Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

Calculate \(g(2.03)\) if \(g(x)=\frac{(\sqrt{x}-\sqrt[3]{x})^{4}}{1-x+x^{2}}\).

Step-by-Step Solution

Answer

The value of \( g(2.03) \) is approximately 0.000202.

Step 1: Understand the Function

The function given is \( g(x) = \frac{(\sqrt{x} - \sqrt[3]{x})^4}{1 - x + x^2} \). This means we need to substitute \( x = 2.03 \) into the function to find \( g(2.03) \).

Step 2: Break Down the Numerator

The numerator is \((\sqrt{x} - \sqrt[3]{x})^4\). Substitute \( x = 2.03 \) into \( \sqrt{2.03} - \sqrt[3]{2.03} \) and then raise the result to the power of 4. Calculate each component separately: \( \sqrt{2.03} \approx 1.425\) and \( \sqrt[3]{2.03} \approx 1.267\). Thus, \( \sqrt{2.03} - \sqrt[3]{2.03} \approx 1.425 - 1.267 = 0.158 \). Raising 0.158 to the 4th power yields approximately \( 0.000625 \).

Step 3: Handle the Denominator

The denominator is \( 1 - x + x^2 \). Insert \( x = 2.03 \) into the expression: \( 1 - 2.03 + 2.03^2 \). First, compute \( 2.03^2 \approx 4.1209 \). Thus, the expression becomes \( 1 - 2.03 + 4.1209 = 3.0909 \).

Step 4: Combine Numerator and Denominator

Now, combine the results from the numerator and the denominator: \( g(2.03) = \frac{0.000625}{3.0909} \approx 0.000202 \).

Step 5: Final Calculation

Ensure the calculations are correct, double-check by performing the division \( 0.000625 \div 3.0909 \) using a calculator to obtain approximately \( 0.000202 \).

Key Concepts

Function Evaluation

In calculus, function evaluation involves finding the value of a function at a specific point. Here, we need to evaluate the function \( g(x) = \frac{(\sqrt{x} - \sqrt[3]{x})^4}{1-x+x^2} \) at \( x = 2.03 \). To do this, we substitute 2.03 into the function, replacing every occurrence of \( x \) with 2.03.

This substitution allows us to determine the output or result of the function for this specific input. Function evaluation is a fundamental tool in calculus and helps us understand how the value of a function changes with different inputs. It’s critical in problem-solving, as it transforms abstract function expressions into concrete numeric results.

Numerical Approximation

Numerical approximation is a technique used to find an approximate value for mathematical expressions that cannot be easily calculated exactly. In our problem, after substituting \( x = 2.03 \), we need to calculate expressions such as \( \sqrt{2.03} \) and \( \sqrt[3]{2.03} \).

These square and cube roots do not result in neat, whole numbers, so we use numerical methods or tools like calculators to approximate them. For example:

• \( \sqrt{2.03} \approx 1.425 \)
• \( \sqrt[3]{2.03} \approx 1.267 \)

This approach allows us to handle otherwise complex or time-consuming calculations practically, providing values that are "close enough" for practical use.

Step-by-Step Calculation

Engaging in step-by-step calculation is crucial for solving calculus problems accurately. It involves breaking down complex expressions into simpler parts and solving each one sequentially.

In our problem, this process involves:

• Firstly, handle the numerator \((\sqrt{x} - \sqrt[3]{x})^4\).
• Substitute \( x = 2.03 \) to calculate \( \sqrt{2.03} - \sqrt[3]{2.03} \), raising the result to the fourth power.
• Also, compute the denominator \(1 - x + x^2\) by plugging in \( x = 2.03 \).

This sequential approach ensures that each calculation follows logically from the last, minimizing the chances of errors and building a clear path to the solution.

Square Root

The square root of a number \( x \) is a value that, when multiplied by itself, gives \( x \). Symbolically, it is written as \( \sqrt{x} \). In our problem, calculating \( \sqrt{2.03} \) is one step toward evaluating the function.Understanding square roots is essential in calculus, as they frequently appear in optimization problems, integrals, and derivative calculations. Here, we use numerical approximation to estimate \( \sqrt{2.03} \approx 1.425 \), as it is not a perfect square, showcasing its importance in problem-solving.

Cube Root

A cube root of a number \( x \) is a number \( y \) such that \( y^3 = x \). It is denoted as \( \sqrt[3]{x} \). Within the provided function, the expression \( \sqrt[3]{2.03} \) is necessary for calculating the output of \( g(2.03) \).Cube roots are fundamental when dealing with volume-related problems or solving for roots of cubic equations. For our problem, finding \( \sqrt[3]{2.03} \approx 1.267 \) is vital.Just like square roots, cube roots are often non-integer numbers, requiring us to use approximations in practical applications. This concept also emphasizes the interconnection of different calculus elements in function evaluation.