Problem 3 - Calculus for Biology and Medicine

Question

In Problems 1-5, find the linear approximation of \(f(x)\) at \(x=0\). $$ f(x)=\frac{1}{1-x} $$

Step-by-Step Solution

Answer

The linear approximation of \(f(x) = \frac{1}{1-x}\) at \(x=0\) is \(L(x) = 1 + x\).

Step 1: Understand Linear Approximation

Linear approximation of a function \(f(x)\) at a point \(x = a\) is given by \(L(x) = f(a) + f'(a)(x - a)\). This is essentially the equation of the tangent line at \(x = a\). Here, \(a = 0\).

Step 2: Find the Function Value at the Given Point

Evaluate \(f(x)\) at \(x = 0\). Given \(f(x) = \frac{1}{1-x}\), we find \(f(0) = \frac{1}{1-0} = 1\).

Step 3: Compute the Derivative of the Function

Find \(f'(x)\) by differentiating \(f(x) = \frac{1}{1-x}\) with respect to \(x\). The derivative is \(f'(x) = \frac{d}{dx}(1-x)^{-1} = (1-x)^{-2} = \frac{1}{(1-x)^2}\).

Step 4: Evaluate the Derivative at the Point

Evaluate \(f'(x)\) at \(x = 0\). Thus, \(f'(0) = \frac{1}{(1-0)^2} = 1\).

Step 5: Formulate the Linear Approximation

Substitute \(f(0)\) and \(f'(0)\) into the linear approximation formula: \[ L(x) = f(0) + f'(0)(x - 0) = 1 + 1 \times x = 1 + x. \] Thus, the linear approximation of \(f(x)\) at \(x = 0\) is \(L(x) = 1 + x\).

Key Concepts

Derivative Calculation

To understand the concept of linear approximation, knowing how to calculate derivatives is essential. Derivatives represent the rate of change of a function concerning one of its variables. In simple terms, it's like finding how steep the hill is at a certain point.

In this exercise, we have the function \( f(x) = \frac{1}{1-x} \). To find the derivative \( f'(x) \), we use the power rule of calculus which helps us differentiate expressions like \( (1-x)^{-1} \). The process involves bringing down the exponent as a coefficient and reducing the original exponent by one:

• Rewrite the function as \((1-x)^{-1}\).
• Apply the power rule: the derivative is \(-1 imes (1-x)^{-2} \cdot (-1) = (1-x)^{-2}\).
• So, \( f'(x) = \frac{1}{(1-x)^2} \).

Finding the derivative efficiently allows us to apply it later in finding tangent lines and linear approximations, making it a vital skill in calculus.

Tangent Line

The concept of a tangent line is crucial when dealing with linear approximations. A tangent line touches the curve of a function at a particular point without cutting through it and represents the best linear approximation of the function at that point.

In a way, it gives us a straight-line version of what the function looks like locally around that point. The equation of a tangent line at a point \( x = a \) is given by the formula:

• \( L(x) = f(a) + f'(a)(x - a) \)

In our exercise, the tangent line is derived by substituting \( a = 0 \), \( f(0) = 1 \), and \( f'(0) = 1 \) into the formula:

• \( L(x) = 1 + 1(x - 0) = 1 + x \)

This line \( L(x) = 1 + x \), closely follows the curve of \( f(x) \) around \( x = 0 \), making it a useful tool for approximation purposes.

Function Evaluation

The process of evaluating a function at a specific point involves substituting the value of \( x \) into the function to find the corresponding output. Evaluating the function is a straightforward affair but serves as the foundation of other operations in calculus.

For our function \( f(x) = \frac{1}{1-x} \), we need to evaluate it at \( x = 0 \) to start constructing our linear approximation. Here's how it works:

• Replace \( x \) with 0: \( f(0) = \frac{1}{1-0} = 1 \)

This evaluation tells us that the function's value at \( x = 0 \) is 1, which fits into our linear approximation formula and informs us about the behavior of the function at that point. Knowing this helps us maintain accuracy in approximating the function's value and behavior near \( x = 0 \).