Problem 37
Question
A measurement error in \(x\) affects the accuracy of the value \(f(x) .\) In each case, determine an interval of the form $$[f(x)-\Delta f, f(x)+\Delta f]$$ that reflects the measurement error \(\Delta x .\) In each problem, the quantities given are \(f(x)\) and \(x=\) true value of \(x \pm|\Delta x|\). $$ f(x)=3 x^{2}, x=2 \pm 0.1 $$
Step-by-Step Solution
Verified Answer
The interval is [10.8, 13.2].
1Step 1: Identify the Problem
You are asked to determine an interval for the function affected by a measurement error in the variable. The function given is \( f(x) = 3x^2 \) and the value of \( x \) is given as \( 2 \pm 0.1 \).
2Step 2: Calculate the Derivative
The derivative of the function \( f(x) = 3x^2 \) with respect to \( x \) is \( f'(x) = 6x \). This derivative will help us approximate the change in \( f \) due to a small change in \( x \).
3Step 3: Compute The Maximum Change in Function
The maximum change, \( \Delta f \), can be found using the formula \( \Delta f = f'(x) \Delta x \). Substitute \( f'(x) = 6x \) and \( \Delta x = 0.1 \), so \( \Delta f = 6 \times 2 \times 0.1 = 1.2 \).
4Step 4: Formulate the Interval for the Function
Compute \( f(x) \) at \( x = 2 \), which is \( f(2) = 3 \times 2^2 = 12 \). Thus, the interval of possible values for \( f(x) \) is given by \([f(x) - \Delta f, f(x) + \Delta f] = [12 - 1.2, 12 + 1.2] = [10.8, 13.2]\).
Key Concepts
DerivativeFunction ApproximationInterval Estimation
Derivative
In mathematics, a derivative is a powerful tool used to understand how a function changes as its input changes. The derivative of a function at a specific point gives us the slope of the tangent line to the function at that point.
It tells us the rate of change of the function concerning its variable. For example, if we have a function, like in our exercise with the formula \( f(x) = 3x^2 \), the derivative \( f'(x) \) is calculated as \( 6x \).
This tells us that for a small change \( \Delta x \) in \( x \), the function \( f(x) \) changes approximately by \( f'(x) \Delta x \). Here’s how derivatives come into play:
It tells us the rate of change of the function concerning its variable. For example, if we have a function, like in our exercise with the formula \( f(x) = 3x^2 \), the derivative \( f'(x) \) is calculated as \( 6x \).
This tells us that for a small change \( \Delta x \) in \( x \), the function \( f(x) \) changes approximately by \( f'(x) \Delta x \). Here’s how derivatives come into play:
- They help to approximate the difference or change in the value of a function given a small error in measurement of its input.
- The derivative provides insights into the function's increasing or decreasing behavior at a particular point.
Function Approximation
Sometimes, the true value of a function is hard to calculate or measure due to small errors or imperfections. Function approximation seeks to estimate the function's value by using derivatives.
This approach leads us to use linear approximations to model these small changes accurately.
For instance, in our example with \( f(x) = 3x^2 \), the derivative allows us to approximate \( f(x) \) when \( x \) is not exactly known. Using the formula \( \Delta f = f'(x) \Delta x \), we estimate how much \( f(x) \) would change across a small interval caused by a measurement error, which is \( \Delta x = 0.1 \).
This approach leads us to use linear approximations to model these small changes accurately.
For instance, in our example with \( f(x) = 3x^2 \), the derivative allows us to approximate \( f(x) \) when \( x \) is not exactly known. Using the formula \( \Delta f = f'(x) \Delta x \), we estimate how much \( f(x) \) would change across a small interval caused by a measurement error, which is \( \Delta x = 0.1 \).
- Function approximation relies on the fact that for small changes, the curve of a function can be approximated as a straight line.
- The accuracy of this approximation depends on how small \( \Delta x \) is.
- This technique is practical when dealing with noisy data or slight inaccuracies in measurements.
Interval Estimation
Interval estimation provides a range within which a true value is expected to lie, considering possible errors or uncertainties in measurement. It accounts for measurement errors by providing a buffer zone around the estimated values.
The goal is to find a generous bracket that captures both the underestimated and overestimated outcomes.
In our case, by flagging the error margin of \( \Delta x = 0.1 \), we maximize and minimize \( f(x) \) within an interval given by \([f(x) - \Delta f, f(x) + \Delta f]\). For \( x = 2 \), this results in the interval \([10.8, 13.2]\).
The goal is to find a generous bracket that captures both the underestimated and overestimated outcomes.
In our case, by flagging the error margin of \( \Delta x = 0.1 \), we maximize and minimize \( f(x) \) within an interval given by \([f(x) - \Delta f, f(x) + \Delta f]\). For \( x = 2 \), this results in the interval \([10.8, 13.2]\).
- Interval estimation is crucial when absolute precision is impractical or impossible.
- It is widely used in statistics to express uncertainty, like margins of error in surveys.
- It offers a sense of reliability, as it reflects the range in which one can be confident the true value lies.
Other exercises in this chapter
Problem 36
Differentiate the functions with respect to the independent variable. $$ f(x)=\ln \sqrt{2 x^{2}-x} $$
View solution Problem 36
Differentiate $$ p(T)=\frac{N k T}{V} $$
View solution Problem 37
Assume that \(f(x)\) and \(g(x)\) are differentiable. Find \(\frac{d}{d x}\left(\frac{f(x)}{g(x)}+1\right)^{2}\).
View solution Problem 37
In Problems \(1-58\), find the derivative with respect to the independent variable. $$ f(x)=\sin (2 x-1) \cos (3 x+1) $$
View solution