Problem 1 - 5 Steps to A 5: AP Calculus AB 2021

Question

Find the linear approximation of \(f(x)=(1+x)^{1 / 4}\) at \(x=0\) and use the equation to approximate \(f(0.1)\).

Step-by-Step Solution

Answer

Answer: The approximate value of f(0.1) is 1.025.

Step 1: Find the derivative

To find the slope of the tangent line, we first need to find the derivative of the function. The derivative of \(f(x) = (1+x)^{\frac{1}{4}}\) is given by: \(f'(x) = \frac{1}{4}(1+x)^{-\frac{3}{4}}\)

Step 2: Evaluate the derivative at \(x=0\)

To find the slope at \(x=0\), plug \(x=0\) into the derivative and solve: \(f'(0) = \frac{1}{4}(1+0)^{-\frac{3}{4}} = \frac{1}{4}\)

Step 3: Find the tangent line equation

Now, we can find the tangent line equation at \(x=0\). The tangent line will have the form: \(L(x) = f(0) + f'(0)(x-0)\). We already found that \(f'(0)\) equals \(\frac{1}{4}\). Now, we need to find \(f(0)\): \(f(0) = (1+0)^{\frac{1}{4}}=1\) So, the tangent line equation is: \(L(x) = 1 + \frac{1}{4}(x-0) = 1 + \frac{1}{4}x\)

Step 4: Approximate \(f(0.1)\) using the tangent line

Now, we can approximate the value of \(f(0.1)\) using the tangent line. Plug in \(x=0.1\) into \(L(x)\): \(L(0.1) =1 + \frac{1}{4}(0.1)=1.025\) Therefore, the linear approximation of \(f(x) = (1+x)^{\frac{1}{4}}\) at \(x=0\) is \(L(x) = 1 + \frac{1}{4}x\), and the approximation to use the equation to approximate \(f(0.1)\) is \(L(0.1) = 1.025\).

Key Concepts

Derivative Calculation

To start our process of linear approximation, we need to understand how the derivative plays a key role. The derivative of a function gives us crucial information: it shows the rate at which the function's value changes as its input changes. In our problem, we begin with the function \( f(x) = (1+x)^{\frac{1}{4}} \). Here, the derivative is computed using the power rule of differentiation which helps in finding the derivative of expressions in the form \( x^n \).
Applying the power rule, we find the derivative:

• The outside function is \((1+x)^{\frac{1}{4}}\), meaning we bring down the power \( \frac{1}{4} \) as a coefficient.
• The inside function (1+x) is left inside, but its power is reduced by one, resulting in \(-\frac{3}{4} \), so we have \((1+x)^{\-\frac{3}{4}} \).
• The final derivative is \( f'(x) = \frac{1}{4}(1+x)^{-\frac{3}{4}} \).

This derivative tells us the slope of the function at any given point, which is essential for finding the tangent line at \( x=0 \).

Tangent Line Equation

Now that we have the derivative, we can determine the tangent line equation. This is a straight line that just touches the curve at a certain point, representing the function's slope at that exact place. Our goal is to find this line at \( x=0 \).
First, we calculate the slope of the tangent line by substituting \( x = 0 \) into the derivative formula:

• Plugging \( x=0 \) into \( f'(x) = \frac{1}{4}(1+x)^{-\frac{3}{4}} \), we get \( f'(0) = \frac{1}{4} \).

Next, we need the actual value of the function at \( x=0 \), which is \( f(0) = 1 \). Now, armed with both the point \((0,1)\) and the slope \(\frac{1}{4}\), we can form the equation of the tangent line:

• The equation is \( L(x) = f(0) + f'(0)(x-0) \).
• Substitute values: \( L(x) = 1 + \frac{1}{4}(x-0) = 1 + \frac{1}{4}x \).

This line provides a good linear model of the function near \( x=0 \).

Approximating Functions

The power of linear approximation lies in its ability to estimate values of functions that may otherwise be difficult to compute directly. Once we have the tangent line equation at a certain point, it becomes a tool to approximate function values near that point. In this problem, we are tasked with approximating \( f(0.1) \). Using our derived tangent line, this task becomes straightforward.

• Insert \( x=0.1 \) into the tangent line equation \( L(x) = 1 + \frac{1}{4}x \).
• Compute as follows: \( L(0.1) = 1 + \frac{1}{4}(0.1) = 1.025 \).

The result, 1.025, is an approximation of \( f(0.1) \). This method assumes the function does not vary wildly between the approximation point and the measured values. It's a preview of the predicted behavior of the function right by \( x=0 \).

Calculus Problem Solving

Solving calculus problems often involves multiple interconnected parts—differentiation, function evaluation, and the application of mathematical principles to reach a solution. Understanding how derivatives form the basis for approximating complex functions is crucial. In the current exercise, we used the following approach:

• Start with finding the derivative, which sets the stage for understanding the behavior of the function.
• Evaluate the derivative and the function at a specific point to construct the tangent line equation.
• Use this linear model to estimate unknown function values near the point.

Through this structured method, calculus problems can be broken down into tackleable steps, making the whole idea less intimidating and more logical. The methodology not only teaches how to perform a task but also why it works, building a deeper understanding of calculus principles.