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 0.00018.

Step 1: Substitute Value into Function

First, substitute 2.03 into the function for x. The function is given by \[ g(x) = \frac{(\sqrt{x} - \sqrt[3]{x})^4}{1-x+x^2}. \]Replacing \(x\) with 2.03, we have:\[ g(2.03) = \frac{(\sqrt{2.03} - \sqrt[3]{2.03})^4}{1-2.03+2.03^2}. \]

Step 2: Calculate Numerator

Evaluate \(\sqrt{2.03}\) and \(\sqrt[3]{2.03}\):- \(\sqrt{2.03} \approx 1.424 \),- \(\sqrt[3]{2.03} \approx 1.270 \).Thus, \( \sqrt{2.03} - \sqrt[3]{2.03} \approx 1.424 - 1.270 = 0.154 \).Now, raise this difference to the power of 4:\[(0.154)^4 \approx 0.00056.\]

Step 3: Calculate Denominator

Evaluate the denominator of the function:\[1 - 2.03 + (2.03)^2.\]Calculate \((2.03)^2:\)\[(2.03)^2 = 4.1209.\]Substituting into the expression gives:\[1 - 2.03 + 4.1209 = 3.0909.\]

Step 4: Compute Function Value

Now divide the calculated numerator by the calculated denominator:\[ g(2.03) = \frac{0.00056}{3.0909} \approx 0.00018. \]

Key Concepts

Function Evaluation

Understanding function evaluation is essential in calculus as it allows us to find the output of a function for given inputs. The process involves plugging the input values into the function. In our exercise, this involves substituting \( x = 2.03 \) into the function.

• Review the structure of the function: \( g(x) = \frac{(\sqrt{x} - \sqrt[3]{x})^4}{1-x+x^2} \).
• Identify where the input value replaces the variable \( x \) in both the numerator and denominator.

Always ensure that substitutions are done precisely. Any small error can lead to a significant change in the outcome. Practicing function evaluations helps deepen comprehension and boosts confidence when handling complex functions.

Substitution Method

The substitution method is a straightforward technique often used to simplify problems, especially when dealing with equations or functions. It involves replacing variables with actual numbers or other expressions.

• This method helps in transitioning from a general form of a function into a specific form with known values.
• It is especially useful when determining the value of a function at a given point, such as \( g(2.03) \) in our example.

Ensure that each part of the function is carefully substituted to avoid mistakes. It acts as a bridge between the abstract function and its numerical evaluation.

Numerical Approximation

Numerical approximation comes into play when exact values cannot be easily calculated or are irrational numbers. For example, in calculating roots like \( \sqrt{2.03} \) and \( \sqrt[3]{2.03} \), we rely on approximations.

• Use calculators or estimation techniques to find these approximations.
• Approximations provide a practical approach to solving real-world problems.

While exact values are ideal, numerical approximations allow us to analyze and interpret functions when perfect precision isn't feasible. They are fundamental in making calculus applicable in real life.

Rational Functions

A rational function is a ratio of two polynomials. It is fundamental in understanding the behavior of many mathematical expressions and functions.

• In our example, the function \( g(x) = \frac{(\sqrt{x} - \sqrt[3]{x})^4}{1-x+x^2} \) is rational, with both the numerator and denominator forming the basis of a ratio.
• The numerator involves the fourth power of the difference between two radical expressions, while the denominator is a quadratic expression.

Understanding rational functions is crucial as it allows us to simplify, analyze, and solve complex equations and predict well-defined behavior of functions.