Problem 2 - Essential Calculus

Question

In Exercises \(1-4,\) find a first-degree polynomial function \(P_{1}\) whose value and slope agree with the value and slope of \(f\) at \(x=c .\) Use a graphing utility to graph \(f\) and \(P_{1} .\) What is \(P_{1}\) called? $$ f(x)=\frac{4}{\sqrt[3]{x}}, \quad c=8 $$

Step-by-Step Solution

Answer

The first-degree polynomial function \(P_{1}\) that has the same value and slope as function \(f\) at \(x=8\) is \(P_{1}(x) = -\frac{1}{12}x + \frac{8}{3}\). Also known as the tangent line at \(x=c\), in this case at \(x=8\).

Step 1: Determine the value of \(f\) at \(x=c\)

First, substitute \(x=c\) into \(f\) to get its value at \(c\). It will be \(f(c) = \frac{4}{\sqrt[3]{8}} = 2\).

Step 2: Determine the slope of \(f\) at \(x=c\)

We first need to find \(f'\), the derivative of \(f(x)\). Using the rule for differentiating \(x^{-n}\), we get \(f'(x) = - \frac{4} {3x^{\frac{4}{3}}}\). Now substitute \(x=c\) into the derivative to get the slope at point \(c\) , which will be: \(f'(c) = -\frac{4}{3*\sqrt[3]{8^4}} = -\frac{1}{12}\).

Step 3: Find the first-degree polynomial function \(P_{1}\)

A first degree polynomial function is of the form \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. We want this polynomial to have the same value and slope as function \(f\) at \(x=c\), therefore we substitute these values into the equation to solve for \(b\). This gives: \(2 = -\frac{1}{12} * 8 + b\). Solving for \(b\) we get \(b = 2 + \frac{2}{3} = \frac{8}{3}\), Hence the polynomial is \(P_{1}(x) = -\frac{1}{12}x + \frac{8}{3}\).

Key Concepts

Polynomial Function

Polynomial functions are a fundamental class of functions in algebra. These functions take on the form of an expression consisting of variables, coefficients, and exponents. The standard form of a polynomial function is written as follows:

• For a single variable: \[ P(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \] where \(a_n, a_{n-1}, \ldots, a_0\) are coefficients and \(n\) is a non-negative integer.
• The degree of the polynomial is the highest power of the variable \(x\) with a non-zero coefficient.

In the context of the exercise you provided, the function we are working with is a first-degree polynomial function, which we will explore further below.

Derivative

The derivative of a function represents its rate of change or the slope at any given point along its curve. It is a crucial concept in calculus. Derivatives provide insights into how a function changes as its input changes.
To find the derivative of a function, you apply rules such as the power rule, product rule, or chain rule, depending on the form of the function. For the exercise you are working on, the function given is \(f(x)=\frac{4}{x^{1/3}}\) and we needed to find its derivative.

• The derivative was calculated as: \[ f'(x) = - \frac{4} {3x^{\frac{4}{3}}} \]
• Evaluating this derivative at \(x=c\), which is 8, provided the slope for constructing the first-degree polynomial approximation.

Graphing Utility

Graphing utilities are powerful tools used to visualize mathematical functions. They provide an interface where you can input functions and analyze their behavior visually. Utilizing graphing utilities, such as calculator software or online graphers, can help confirm solutions and understand function structures.

• They allow for inputting complex functions and automatically generate their graphs.
• They can illustrate both the original function and its approximation (in our case, the first-degree polynomial function).

In this exercise, graphing the function \(f(x)=\frac{4}{\sqrt[3]{x}}\) alongside the first-degree polynomial \(P_1(x)\) helps visualize how well the linear approximation fits at \(x = c = 8\). This can be valuable to see the agreement of values and slopes at that point.

First-Degree Polynomial

A first-degree polynomial is the simplest form of a polynomial, described by the equation of a straight line: \[ y = mx + b \] where \(m\) is the slope, and \(b\) is the y-intercept. This is often referred to as a linear function. Importantly, such a polynomial can be used for linear approximations in calculus.

• In the context of tangent line approximation, the objective is to match both the slope and the value of a more complex function at a particular point.
• For this problem, we found \(P_{1}(x) = -\frac{1}{12}x + \frac{8}{3}\), which serves as the tangent line approximation for our function at \(x = 8\).

This function simplifies the behavior of the original function at a specific interval, offering a linear model with matching initial conditions.