Problem 29 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the function and point of approximation

We are given the function \( f(x) = \sqrt{1 + x^2} \), and we need to find its linear approximation at the point \( a = 0 \).

Step 2: Find the value of the function at the approximation point

Substitute \( a = 0 \) into the function: \( f(0) = \sqrt{1 + 0^2} = \sqrt{1} = 1 \).

Step 3: Calculate the derivative of the function

The derivative of \( f(x) = \sqrt{1 + x^2} \) is \( f'(x) = \frac{x}{\sqrt{1 + x^2}} \). This is found using the chain rule.

Step 4: Evaluate the derivative at the approximation point

Substitute \( x = 0 \) into the derivative: \( f'(0) = \frac{0}{\sqrt{1 + 0^2}} = 0 \).

Step 5: Construct the linear approximation

Use the linear approximation formula \( L(x) = f(a) + f'(a)(x - a) \). Substituting in our known values: \( L(x) = 1 + 0\cdot(x - 0) = 1 \).

Key Concepts

Understanding Derivatives

At the heart of calculus, the derivative tells us the rate at which a function is changing at any point. It's like a speedometer for a curve. Essentially, it measures how much the function's output changes for a tiny change in the input. When we say "taking the derivative of a function," we're finding another function that describes these rates of change. For the function \( f(x) = \sqrt{1 + x^2} \), the derivative \( f'(x) \) can be found using some foundational rules of differentiation. In our case, the derivative is calculated as \( f'(x) = \frac{x}{\sqrt{1 + x^2}} \). This result takes into account the changes in both the numerator and the denominator through a more complex function involving \( x \). Using the derivative, we can predict how the function behaves near a specific point, which is crucial for linear approximation.

Applying the Chain Rule

The chain rule is a powerful tool in calculus that helps us differentiate composite functions. A composite function is like a cake made of layers; you need to handle each layer separately. The chain rule works by breaking down these layers and finding how each part changes. In our example, \( f(x) = \sqrt{1 + x^2} \), the expression inside the square root, \( 1 + x^2 \), is an inner function. The chain rule tells us we need to differentiate both the outer function (the square root) and the inner function \( 1 + x^2 \). The derivative \( f'(x) = \frac{x}{\sqrt{1 + x^2}} \) is derived by differentiating the outer square root function and then multiplying by the derivative of the inner function, \( 2x \). This approach simplifies the process of dealing with complex derivatives by dividing them into manageable parts, like handling one layer of the cake at a time.

Point of Approximation

The point of approximation is where we zoom in on a curve to approximate it with a line. It's the specific value of \( x \) around which we are approximating the function. This point is denoted by \( a \) in the formula \( L(x)=f(a)+f^{\prime}(a)(x-a) \). In our exercise, the approximation point is \( a = 0 \). This choice simplifies the calculation: when you substitute \( a = 0 \) into the function and its derivative, you get \( f(0) = 1 \) and \( f'(0) = 0 \), respectively. This point is where our tangent line touches the curve, letting us estimate the function's values nearby. With linear approximation, understanding the behavior of the function at this point requires both the value of the function and its rate of change (derivative). The linear approximation formula then gives us a straight line that closely resembles our function's curve near \( a = 0 \).