Problem 1 - Calculus for Biology and Medicine

Question

Use the formula $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ to approximate the value of the given function. Then compare your result with the value you get from a calculator. $$ \sqrt{65} \text { ; let } f(x)=\sqrt{x}, a=64 \text { , and } x=65 $$

Step-by-Step Solution

Answer

The linear approximation of \( \sqrt{65} \) is 8.0625, closely matching the calculator value 8.0623.

Step 1: Identify the Function and Its Derivative

We are given the function \( f(x) = \sqrt{x} \). First, we need to find its derivative. The derivative of \( \sqrt{x} \) is \( f'(x) = \frac{1}{2\sqrt{x}} \).

Step 2: Evaluate the Function at a

Substitute \( a = 64 \) into the function to find \( f(a) \): \( f(64) = \sqrt{64} = 8 \).

Step 3: Evaluate the Derivative at a

Substitute \( a = 64 \) into the derivative to find \( f'(a) \): \( f'(64) = \frac{1}{2\sqrt{64}} = \frac{1}{16} \).

Step 4: Substitute into the Linear Approximation Formula

Use the formula \( f(x) \approx f(a) + f'(a)(x-a) \). Substitute \( f(a) = 8 \), \( f'(a) = \frac{1}{16} \), and \( x = 65 \): \[ f(65) \approx 8 + \frac{1}{16}(65 - 64) \].

Step 5: Calculate the Approximation

Compute the approximation: \[ f(65) \approx 8 + \frac{1}{16}(1) = 8 + \frac{1}{16} = 8.0625 \].

Step 6: Calculate the Exact Value Using a Calculator

Enter \( \sqrt{65} \) into a calculator to find the exact value. This results in approximately 8.0623.

Step 7: Compare the Two Values

Compare the linear approximation \( 8.0625 \) with the calculator result \( 8.0623 \). The approximation is very close to the exact value.

Key Concepts

Derivative Calculation

To approximate the value of a function using linear approximation, understanding the derivative is essential. It helps us determine how the function behaves at small changes around a specific point.

The function given here is the square root of x, written as \( f(x) = \sqrt{x} \). To find the derivative of \( \sqrt{x} \), we start by recognizing that the derivative of \( x^{1/2} \) involves using the power rule. So, the derivative is:

• \( f'(x) = \frac{1}{2\sqrt{x}} \)

This derivative tells us how quickly \( \sqrt{x} \) changes at any point \( x \). It's crucial for linear approximation because it allows us to compute a slope that gives us an approximate line near the value we're interested in. Knowing how to calculate derivatives helps simplify real-world problems and provides insight into how functions behave.

Function Evaluation

Function evaluation involves calculating the value of a function at a specific point. It's a straightforward step that sets the stage for linear approximation.

In this exercise, we're asked to evaluate the function \( f(x) = \sqrt{x} \) at \( x = 64 \), which is labeled as \( a \). Using basic arithmetic, we determine:

• \( f(64) = \sqrt{64} = 8 \)

Through evaluation, we verify that the real value at this point is \( 8 \). Function evaluation doesn't just provide necessary numbers for the approximation; it reaffirms our understanding of the standard operations and how they reflect the change in values.

Value Comparison

To assess the accuracy of our approximation, comparing its result to a calculator-generated value is indispensable. This exercise illustrates the efficacy of linear approximation compared to the exact computation.

After calculating the linear approximation for \( \sqrt{65} \), we obtained 8.0625. Plugging \( \sqrt{65} \) into a calculator yields approximately 8.0623. Here's how they compare:

• Approximation: 8.0625
• Exact (Calculator): 8.0623

The close match between the two values shows that linear approximation can provide a reliable estimation, especially useful for quick calculations when precision is not critical. By mastering approximation techniques, we empower ourselves to tackle complex problems more efficiently.