Problem 1 - Calculus for Biology and Medicine

Question

Find the linear approximation of \(f(x)\) at \(x=0 .\) $$ f(x)=e^{x+1} $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Concept of Linear Approximation

Linear approximation is a method to approximate the value of a function near a certain point using the tangent line at that point. The formula for linear approximation of the function \( f(x) \) at \( x = a \) is given by \( L(x) = f(a) + f'(a)(x-a) \).

Step 2: Identify the Function and the Point of Approximation

The function provided is \( f(x) = e^{x+1} \) and we need to find the linear approximation at \( x = 0 \).

Step 3: Calculate \( f(a) \)

Evaluate the function at the point of approximation. Here \( a = 0 \), so compute \( f(0) = e^{0+1} = e \).

Step 4: Determine the Derivative \( f'(x) \)

Find the derivative of the function \( f(x) = e^{x+1} \). The derivative is given by the chain rule: \( f'(x) = e^{x+1} \cdot (1) = e^{x+1} \).

Step 5: Calculate \( f'(a) \)

Evaluate the derivative at the point \( a = 0 \). This gives \( f'(0) = e^{0+1} = e \).

Step 6: Substitute into Linear Approximation Formula

Substitute \( f(0) = e \), \( f'(0) = e \), and \( a = 0 \) into the linear approximation formula: \( L(x) = e + e \cdot (x - 0) \). Simplifying, we get \( L(x) = e(1 + x) \).

Key Concepts

Tangent Line

In mathematics, the concept of a tangent line is essential for understanding linear approximation. A tangent line touches a curve at a single point and has the same slope as the curve at that point. This is why it's often used to approximate the behavior of the curve near that point.
To find a tangent line for a function at a particular point, you need two things:

• The value of the function at that point.
• The derivative of the function at that point.

For the function given, \( f(x) = e^{x+1} \), and the point \( x = 0 \), we calculate the tangent line using the values \( f(0) \) and \( f'(0) \). The linear approximation formula, \( L(x) = f(a) + f'(a)(x-a) \), is essentially the equation of this tangent line.

Derivative

A derivative is a fundamental tool in calculus that measures how a function changes as its input changes. In simpler terms, it tells us the slope or rate of change of a function at a particular point. Understanding derivatives is crucial for linear approximation because it helps determine the slope of the tangent line.

For the function \( f(x) = e^{x+1} \), we apply the rules of differentiation. Specifically, the chain rule, which we will discuss next, helps us differentiate composite functions. The derivative \( f'(x) \) is calculated as \( f'(x) = e^{x+1} \times 1 = e^{x+1} \). Evaluating this at \( x = 0 \) gives us \( f'(0) = e \), indicating the slope of the tangent line at that point.

Chain Rule

The chain rule is a technique in calculus for dealing with composite functions, where one function is applied to the result of another. It is particularly useful when differentiating expressions like \( e^{x+1} \), which is a composition of an exponential and a linear function.

To use the chain rule, we take the derivative of the outer function first, keeping the inner function the same, and then multiply it by the derivative of the inner function. For our function \( f(x) = e^{x+1} \):

• Outer function: \( e^u \) where \( u = x + 1 \)
• Inner function: \( u = x + 1 \)
• Derivative of the outer function: \( e^u \)
• Derivative of the inner function: \( \frac{d}{dx}(x+1) = 1 \)

So, \( f'(x) = e^{x+1} \times 1 = e^{x+1} \). This result is then used to find the slope of the tangent line for the linear approximation.