Problem 27 - Calculus for Biology and Medicine

Question

Calculate the linear approximation for \(f(x)\) : $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ \(f(x)=(1+x)^{-n}\) at \(a=0 .\) (Assume that \(n\) is a positive integer.)

Step-by-Step Solution

Answer

The linear approximation of \(f(x)=(1+x)^{-n}\) at \(a=0\) is \(f(x) \approx 1 - nx.\)

Step 1: Identify the Function and Point

We are tasked with finding the linear approximation of the function \(f(x)=(1+x)^{-n}\) at \(a=0\). Our formula is \(f(x) \approx f(a) + f'(a)(x-a)\). Here, \(a=0\).

Step 2: Calculate the Function Value at a

First, we need to calculate the value of the function at \(a=0\). So, we substitute \(a=0\) into the function: \[ f(0) = (1+0)^{-n} = 1^{-n} = 1. \]

Step 3: Differentiate the Function

Next, differentiate the given function \(f(x) = (1+x)^{-n}\) to find \(f'(x)\). Using the chain rule: \[ f'(x) = -n(1+x)^{-(n+1)}. \]

Step 4: Calculate the Derivative at a

Now, we substitute \(a=0\) into the derivative to find \(f'(0)\):\[ f'(0) = -n(1+0)^{-(n+1)} = -n. \]

Step 5: Construct the Linear Approximation

Using the linear approximation formula:\[ f(x) \approx f(0) + f'(0)(x - 0), \]we substitute \(f(0)\) and \(f'(0)\) to get:\[ f(x) \approx 1 - nx. \]

Key Concepts

Linear Approximation

Linear approximation is a technique used in calculus to estimate the value of a function near a particular point using the tangent line. The method relies on the concept that close to this point, the function behaves nearly linearly; thus, the tangent could represent the function well.

The formula for linear approximation is:

• \( f(x) \approx f(a) + f'(a)(x-a) \)

Here, \( f(a) \) is the value of the function at the point \( a \), and \( f'(a) \) is the derivative of the function at that same point. The term \( (x-a) \) represents the distance from the point \( a \) to the desired \( x \) value.

This helps simplify complex functions by transforming them into a linear form which is much easier to work with, especially for small ranges near \( a \). In our example, we used linear approximation for the function \( f(x)=(1+x)^{-n} \) at \( a=0 \), resulting in the simpler form \( f(x) \approx 1 - nx \). This gives a quick way to approximate the function for values of \( x \) close to zero.

Differentiation

Differentiation is a fundamental concept in calculus that refers to finding the derivative of a function. A derivative represents the rate at which a function is changing at any point. It's essentially the slope of the tangent line to the function at a given point.

For the function \( f(x) = (1+x)^{-n} \), we use differentiation to determine how changes in \( x \) affect \( f(x) \). The derivative in this case is found to be:

• \( f'(x) = -n(1+x)^{-(n+1)} \)

Knowing how to differentiate a function allows us to perform linear approximations and understand the behavior of the function in a local neighborhood. Differentiation lays the foundation for getting insights into increasing or decreasing trends within a function, critical points, and curves' concavity.

Chain Rule

The chain rule is a powerful differentiation tool that simplifies finding derivatives of composite functions. In essence, it allows us to differentiate a function composed of two or more nested functions.

When utilizing the chain rule, the derivative of a composite function \( f(g(x)) \) is given by:

• \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \)

In the scenario we examined, we applied the chain rule to the function \( f(x) = (1+x)^{-n} \). Here, the outer function is \( u^{-n} \) and the inner function is \( u = 1+x \). By taking the derivative of the outer function with respect to the inner function and multiplying it by the derivative of the inner function, we obtained:

• \( f'(x) = -n(1+x)^{-(n+1)} \)

This application of the chain rule allowed us to differentiate a more complex expression efficiently, demonstrating how composite functions can be broken down into manageable pieces for differentiation.