Problem 18 - Calculus for Biology and Medicine

Question

Calculate the linear approximation for \(f(x)\) : $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ \(f(x)=\ln (1+2 x)\) at \(a=0\)

Step-by-Step Solution

Answer

The linear approximation is \( f(x) \approx 2x \).

Step 1: Understand the problem

The goal is to find the linear approximation of a function, essentially the tangent line at a given point. The function provided is \( f(x) = \ln(1+2x) \), and we need to approximate this function near the point \( a = 0 \).

Step 2: Calculate the function value at \( a = 0 \)

Substitute \( a = 0 \) into the function \( f(x) = \ln(1+2x) \) to find \( f(a) \):\[f(0) = \ln(1 + 2 \cdot 0) = \ln(1) = 0\]

Step 3: Find the derivative of the function

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

Step 4: Calculate the derivative value at \( a = 0 \)

Substitute \( a = 0 \) into the derivative to find \( f'(a) \):\[f'(0) = \frac{2}{1+2 \cdot 0} = \frac{2}{1} = 2\]

Step 5: Apply the linear approximation formula

Substitute \( f(a) \) and \( f'(a) \) into the linear approximation formula:\[f(x) \approx f(a) + f'(a)(x-a)\]\[\therefore f(x) \approx 0 + 2(x-0) = 2x\]

Step 6: Interpret the result

The linear approximation for the function \( f(x) = \ln(1+2x) \) at \( a = 0 \) is \( f(x) \approx 2x \). This means near \( x = 0 \), \( f(x) \) is approximately \( 2x \).

Key Concepts

Derivative

In mathematics, the derivative represents the rate at which a function is changing at any given point. It's like looking at how the slope of a function changes. In simpler terms, imagine you're walking on a hill; the steeper the slope, the harder it is to climb. This slope is what we refer to as the derivative.
The derivative of a function gives us a formula that tells us the slope of a tangent line to the function at any point. This is crucial because it helps us understand the behavior of the function in a small neighborhood around that point. For the function we were working with, which is \(f(x) = \ln(1+2x)\), we used the chain rule (explained later) to find its derivative:
\[f'(x) = \frac{2}{1+2x}\]
This tells us how steep the curve is at any point \(x\).
The derivative can help to approximate the function around the point of interest such as when we calculated at \(a = 0\), where the derivative \(f'(0)\) was \(2\). This value indicates that at \(x=0\), the curve is rising at a rate of \(2\). This rate is crucial for building the linear approximation, which leads us to the concept of the tangent line.

Tangent Line

The tangent line to a curve at a given point is the line that just "touches" the curve at that point and reflects the immediate direction of the curve. You can think of it like a tangent line on a circle, which just barely meets the circle at one point. It's essentially the best linear approximation to the curve at that point.
When we compute a linear approximation for a function using its derivative, we're essentially finding the equation of the tangent line.

• The equation for a tangent line at point \(a\) on the function \(f(x)\) is: \[f(x) \approx f(a) + f'(a)(x-a)\]
• This formula uses \(f(a)\), the actual function value at \(a\), and \(f'(a)\), the derivative (slope) at \(a\).

The tangent line is powerful because it simplifies complex functions into simple linear expressions like in our example, where we've simplified into \(f(x) \approx 2x\) around \(x=0\).
This approximation means we replace our curve \(\ln(1+2x)\) with this simpler, easy-to-compute line just around \(x=0\), giving us a quick and easy way to estimate the function’s values.

Chain Rule

The chain rule is a fundamental technique in calculus for finding the derivative of composite functions. Think of it as a rule that helps take apart composite functions like peeling layers of an onion.
Suppose you have a function \( h(x) \) that is a composition of two functions, say \( f(g(x)) \). The chain rule tells us how to take the derivative of such a composite function. It states:
\[(h(x))' = f'(g(x)) \cdot g'(x)\]
This means you first differentiate the outer function \( f \) with respect to its argument \( g(x) \), and then multiply it by the derivative of the inner function \( g(x) \).
In our exercise, the function \( f(x) = \ln(1+2x) \) is a composition where the outer function is the natural logarithm \(\ln\) and the inner function is \(1+2x\).
Using the chain rule, we applied:

• \( f'(x) = \frac{1}{1+2x} \) for the logarithmic part, because the derivative of \( \ln(u) \) is \( \frac{1}{u} \).
• Then multiplied by the derivative of the inner function \( (1+2x)' = 2 \).

This resulted in the final derivative:
\[f'(x) = \frac{2}{1+2x}\]
The chain rule allows navigation through complex functions seamlessly to calculate a derivative and enables us to apply these results in practical applications like approximation.