a) + f'(a)(x - a) \). Substituting the values, \( L(x) = 2 + \frac{1}{12}(x - 8) \). Simplify to get the linearization: \( L(x) = 2 + \frac{1}{12}x - \frac{8}{12} \).

Step 6: Simplify the Linearization Equation

Simplify \( L(x) = 2 + \frac{1}{12}x - \frac{8}{12} \) to \( L(x) = \frac{16}{12} + \frac{1}{12}x = \frac{4}{3} + \frac{1}{12}x \). Thus, the linearization is \( L(x) = \frac{1}{12}x + \frac{4}{3} \).

Key Concepts

Function Evaluation

Function evaluation is the process of determining the output of a function at a specific point. For the problem at hand, the function is given by \( f(x) = \sqrt[3]{x} \). To evaluate this at our chosen integer center, \( a = 8 \), we simply plug 8 into the function. So, \( f(8) = \sqrt[3]{8} \).
This calculation results in \( f(8) = 2 \), which means when \( x \) is 8, the function outputs 2. Finding a value that simplifies easily helps later steps. Therefore, choosing an integer like 8 is advantageous.
Choosing a nearby integer such as 8 allows us to focus on straightforward math, avoiding unnecessary complexity while evaluating.

Derivative Calculation

The derivative calculation is essential for linearization, which involves finding not just the value of a function at a particular point, but also how it changes around that point. For \( f(x) = \sqrt[3]{x} \), we calculate the derivative as \( f'(x) = \frac{1}{3}x^{-\frac{2}{3}} \).
This derivative tells us how fast the function \( f(x) \) changes at any given value of \( x \). After calculating the general form of the derivative, the next step is evaluating it at the chosen integer center.
For \( x = 8 \), the derivative becomes \( f'(8) = \frac{1}{3} \, 8^{-\frac{2}{3}} = \frac{1}{12} \). This results in a straightforward value that helps shape the linearization equation.

Integer Center

Choosing an integer center close to \( x_0 = 8.5 \) simplifies the entire process of linearization. The idea is to select a value \( a \) that makes calculation easy, both for the function and its derivative.
In this exercise, the integer center selected is \( a = 8 \). This is because it is the nearest whole number to 8.5, and computing \( \sqrt[3]{8} \) is simple since we get a neat result of 2.
The integer center serves as the pivotal point around which the approximation of the linear function is made, ultimately streamlining the mathematics involved in deriving the linearization.

Simplifying Expressions

Once we have the linearization formula \( L(x) = f(a) + f'(a)(x - a) \), we can substitute the evaluated values. Initially, our result was \( L(x) = 2 + \frac{1}{12}(x - 8) \).
Subsequent simplification involves distributing and combining like terms in the linear equation. So, \( L(x) = 2 + \frac{1}{12}x - \frac{8}{12} \).

• First, reduce \( \frac{8}{12} \) to \( \frac{2}{3} \), yielding \( L(x) = 2 + \frac{1}{12}x - \frac{2}{3} \).
• Then, simplify \( 2 - \frac{2}{3} \) to get \( \frac{4}{3} \).

This results in the final simplified linearization \( L(x) = \frac{1}{12}x + \frac{4}{3} \). Simplification ensures the expression is neat and easy to interpret.