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Chapter 2: Problem 49

Find \(\Delta y\) for the given values of \(x_{1}\) and \(x_{2}\). $$ y=\frac{3}{x+1}, x_{1}=2.34, x_{2}=2.31 $$

Short Answer

Expert verified
\( \Delta y = 0.008 \)

Step by step solution

01

Determine the Expression for \( y \)

Given the function \( y = \frac{3}{x+1} \), we will use this to find \( y \) at both \( x_1 \) and \( x_2 \).
02

Calculate \( y_1 \)

Substitute \( x_1 = 2.34 \) into the function: \[ y_1 = \frac{3}{2.34 + 1} = \frac{3}{3.34} \approx 0.898 \]
03

Calculate \( y_2 \)

Substitute \( x_2 = 2.31 \) into the function: \[ y_2 = \frac{3}{2.31 + 1} = \frac{3}{3.31} \approx 0.906 \]
04

Compute \( \Delta y \)

Find the change in \( y \) by calculating \( \Delta y = y_2 - y_1 \): \[ \Delta y = 0.906 - 0.898 = 0.008 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Evaluation
Function evaluation is the process of determining the output of a mathematical function for specific inputs. In simple terms, when you evaluate a function, you replace the variable in the function with a given number and solve. For example, consider the function given in our problem, \( y = \frac{3}{x+1} \). Here is how function evaluation works when we are given specific values:
  • To find \( y_1 \), we use the first value: substitute \( x_1 = 2.34 \) into the function. This means we replace every \( x \) in \( y = \frac{3}{x+1} \) with 2.34, resulting in \( y_1 = \frac{3}{2.34+1} \).
  • To find \( y_2 \), we utilize the second value: substitute \( x_2 = 2.31 \) into the function, giving \( y_2 = \frac{3}{2.31+1} \).
Function evaluation is a fundamental skill in calculus and algebra. It helps explore the behavior of functions at different points.
Difference Quotient
The difference quotient is a method for finding the average rate of change of a function over an interval. It's an essential concept in calculus, forming the foundation for derivatives, which describe how a function changes at any given point. To understand this, think of the difference quotient as a bridge between algebra and calculus.
  • In our exercise, the difference quotient helps determine how \( y \) changes between two points, \( x_1 \) and \( x_2 \).
  • The formula for the difference quotient is \( \frac{y_2 - y_1}{x_2 - x_1} \), where \( y_1 \) and \( y_2 \) are the function values at \( x_1 \) and \( x_2 \), respectively.
  • This gives insight into how quickly or slowly the function \( y = \frac{3}{x+1} \) responds to changes in \( x \).
By calculating \( \Delta y \), which represents the change in \( y \) values, we highlight the useful application of the difference quotient in understanding these changes.
Variable Substitution
Variable substitution involves replacing a variable in an expression or equation with a specific value or another expression. It is a crucial tool in mathematical problem-solving, simplifying complex equations and allowing for function evaluation. Here's a breakdown of variable substitution at work in our exercise:
  • First, we cure our function \( y = \frac{3}{x+1} \) with specific values; these are \( x_1 = 2.34 \) and \( x_2 = 2.31 \).
  • We perform substitution to find \( y_1 \) and \( y_2 \): for instance, plugging \( 2.34 \) and \( 2.31 \) in place of \( x \) to calculate the corresponding \( y \) values.
Such substitution makes it possible to evaluate functions at specific points and analyze the resulting behavior. It's especially helpful in calculus when exploring the limits, derivatives, and integrals of functions.

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Most popular questions from this chapter

A wire of length 8 centimeters is such that the mass between its left end and a point \(x\) centimeters to the right is \(x^{3}\) grams (Figure 12). (a) What is the average density of the middle 2 -centimeter segment of this wire? Note: Average density equals mass/length. (b) What is the actual density at the point 3 centimeters from the left end?

Prove that the derivative of an odd function is an even function and that the derivative of an even function is an odd function.

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