Problem 22 - Calculus

Question

Let \(y=f(x)\) with \(f(3)=5\) and \(f^{\prime}(3)=9.\) (a) For small \(\Delta x,\) write an approximate equation relating \(\Delta y\) and \(\Delta x\) near \(x=3.\) (b) Estimate \(f(2.9).\)

Step-by-Step Solution

Answer

(a) \( \Delta y \approx 9 \cdot \Delta x \). (b) \(f(2.9) \approx 4.1\).

Step 1: Introduction to Essential Information

We are given that a function \(y=f(x)\) has \(f(3) = 5\) and the derivative at this point is \(f'(3) = 9\). We need to approximate \( \Delta y \) in terms of \( \Delta x\) near \(x=3\), and then estimate the function value at \(x=2.9\).

Step 1: Linear Approximation Formula

The linear approximation or the differential of \(f(x)\) at \(x=3\) can be written as \( \Delta y \approx f'(3) \cdot \Delta x \). This approximation uses the idea that small changes in \(y\) can be estimated using the derivative of \(f\) at that point.

Step 2: Determine \( \Delta x \)

For our estimation, we consider \( \Delta x = x - 3 \). Since we want to estimate \(f(2.9)\), we'll find \( \Delta x = 2.9 - 3 = -0.1\).

Step 3: Calculate \( \Delta y \)

Using the formula from Step 1, substitute the values we know: \( \Delta y \approx f'(3) \cdot \Delta x = 9 \times (-0.1) = -0.9\).

Step 4: Estimate \( f(2.9) \)

To find \(f(2.9)\), we use \( \Delta y \approx f(2.9) - f(3)\). Therefore, \(f(2.9) \approx f(3) + \Delta y = 5 - 0.9 = 4.1\).

Key Concepts

Differential Calculus

Differential calculus is a fundamental branch of mathematics that focuses on the concept of the derivative. The derivative represents how a function changes as its input changes, and it plays a crucial role in understanding the behavior and properties of functions. By evaluating the derivative at specific points, we can gain insights into the function's rate of change and create linear models to estimate function values.

• It revolves around the idea of limits and rates of change.
• The derivative, denoted as \(f'(x)\) or \(\frac{dy}{dx}\), measures how sensitive a function \(f(x)\) is to changes in \(x\).
• Using derivatives, we can perform linear approximations, which are useful for making quick estimates.

Understanding differential calculus allows us to approximate changes in a function, as seen in this exercise. When given a specific point and its derivative, we can use calculus to predict nearby function values without evaluating the entire function.

Function Estimation

Function estimation is an incredibly useful application of calculus that allows us to predict or estimate values of a function at points near those we already know. In many practical scenarios, calculating the exact value of a function can be complex or time-consuming. That's where estimation comes in handy.

• The key technique is using a known value and its corresponding derivative.
• For small changes \(\Delta x\), we can approximate \(\Delta y\) using the linear approximation formula: \(\Delta y \approx f'(x_0) \cdot \Delta x\).
• This translates to \(f(x_1) \approx f(x_0) + f'(x_0) \cdot \Delta x\), providing an estimated value for \(f(x_1)\).

In our given exercise, the function \(f(x)\) was estimated at \(x = 2.9\) by adjusting the known value at \(x = 3\) using this standard formula. This technique helps significantly reduce computational effort while providing a reasonably accurate approximation.

Derivative Approximation

Derivative approximation is a method used to estimate changes in a function without determining its exact form. By leveraging the derivative, you can predict how much a function will increase or decrease with small changes in its input. This helps in situations where calculating the function directly is challenging.

• The derivative \(f'(x)\) serves as the slope of the tangent line at a particular point.
• Using this slope, we can approximate changes: \(\Delta y \approx f'(x) \cdot \Delta x\).
• Useful for problems where precise computations are impractical but an approximate result suffices.

For the exercise, knowing that \(f'(3) = 9\) and using \(\Delta x = -0.1\), the change in \(y\) was approximated. This method capitalizes on the power of derivatives to simplify and solve complex problems efficiently.