Problem 17 - Calculus for Biology and Medicine

Question

Approximate \(f(x)\) at a by the linear approximation $$L(x)=f(a)+f^{\prime}(a)(x-a)$$ $$ f(x)=\ln (1+2 x) \text { at } a=0 $$

Step-by-Step Solution

Answer

The linear approximation of \(f(x)\) at \(x = 0\) is \(L(x) = 2x\).

Step 1: Understand the Linear Approximation Formula

The linear approximation formula is given by \[ L(x) = f(a) + f'(a)(x-a) \] where \(L(x)\) is the linear approximation of \(f(x)\) near \(x = a\), \(f(a)\) is the value of the function at \(a\), \(f'(a)\) is the derivative of the function evaluated at \(a\), and \(x-a\) is the difference from \(a\). First, we need to find \(f(a)\) and \(f'(a)\).

Step 2: Evaluate f(a)

Start by evaluating the function \(f(x) = \ln(1+2x)\) at \(a = 0\):\[ f(0) = \ln(1 + 2 \times 0) = \ln(1) = 0 \] Thus, \(f(a) = 0\).

Step 3: Find the Derivative f'(x)

Calculate the derivative of the function \(f(x) = \ln(1+2x)\). Using the chain rule, we have:\[ f'(x) = \frac{d}{dx}[\ln(1 + 2x)] = \frac{1}{1+2x} \cdot (2) = \frac{2}{1+2x} \]

Step 4: Evaluate the Derivative at a

Now evaluate the derivative \(f'(x)\) at \(a = 0\):\[ f'(0) = \frac{2}{1 + 2 \times 0} = \frac{2}{1} = 2 \]So, \(f'(a) = 2\).

Step 5: Construct the Linear Approximation

Using the values calculated, substitute in the linear approximation formula:\[ L(x) = f(a) + f'(a)(x-a) = 0 + 2(x-0) = 2x \]Thus, the linear approximation of \(f(x)\) at \(a = 0\) is \(L(x) = 2x\).

Key Concepts

Derivative

When we talk about derivatives, we are referring to a fundamental concept in calculus that describes how a function changes as its input changes. At its core, the derivative is the rate of change or the slope of the function at any given point. In our exercise, calculating the derivative of the function is crucial for forming a linear approximation.
In the function given by the exercise, \(f(x) = \ln (1+2x)\), we use the rule of differentiation to find its derivative. Applying differentiation rules correctly, especially for composite functions like this one, helps us understand how tiny changes in input cause changes in the function's output. The derivative \(f'(x)\) ends up being \(-\frac{2}{1+2x}\).
This derivative tells us how the function \(f(x) = \ln (1+2x)\) behaves at any point, providing insight into its increasing or decreasing nature.

Chain Rule

The chain rule is a powerful differentiation tool used when dealing with composite functions. A composite function is essentially a function within another function, and that's exactly what we have with \(f(x) = \ln(1+2x)\).
To apply the chain rule, we break the composite function into its outer and inner components. In our example, the outer function is \(\log(x)\) and the inner function is \(1+2x\). The rule tells us to differentiate the outer function and multiply it by the derivative of the inner function.
This process is crucial as it allows us to handle complex functions with precision. By applying the chain rule correctly, we arrive at the derivative \(f'(x) = \frac{2}{1+2x}\), essential for evaluating the linear approximation.

Function Evaluation

Function evaluation means finding the specific output of a function for a particular input value. In our exercise, evaluating \(f(x) = \ln(1+2x)\) at the point \(a = 0\) is necessary to start building the linear approximation.
When we substitute \(a = 0\) into our function, we calculate \(f(a) = \ln(1 + 2 \times 0) = \ln(1) = 0\). Thus, \(f(0) = 0\). This evaluation provides the baseline value for our linear approximation at \(a = 0\).
Function evaluation is a straightforward step, but it's an important building block in constructing the linear approximation model. With \(f(a)\) known, along with \(f'(a)\), we can accurately predict the function's behavior near the point we are interested in.