Problem 14 - Calculus - AP Edition

Question

Confirm that the stated formula is the local linear approximation of \(f\) at \(x_{0}=1,\) where \(\Delta x=x-1\). $$f(x)=(4+x)^{3} ;(5+\Delta x)^{3} \approx 125+75 \Delta x$$

Step-by-Step Solution

Answer

The formula \((5+\Delta x)^3 \approx 125 + 75\Delta x\) is correct for the local linear approximation at \(x_0=1\).

Step 1: Identify the Function and Linear Approximation

The function given is \(f(x) = (4+x)^3\). We need to find the local linear approximation at \(x_0 = 1\). This approximation is in the form \(f(x_0) + f'(x_0)\Delta x\).

Step 2: Calculate the Derivative

First, find the derivative of \(f(x) = (4+x)^3\). Use the chain rule: \(f'(x) = 3(4+x)^2\times (1) = 3(4+x)^2\).

Step 3: Evaluate the Derivative at x_0

Substitute \(x_0 = 1\) into the derivative to find \(f'(1)\): \(f'(1) = 3(4+1)^2 = 3(5)^2 = 75\).

Step 4: Construct the Linear Approximation

The linear approximation formula is \(f(x_0) + f'(x_0)\Delta x\). At \(x_0 = 1\), we have \(f(1) = (4+1)^3 = 5^3 = 125\). Therefore, the approximation is \(125 + 75\Delta x\).

Step 5: Verify the Given Approximation

The given approximation \((5+\Delta x)^3 \approx 125+75 \Delta x\) matches the one we calculated. Thus, the formula is indeed the local linear approximation of \(f\) at \(x_0 = 1\).

Key Concepts

Derivative

In calculus, a derivative represents how a function changes as its input changes. It is a core concept that essentially gives the slope of the function at any point, indicating if the function is increasing or decreasing, and how steeply.
To find the derivative of a function like \(f(x) = (4+x)^3\), we use calculus techniques. In this case, the chain rule helps us differentiate complex functions.

• The derivative of \((4+x)^3\) is found by multiplying the derivative of the inner function (which is just 1 in this example since it's a simple linear function) by the derivative of the outer function \((4+x)^3\). This process gives us \(f'(x) = 3(4+x)^2\).
• Once we have \(f'(x)\), we substitute \(x_0 = 1\) to find \(f'(1)\). So \(f'(1) = 3(5)^2 = 75\).

Derivatives are foundational in approximating and predicting the behavior of functions, especially when analyzing changes over small intervals.

Chain Rule

The chain rule is a fundamental differentiation rule used to find the derivative of composite functions. Understanding how to apply the chain rule is crucial when working with functions nested within one another.
When you see a function within a function, like \((4+x)^3\), the chain rule is your go-to tool for differentiation. Here's how it works:

• Identify the outer function: In this case, the outer function is \(u^3\), where \(u = 4+x\).
• Differentiate the outer function: The derivative of \(u^3\) with respect to \(u\) is \(3u^2\).
• Differentiate the inner function: The derivative of \(4+x\) is simply \(1\).
• Multiply the derivatives: Multiply the derivative of the outer function by the derivative of the inner function to get \(3(4+x)^2 \times 1\).

By using the chain rule, we effectively break down complex differentiations into more manageable parts, helping us compute derivatives precisely and efficiently.

Linear Approximation Formula

Linear approximation, also known as tangent line approximation, allows us to estimate the value of a function at a particular point using the slope of the tangent line. It is especially useful when calculating the exact value is too complex.
The linear approximation formula is given by:\[ f(x) \approx f(x_0) + f'(x_0)(x-x_0) = f(x_0) + f'(x_0)\Delta x\]

• Here, \(x_0\) is where we're approximating, and \(\Delta x = x - x_0\) is a small change in \(x\).
• First, calculate \(f(x_0)\), the function value at the point of approximation, giving us \(125\) when \(x_0 = 1\).
• Next, calculate \(f'(x_0)\), the derivative at \(x_0\), which we found to be \(75\).
• Combine them to form the approximation: \(125 + 75\Delta x\).

By using this formula, we effectively transform a curved function into a simpler linear form at a specific point, allowing for quick calculations and insights into the behavior of the function around that point.