Problem 22 - Calculus for Biology and Medicine

Question

Calculate the linear approximation for \(f(x)\) : $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ \(f(x)=e^{2 x}\) at \(a=0\)

Step-by-Step Solution

Answer

The linear approximation of \( f(x) = e^{2x} \) at \( a = 0 \) is \( f(x) \approx 1 + 2x \).

Step 1: Identify the function and point of approximation

We are given the function \( f(x) = e^{2x} \) and need to approximate it around \( a = 0 \). The linear approximation formula we will use is:\[ f(x) \approx f(a) + f^{\prime}(a)(x-a) \]

Step 2: Evaluate the function at \(a\)

Substitute \( a = 0 \) into the function \( f(x) = e^{2x} \) to find \( f(a) \):\[ f(0) = e^{2 \times 0} = e^0 = 1 \]

Step 3: Find the derivative of the function

Differentiate the function \( f(x) = e^{2x} \). Using the chain rule, we have:\[ f^{\prime}(x) = \frac{d}{dx}[e^{2x}] = 2e^{2x} \]

Step 4: Evaluate the derivative at \(a\)

Substitute \( a = 0 \) into the derivative to find \( f^{\prime}(a) \):\[ f^{\prime}(0) = 2e^{2 \times 0} = 2e^0 = 2 \]

Step 5: Substitute values into the linear approximation formula

Now that we have \( f(a) = 1 \) and \( f^{\prime}(a) = 2 \), substitute these values into the linear approximation formula:\[ f(x) \approx 1 + 2(x-0) = 1 + 2x \]

Key Concepts

Derivative

When working with functions, derivatives are essential tools in calculus used to measure how a function changes as its input changes. In simpler terms, you can think of the derivative as the "rate of change" or "slope" of the function at any given point. For a straight line, the slope remains constant, while for curves, it varies at every point.

In our problem, we encountered the function \( f(x) = e^{2x} \). Finding the derivative of this function means determining how \( e^{2x} \) changes with respect to \( x \). We perform this differentiation using the chain rule, a powerful method that helps us tackle complex functions that involve "functions within functions." This leads directly into our next topic.

Chain Rule

The chain rule is a critical calculus concept used to find the derivative of composite functions. These are functions where one function is nested inside another, often written in the form \( f(g(x)) \), such as our exercise's \( e^{2x} \).

• Outer function: In this case, \( f(u) = e^u \).
• Inner function: And \( u = 2x \).

To solve using the chain rule, we first find the derivative of the outer function with respect to the inner function. Next, multiply this by the derivative of the inner function with respect to \( x \). For \( f(x) = e^{2x} \), it results in: \[f^{\prime}(x) = \frac{d}{dx}[e^{2x}] = e^{2x} \cdot \frac{d}{dx}[2x] = 2e^{2x}.\]This calculation is essential for determining the linear approximation of \( e^{2x} \) at any point, as it helps us understand how the function's value changes around that point.

Exponential Function

Exponential functions like \( e^{x} \) are fundamental in mathematics, with applications spanning finance, science, and engineering. These functions exhibit constant relative growth, which means they multiply by a fixed factor over equal increments of input, resulting in rapid growth or decay.In this exercise, we have the exponential function \( f(x) = e^{2x} \). This is a specific type of exponential function where the exponent itself is a linear expression. An exponential function is distinguished by its base, which is usually the Euler's number \( e \), an irrational number approximately equal to 2.718.A unique property of exponential functions is that their derivative is directly related to their original form. For \( e^{x} \), the derivative is \( e^{x} \) itself. Similarly, with \( e^{2x} \), the derivative involves both the function and the additional constant \( 2 \), as derived using the chain rule. The exponential function's properties enable it to model real-world situations where growth processes are dependent on current value, such as compound interest or population growth.