Problem 56 - Calculus Volume 1

Question

Compute the values given within 0.01 by deciding on the appropriate \(f(x)\) and \(a\), and evaluating \(L(x)=f(a)+f^{\prime}(a)(x-a)\). Check your answer using a calculator. [T] \((2.001)^{6}\)

Step-by-Step Solution

Answer

The value is approximately 64.192.

Step 1: Choose the Appropriate Function and Point

For the calculation of powers such as \(2.001^6\), the function \(f(x) = x^6\) is appropriate. We choose \(a = 2\) because it is a close, more straightforward value around which to evaluate the function.

Step 2: Evaluate the Function at the Point

Calculate \(f(a) = 2^6.\) Therefore, \(f(2) = 64\) since multiplying \(2 imes 2 imes 2 imes 2 imes 2 imes 2 = 64.\)

Step 3: Find the Derivative of the Function

Compute the derivative \(f'(x) \). Since \(f(x) = x^6\), its derivative is \(f'(x) = 6x^5\).

Step 4: Evaluate the Derivative at the Point

Calculate \(f'(a) = 6 imes (2)^5\). \This results in \(f'(2) = 6 imes 32 = 192\) since \(2^5 = 32\).

Step 5: Use Linear Approximation Formula

Substitute into the linear approximation formula \(L(x) = f(a) + f'(a)(x-a).\)Substitute \(f(a) = 64\), \(f'(a) = 192\), and \(x = 2.001\) to get \\(L(2.001) = 64 + 192 imes (2.001 - 2) = 64 + 192 imes 0.001.\)

Step 6: Compute the Value

Calculate \(L(2.001) = 64 + 192 imes 0.001 = 64 + 0.192 = 64.192.\)

Step 7: Verification with Calculator

Using a calculator or computational tool, evaluate \(2.001^6\).You should find that it is approximately \(64.192016\), verifying that the linear approximation \(L(x)\) computed matches closely with the calculator's output.

Key Concepts

Derivative

A derivative is a fundamental concept in calculus used to understand how a function changes as its input changes. Think of it as a tool that gives you the "rate of change" or the "slope" of a function at any given point. In our linear approximation problem, we started with the function \(f(x) = x^6\). To find the derivative of this function, we use the power rule of differentiation.

• The power rule states that if \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\).

Applying this rule to our function, we get \(f'(x) = 6x^5\). This derivative helps us understand how \(f(x)\) behaves near the point \(a = 2\) that we selected. Evaluating the derivative at \(a = 2\), we plug into \(f'(a) = 6 \times (2)^5 = 192\), indicating the steepness of the function \(x^6\) around this point. Understanding derivatives is crucial for accurately using the linear approximation formula.

Function Evaluation

Function evaluation involves plugging a specific value into a function and computing the result. In our example, we evaluated the function \(f(x) = x^6\) at \(a = 2\).

• We found \(f(2) = 2^6 = 64\).
• This step is about computing the actual value of the function at our chosen point.

Choosing \(a = 2\) simplifies the computation since calculating powers of 2 is straightforward. Performing this evaluation accurately is vital because it forms the base value for our linear approximation, \(L(x) = f(a) + f'(a)(x-a)\). In this scenario, knowing that \(f(2) = 64\) lays the groundwork for computing the approximation of \(2.001^6\). Function evaluation sets the stage for the next calculations.

Power Calculation

When calculating powers, we are multiplying a number by itself a certain number of times. In our task of approximating \((2.001)^6\), the calculation of \(2^6\) was a straightforward first step. Calculating powers efficiently is critical:

• Powers of integers, especially small bases like 2, are often easier to handle and compute accurately.
• For example, \(2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64\).

Understanding power calculations is not just about knowing multiplication; it's about applying it effectively in approximations and problem-solving. In this context, knowing \(2^6\) helps us start our approximation, as it served as the value for \(f(a)\) in the linear approximation formula.

Verification with Calculator

To ensure our linear approximation is accurate, verifying with a calculator is essential. By using technology to check our work, we can verify that the approximation isn't just mathematical in theory, but practical in computation. In our example:

• After calculating \(L(2.001) = 64.192\), we verified this by computing \((2.001)^6\) with a calculator.
• The calculator gave us approximately 64.192016, very close to our approximation.

This step reassures us that our manual calculations are correct. It serves as a "reality check," confirming that our linear approximation closely matches observational results. Calculators are invaluable in verifying exactness, particularly when decimal placement and precision are crucial.