Problem 26 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The linear approximation of \( f(x) \) at \( a = -\frac{1}{2} \) is \( L(x) = 2x + 2 \).

Step 1: Identify the Function and Point

We are given the function \( f(x) = e^{2x+1} \), and we want to find its linear approximation at \( a = -\frac{1}{2} \).

Step 2: Find the Value of f(a)

Substitute \( a = -\frac{1}{2} \) into the function \( f(x) \):\[f\left(-\frac{1}{2}\right) = e^{2(-\frac{1}{2})+1} = e^{0} = 1.\]

Step 3: Calculate the Derivative of f(x)

The derivative of \( f(x) = e^{2x+1} \) is found using the chain rule. The derivative of \( e^{u} \) with respect to \( x \) is \( e^{u} \cdot \frac{du}{dx} \) where \( u = 2x + 1 \).Thus:\[f'(x) = e^{2x+1} \cdot 2.\]

Step 4: Evaluate the Derivative at a

Substitute \( a = -\frac{1}{2} \) into \( f'(x) \):\[f'\left(-\frac{1}{2}\right) = e^{0} \cdot 2 = 2.\]

Step 5: Construct the Linear Approximation L(x)

Substitute \( f(a) \) and \( f'(a) \) into the linear approximation formula:\[L(x) = f(a) + f'(a)(x-a) = 1 + 2\left(x + \frac{1}{2}\right).\]

Step 6: Simplify the Expression for L(x)

Expand and simplify the expression obtained for \( L(x) \):\[L(x) = 1 + 2x + 1 = 2x + 2.\]

Key Concepts

Derivative Calculation

Calculating the derivative is essential for determining how a function changes as its input changes. When we have a function like \( f(x) = e^{2x+1} \), we need to find its derivative to construct a linear approximation. The derivative provides the slope of the tangent line at any point on the function. For exponential functions, the derivative generally involves applying some rules, such as the chain rule.The derivative of an exponential function \( e^{u} \) with respect to \( x \) is \( e^{u} \cdot \frac{du}{dx} \). Here, the exponent \( u \) is more complex, \( 2x + 1 \), so it's crucial to find how the inner function \( u \) changes with \( x \). By identifying \( u = 2x + 1 \), we can easily compute \( \frac{du}{dx} = 2 \). Thus, the derivative of the function is:\[ f'(x) = e^{2x+1} \cdot 2. \]This derivative formula is used to analyze and predict the behavior of \( f(x) \), particularly for applications like linear approximation.

Chain Rule

The chain rule is a powerful differential technique in calculus that helps us find the derivative of composite functions. A composite function is a function within another function. When you encounter expressions like \( f(x) = e^{2x+1} \), it’s helpful to see the two layers: the exponential function and its inner linear expression \( 2x + 1 \).To apply the chain rule effectively:

• Identify the outer function \( g(u) = e^{u} \) and the inner function \( u = 2x + 1 \).
• Compute the derivative of the outer function, \( \frac{dg}{du} = e^{u} \).
• Compute the derivative of the inner function, \( \frac{du}{dx} = 2 \).
• Multiply these results: \( \frac{df}{dx} = e^{2x+1} \cdot 2 \).

This method ensures you're considering the rate of change of the entire composite function, providing a complete slope calculation necessary for linearization and other analyses.

Exponential Function

An exponential function is a mathematical expression where a constant base is raised to a variable exponent. In \( f(x) = e^{2x+1} \), \( e \) is the base of the natural logarithm, approximately equal to 2.718. Exponential functions like these arise frequently in real-world scenarios, including population growth and radioactive decay.The function \( e^{2x+1} \) presents a specific type of exponential function with a linear exponent, \( 2x + 1 \). This structure influences how the function increases or decreases with \( x \), leading to dynamically changing outputs. Exponential functions are especially noteworthy for their characteristics:

• The rate of growth (or decay) is proportional to its current value.
• They exhibit rapid changes, which can make them suitable for representing phenomena with accelerated growth.
• They always remain positive as \( e^{anything} > 0 \).

Understanding this behavior is key to approximating the function using linearization techniques, helping to simplify analysis and calculations in many fields.

Simplifying Expressions

Simplifying expressions is a fundamental algebraic procedure that makes handling mathematical expressions more manageable. Once you've formed a function's linear approximation, simplifying ensures you have the most streamlined version.For example, in the linear approximation \( L(x) = 1 + 2(x + \frac{1}{2}) \), notice how the expanded form could initially look cumbersome. This expression requires:

• Distributing the multiplication: \( 2(x + \frac{1}{2}) = 2x + 1 \).
• Adding up terms: \( 1 + 2x + 1 \), leading to \( 2x + 2 \).

By simplification, you arrive at \( L(x) = 2x + 2 \). This compact version is not only easier to work with but also provides clearer insight into connection terms, like slopes and intercepts, helping understand the function's behavior at a given point more intuitively.