Problem 16 - Calculus for Biology and Medicine

Question

Approximate \(f(x)\) at a by the linear approximation $$L(x)=f(a)+f^{\prime}(a)(x-a)$$ $$ f(x)=\frac{1}{(1-x)^{2}} \text { at } a=0 $$

Step-by-Step Solution

Answer

The linear approximation of \( f(x) \) at \( a=0 \) is \( L(x) = 1 + 2x \).

Step 1: Understand the Exercise

We need to approximate the function \( f(x) = \frac{1}{(1-x)^2} \) at \( a = 0 \) using linear approximation. The formula for the linear approximation is \( L(x) = f(a) + f'(a)(x-a) \). Here, we need to find \( f(a) \) and \( f'(a) \) and use these values in the formula.

Step 2: Evaluate the Function at a=0

First, we find \( f(a) \), which is the value of the function at \( x = 0 \). Thus, \( f(0) = \frac{1}{(1-0)^2} = 1 \).

Step 3: Calculate the Derivative of f(x)

To find \( f'(x) \), we need to differentiate \( f(x) = \frac{1}{(1-x)^2} \). Using the chain rule, the derivative is \( f'(x) = \frac{d}{dx} ((1-x)^{-2}) = -2(1-x)^{-3} \times (-1) = \frac{2}{(1-x)^3} \).

Step 4: Evaluate the Derivative at a=0

Next, we calculate \( f'(0) \), the value of the derivative at \( x = 0 \). So, \( f'(0) = \frac{2}{(1-0)^3} = 2 \).

Step 5: Form the Linear Approximation

Now we have both \( f(0) = 1 \) and \( f'(0) = 2 \). We can substitute these into the linear approximation formula:\[ L(x) = f(0) + f'(0)(x - 0) = 1 + 2x \].

Key Concepts

Derivative Calculation

Calculating the derivative of a function is like finding the rate at which the function's output changes with respect to changes in its input. This "rate of change" is what we label as the derivative. In the given exercise, the function is \( f(x) = \frac{1}{(1-x)^2} \). This expression can be a little tricky due to the powers and the fraction involved.

To calculate its derivative, we usually need to apply some rules of differentiation. Here, we used:

• The power rule: Differentiating \( g(x)^n \) gives \( ng(x)^{n-1}g'(x) \).
• Identifying the terms: For our function, \( n = -2 \) and the function within the power is \( (1-x) \).

In this case, we have to differentiate \( (1-x)^{-2} \) which results in \( -2(1-x)^{-3} \) when you multiply by the derivative of the inner function. This results in the derivative \( f'(x) = \frac{2}{(1-x)^3} \) after simplification.

Chain Rule

The chain rule is an essential tool when dealing with compositions of functions, making it indispensable in the exercise. It allows us to find derivatives when functions are nested within each other. Imagine peeling layers of an onion to get to the core.

For our exercise, we have the function \( (1-x)^{-2} \), a composition where one function is raised to a power. The outer layer is \( (u)^{-2} \) and the inner layer is \( u = 1-x \).

• Applying the chain rule: First, differentiate the outer function \( u^{-2} \) with respect to \( u \), obtaining \( -2u^{-3} \).
• Differentiating the inner function: Compute the derivative of \( u = 1-x \) with respect to \( x \), which is simply \( -1 \).
• Combine them: Multiply the derivative of the outer function by the derivative of the inner function to get \( f'(x) = -2(1-x)^{-3}(-1) = \frac{2}{(1-x)^3} \).

This result leads us to a simplified derivative that we use in the linear approximation.

Function Evaluation

Function evaluation is the process of finding the output of a function for a specific input. In our exercise, we were tasked to evaluate the function \( f(x) = \frac{1}{(1-x)^2} \) at \( a = 0 \). This is a critical step in linear approximation as it provides the starting value to build from.

For the given function:

• Evaluate at \( x = 0 \): Substitute \( x = 0 \) into the function, resulting in \( f(0) = \frac{1}{(1-0)^2} = 1 \).

Similarly, evaluating the derivative at this point helps determine the slope of the tangent line at \( x = 0 \). Thus, substituting \( x = 0 \) into the derivative \( f'(x) = \frac{2}{(1-x)^3} \) gives us \( f'(0) = 2 \).

With these evaluations, we were able to construct the linear approximation \( L(x) = 1 + 2x \) around the point \( x = 0 \). This approximation allows us to predict what the function looks likes near \( x=0 \).