Chapter 2: Problem 31
Find \(\Delta y\) and \(f^{\prime}(x) \Delta x\). Round to four and two decimal places, respectively. For \(y=f(x)=1 / x, x=1,\) and \(\Delta x=0.2\)
Short Answer
Expert verified
\( \Delta y = -0.1667 \), \( f^{}(x) \Delta x = -0.20 \).
Step by step solution
01
Understanding the Problem
We need to find the change in the function value, \( \Delta y \), and the product of the derivative, \( f'(x) \), with \( \Delta x \), given the function \( y = f(x) = \frac{1}{x} \), \( x = 1 \), and \( \Delta x = 0.2 \).
02
Calculate \( \Delta y \)
First, calculate \( f(x + \Delta x) \) and \( f(x) \). Then \( \Delta y = f(x + \Delta x) - f(x) \).Compute:\( f(1 + 0.2) = \frac{1}{1 + 0.2} = \frac{1}{1.2} \approx 0.8333\)\( f(1) = \frac{1}{1} = 1\)Thus,\( \Delta y = 0.8333 - 1 = -0.1667\).
03
Calculate the Derivative \( f^{}(x) \)
The derivative of \( f(x) = \frac{1}{x} \) is given by \( f^{}(x) = -\frac{1}{x^2} \).
04
Evaluate \( f^{}(x) \) at \( x = 1 \)
Substitute \( x = 1 \) into the derivative:\( f^{}(1) = -\frac{1}{1^2} = -1\).
05
Calculate \( f^{}(x) \Delta x \)
Multiply the derivative \( f^{}(x) \) by \( \Delta x = 0.2 \):\( f^{}(1) \Delta x = -1 \times 0.2 = -0.2\).
06
Rounding the Results
Round \( \Delta y \) to four decimal places and \( f^{}(x) \Delta x \) to two decimal places:- \( \Delta y = -0.1667 \) (already rounded to four decimal places)- \( f^{}(x) \Delta x = -0.20 \) (already rounded to two decimal places)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative
In calculus, the derivative is a concept that represents an instantaneous rate of change. We often describe it as the slope of a function at a given point. The derivative tells us how a function's output changes as we vary the input. For a function \( f(x) \), its derivative is often denoted as \( f'(x) \). If we have a function \( y = f(x) = \frac{1}{x} \), the derivative is \( f'(x) = -\frac{1}{x^2} \). This tells us that for small changes in \( x \), the change in \( y \) is approximately \( -\frac{1}{x^2} \) times the change in \( x \). The derivative is crucial in many applications, such as understanding motion, optimizing functions, and in data analysis processes. Each derivative gives precise insights into how variables interact, making it a powerful tool for understanding dynamic systems.
Function Notation
Function notation is a way of presenting a function in a form that provides clarity in its usage. By using function notation, we express dependencies between variables clearly. In mathematical terms, if \( y \) is a function of \( x \), then we can write this as \( y = f(x) \). For instance, if \( y = \frac{1}{x} \), we indicate that for every input \( x \), there is an output \( y \) determined by the formula \( f(x) = \frac{1}{x} \). This notation allows us to easily substitute values for \( x \) and calculate corresponding outputs. For example, to find the value when \( x = 1.2 \), we calculate \( f(1.2) = \frac{1}{1.2} \). This framework thus simplifies the process of substituting values, visualizing relationships, and communicating how a function works. Function notation is a cornerstone in the study and application of calculus.
Change in Function Value
Change in function value, often denoted as \( \Delta y \), describes how the value of a function changes as we make a small change to the input. It's an important concept for understanding how functions behave over small intervals. To determine \( \Delta y \), we calculate the function at two points and measure the difference. For a function \( f(x) = \frac{1}{x} \), suppose we change \( x \) from 1 to 1.2. We find \( f(1.2) = 0.8333 \) and \( f(1) = 1 \). Thus, \( \Delta y = 0.8333 - 1 = -0.1667 \). The negative value indicates the function's output decreases as \( x \) increases. In calculus, understanding \( \Delta y \) helps us quantify the function's response to input changes, especially in approximation and error analysis. This concept is widely utilized in fields such as physics, economics, and engineering.