Problem 1
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? $$ f(x)=\frac{4}{\sqrt{x}}, \quad c=1 $$
Step-by-Step Solution
Verified Answer
The first-degree polynomial function \(P_1(x)\) is \(P_1(x) = 4 - 2 * (x - 1) = 6 - 2x\). This function is called the tangent line or linear approximation to \(f(x)\) at \(x=c\).
1Step 1: Find the Function Value at a Given Point
Substitute \(x=c\) into the function \(f (x) = \frac{4}{\sqrt{x}}\) to find \(f(c)\). This gives \(f(1) = \frac{4}{\sqrt{1}} = 4\).
2Step 2: Find the Function's Derivative
To find the slope of the function at the point \(x=c\), derive the function. The derivative of \(f(x) = \frac{4}{\sqrt{x}}\) is \(f'(x) =\frac{-2}{x^{3/2}}\).
3Step 3: Find the Derivative Value at the Given Point
Substitute \(x=c\) into the derivative \(f'(x) =\frac{-2}{x^{3/2}}\) to find \(f'(c)\). This gives \(f'(1) = \frac{-2}{1^{3/2}} = -2\).
4Step 4: Define the First-Degree Polynomial
The first-degree polynomial function \(P_1\) that matches \(f\) in both value and slope at \(x=c\) is \(P_1 (x) = f(c) + f'(c) * (x-c)\). Substitute \(c=1, f(c) = 4\) and \(f'(c) = -2\) into the formula to find \(P_1 (x)\). This gives \(P_1(x) = 4 - 2 * (x - 1)\).
Key Concepts
First-Degree PolynomialFunction DerivativeFunction ValueSlope Calculation
First-Degree Polynomial
A first-degree polynomial is essentially a linear function of the form \( P(x) = mx + b \), where \( m \) and \( b \) are constants. This polynomial is called 'first-degree' because the highest power of \( x \) is one. It represents a straight line when graphed.
In the context of linear approximation, we use a first-degree polynomial to approximate the value and slope of a more complex function \( f(x) \) at a particular point \( x = c \).
Specifically, the goal is to create a polynomial, \( P_1(x) \), that matches the original function \( f(x) \) in both value and slope at the specified point. In practical terms, \( P_1(x) \) provides an easy way to estimate the behavior of the function nearby \( x = c \).
In the context of linear approximation, we use a first-degree polynomial to approximate the value and slope of a more complex function \( f(x) \) at a particular point \( x = c \).
Specifically, the goal is to create a polynomial, \( P_1(x) \), that matches the original function \( f(x) \) in both value and slope at the specified point. In practical terms, \( P_1(x) \) provides an easy way to estimate the behavior of the function nearby \( x = c \).
Function Derivative
Derivatives are essential in understanding how functions behave. The derivative of a function \( f(x) \), denoted as \( f'(x) \), represents the rate of change or "slope" of the function at any given point. Calculating derivatives involves using rules such as the power rule, product rule, and quotient rule.
In this exercise, the derivative of the function \( f(x) = \frac{4}{\sqrt{x}} \) was calculated. The derivative \( f'(x) = \frac{-2}{x^{3/2}} \) tells us how steep the curve of \( f(x) \) is at different values of \( x \).
This slope information is crucial for constructing our first-degree polynomial, as the derivative value at \( x = c \) defines how steep the approximating line should be at that point.
In this exercise, the derivative of the function \( f(x) = \frac{4}{\sqrt{x}} \) was calculated. The derivative \( f'(x) = \frac{-2}{x^{3/2}} \) tells us how steep the curve of \( f(x) \) is at different values of \( x \).
This slope information is crucial for constructing our first-degree polynomial, as the derivative value at \( x = c \) defines how steep the approximating line should be at that point.
Function Value
A function value represents the output of a function for a particular input. For the function \( f(x) = \frac{4}{\sqrt{x}} \), the function value at \( x = c \) is \( f(c) \).
In the given problem, substituting \( c = 1 \) into the function gives \( f(1) = 4 \). The function value at this point is an essential component of the linear approximation, as it serves as the starting point or y-intercept for our approximating polynomial \( P_1(x) \).
When approximating a function with a linear polynomial, ensuring that the function value at a point agrees helps maintain the accuracy and real-world applicability of the approximation.
In the given problem, substituting \( c = 1 \) into the function gives \( f(1) = 4 \). The function value at this point is an essential component of the linear approximation, as it serves as the starting point or y-intercept for our approximating polynomial \( P_1(x) \).
When approximating a function with a linear polynomial, ensuring that the function value at a point agrees helps maintain the accuracy and real-world applicability of the approximation.
Slope Calculation
Calculating the slope is a key step in formulating a linear approximation. The slope of a function at a point \( x = c \) is determined using the derivative \( f'(c) \). In this exercise, the slope was found by evaluating the derivative at \( x = 1 \), resulting in \( f'(1) = -2 \).
The slope signifies the angle or steepness of the tangent line to the function at that particular point. This value is used in forming the first-degree polynomial \( P_1(x) \), as it determines how the approximating line should rise or fall as \( x \) changes.
Thus, by knowing both the slope and the function value, we can accurately draw the tangent line that best represents the behavior of the function near \( x = c \).
The slope signifies the angle or steepness of the tangent line to the function at that particular point. This value is used in forming the first-degree polynomial \( P_1(x) \), as it determines how the approximating line should rise or fall as \( x \) changes.
Thus, by knowing both the slope and the function value, we can accurately draw the tangent line that best represents the behavior of the function near \( x = c \).
Other exercises in this chapter
Problem 1
Use the Integral Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{1}{n+1} $$
View solution Problem 1
In Exercises 1 and 2 , state where the power series is centered. $$ \sum_{n=0}^{\infty} n x^{n} $$
View solution Problem 1
In Exercises \(1-10,\) use the definition to find the Taylor series (centered at \(c\) ) for the function. $$ f(x)=e^{2 x}, \quad c=0 $$
View solution Problem 1
In Exercises \(1-16,\) determine the convergence or divergence of the series. \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\)
View solution