Problem 25 - Calculus for Biology and Medicine

Question

Calculate the linear approximation for \(f(x)\) : $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ \(f(x)=e^{x-1}\) at \(a=1\)

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

The goal is to find the linear approximation of the function \( f(x) = e^{x-1} \) at the point \( a = 1 \). This involves calculating \( f(a) \), \( f'(x) \), and \( f'(a) \), and then using the formula \( f(x) \approx f(a) + f'(a)(x-a) \) to find the linear approximation.

Step 2: Calculate \( f(a) \)

Evaluate the function at \( a = 1 \). Since \( f(x) = e^{x-1} \), we have \( f(1) = e^{1-1} = e^0 = 1 \).

Step 3: Find the Derivative \( f'(x) \)

The derivative of \( f(x) = e^{x-1} \) with respect to \( x \) is \( f'(x) = e^{x-1} \). This is determined by using the chain rule, where the derivative of \( e^{u} \) is \( e^{u} \cdot u' \), and here \( u = x-1 \), so \( u' = 1 \).

Step 4: Calculate \( f'(a) \)

Substitute \( a = 1 \) into the derivative \( f'(x) = e^{x-1} \). We get \( f'(1) = e^{1-1} = e^0 = 1 \).

Step 5: Apply the Linear Approximation Formula

Insert the values \( f(1) = 1 \) and \( f'(1) = 1 \) into the linear approximation formula: \( f(x) \approx f(a) + f'(a)(x-a) \). This gives \( f(x) \approx 1 + 1(x-1) \). Therefore, the linear approximation is \( f(x) \approx x \).

Key Concepts

Derivative

Derivatives are fundamental in calculus and are often described as the rate of change or the slope of a function at a specific point. In the context of linear approximation, the derivative provides crucial information about how the function behaves in the immediate vicinity of a given point. The formula for linear approximation is given as: \[ f(x) \approx f(a) + f'(a)(x-a) \]. Here, \( f'(a) \) is the derivative of the function at the point \( a \). This expression uses the concept of a tangent line to estimate the value of the function \( f(x) \) around \( a \). Calculating the derivative involves finding the slope of this tangent line.
For the function \( f(x) = e^{x-1} \), taking the derivative \( f'(x) \) results in \( e^{x-1} \). This shows how swiftly the function's value is changing per unit change in \( x \). Understanding this enables you to approximate complicated functions more easily in a small neighborhood of any given point.

Chain Rule

The chain rule is a powerful tool in calculus used to find the derivative of composite functions. When you have a function that is composed of multiple functions, the chain rule helps break it down more conveniently. The rule is expressed as follows: if you have a function \( g(x) = u(v(x)) \), then the derivative \( g'(x) \) is \( u'(v(x)) \cdot v'(x) \).

In the exercise with the function \( f(x) = e^{x-1} \), you encounter a composition of functions where \( u = x-1 \). The derivative is computed using the chain rule. Since the derivative of \( e^u \) is \( e^u \), and the derivative of \( u = x-1 \) is just \( 1 \), you multiply them together to get the final result \( f'(x) = e^{x-1} \cdot 1 = e^{x-1} \). This application of the chain rule is essential for simplifying and handling complex functions' derivatives.

Exponential Function

The exponential function, most commonly represented as \( e^x \), is a key mathematical function. It features prominently in many different fields, including natural sciences and financial contexts. One of its unique properties is that its derivative is the same as the function itself—this means that \( \frac{d}{dx} e^x = e^x \).

In the given problem, this property of the exponential function simplifies the differentiation process. Despite the shift in the exponent (\( x-1 \) instead of just \( x \)), the characteristic properties remain. The function \( f(x) = e^{x-1} \) behaves predictably based on these rules. Such stability makes exponential functions immensely useful for modeling growth processes, decay, and other natural phenomena. Through linear approximation, we see how these continuous growth models, like \( e^x \), can be approximated linearly for small changes in \( x \). This allows for easier calculation and interpretation of results in an otherwise complex mathematical landscape.