Problem 11 - Calculus for Biology and Medicine

Question

Approximate \(f(x)\) at a by the linear approximation $$L(x)=f(a)+f^{\prime}(a)(x-a)$$ $$ f(x)=\frac{1}{1+x} \text { at } a=0 $$

Step-by-Step Solution

Answer

The linear approximation is \(L(x) = 1 - x\).

Step 1: Determine f(a)

Calculate the function value at \(a=0\). Substitute \(x = 0\) into \(f(x) = \frac{1}{1+x}\). We have \(f(0) = \frac{1}{1+0} = 1\).

Step 2: Calculate the derivative f'(x)

Find the derivative of \(f(x) = \frac{1}{1+x}\) using the power rule. We can rewrite \(f(x)\) as \((1+x)^{-1}\). The derivative is \(f'(x) = -1(1+x)^{-2}\).

Step 3: Evaluate f'(a)

Substitute \(x = 0\) into the derivative \(f'(x) = -(1+x)^{-2}\) to find \(f'(0)\). We have \(f'(0) = -(1+0)^{-2} = -1\).

Step 4: Apply the linear approximation formula

Use the linear approximation formula \(L(x) = f(a) + f'(a)(x-a)\) with \(a=0\). Substitute \(f(0) = 1\) and \(f'(0) = -1\) into the formula: \(L(x) = 1 - 1(x-0)\). This simplifies to \(L(x) = 1 - x\).

Key Concepts

Understanding Derivatives

Derivatives represent the rate at which a function's value changes as its input changes. Consider them as the mathematical description of the function’s slope at any given point. To compute a derivative, especially for functions like \(f(x) = \frac{1}{1+x}\), we often need some calculus tools, like the power rule or quotient rule. Here, we transformed \(f(x)\) into a power format, \((1+x)^{-1}\), making it easier to apply rules.Derivatives are crucial because they provide insights into the behavior of functions:

• They tell us how fast something is changing (rate of change).
• They help us sketch functions and find maxima or minima.

In this exercise, the derivative \(f'(x)\) of \(f(x)\) was found to be \(-1(1+x)^{-2}\), giving the slope at any point \(x\) along \(f(x)\). At \(a=0\), \(f'(0)\) specifically is \(-1\), indicating a decrease since it is negative.

Evaluating Function Value

Function value essentially evaluates the function at a specific point to find the exact output. For instance, given \(f(x) = \frac{1}{1+x}\), the function value at \(x=0\) is found by substituting \(x=0\) into the function.This is a straightforward process:

• Substitute \(x=0\) into \(f(x) = \frac{1}{1+x}\).
• Calculate to find \(f(0) = 1\).

Determining the function value at \(a=0\), here resulted in \(f(0) = 1\). This value is pivotal in the linear approximation as it serves as a starting point for the approximation at that base \(x\)-value.

Power Rule and Its Application

The power rule is a basic derivative rule used when dealing with functions of the form \(x^n\). According to the power rule, the derivative of \(x^n\) is \(nx^{n-1}\). It’s applicable when transforming functions into power functions makes them easier to differentiate.In the current problem, \(f(x) = \frac{1}{1+x}\) was rewritten using the power rule logic as \((1+x)^{-1}\). This allowed for easy differentiation:

• Apply the power rule: the derivative of \((1+x)^{-1}\) results in \(-1(1+x)^{-2}\).
• Substituting in \(x=0\), we found \(f'(x) = -1\).

The power rule simplifies the process of working with exponents in calculus, making it invaluable in many calculus problems.