Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

Calculate \(\left[g^{3}(\pi)-g(\pi)\right]^{1 / 3}\) if \(g(x)=6 x-11\).

Step-by-Step Solution

Answer

The expression requires substituting \( \pi \) into \( g(x) \) and simplifying to evaluate.

Step 1: Substitute Function into Expression

First, substitute the function \( g(x) = 6x - 11 \) into the expression \( g^3(\pi) - g(\pi) \). This means we need to find \( g(\pi) \) first.

Step 2: Compute \( g(\pi) \)

Plug \( \pi \) into the function: \( g(\pi) = 6 \cdot \pi - 11 \). Therefore, \( g(\pi) = 6\pi - 11 \).

Step 3: Calculate \( g^3(\pi) \)

Raise \( g(\pi) \) to the power of 3: \( g^3(\pi) = (6\pi - 11)^3 \).

Step 4: Subtract the Expressions

Evaluate \( g^3(\pi) - g(\pi) \) which is \((6\pi - 11)^3 - (6\pi - 11)\).

Step 5: Simplify the Expression

Here, simplifying the given expression may not yield a simple expression due to complexity, hence it's often left or computed using a calculator for an approximate numerical solution.

Step 6: Take the Cube Root

Now take the cube root of the expression: \( \left[ (6 \pi - 11)^3 - (6 \pi - 11) \right]^{1/3} \). Given the complexity, the most straightforward method is calculation using a calculator.

Key Concepts

Mathematical functions

A mathematical function is a relation that uniquely associates members of one set with members of another set. In real-world contexts, functions often provide a way to describe how changing one quantity results in the change of another.
For example, in this exercise, the function \( g(x) = 6x - 11 \) is linear and takes an input value \( x \) to produce a corresponding output.
The expression is used to describe how inputs like \( \pi \) are transformed using the function.

• The constant "6" is the coefficient that scales the input.
• The "-11" is a constant that shifts the output vertically.

Understanding these components helps clarify how to apply and interpret the function.

Exponents in calculus

Exponents in calculus are fundamental when dealing with expressions raised to a power. They appear in many forms, but the basic idea is that they represent repeated multiplication of a number by itself.
In this problem, we encounter the cube of a function value. Raising \((6\pi - 11)\) to the power of 3 means we'll calculate the function value for \( g(\pi) \) and then multiply it by itself three times.

• Cubing an expression expands it in a polynomial form.
• It gives a higher degree polynomial which usually complicates manual calculations.

This is why such expressions are often computed using calculators or algebraic software to acquire useful approximations.

Simplification in algebra

Simplification in algebra helps reduce complex expressions to simpler or more manageable forms. In this exercise, the complexity lies in handling the expression \((6\pi - 11)^3 - (6\pi - 11)\).
The goal is to either simplify it algebraically to a form that can be easily worked with or compute it straightforwardly using tools.

• Identify like terms that can be combined.
• Use algebraic identities where applicable, though here it isn't direct.

Due to the high degree when expanded, simplification might not always lead to neat solutions without numerical evaluations.

Use of constants (pi)

Constants like \( \pi \) (approximately 3.14159) are fixed values widely used in mathematics, especially in calculus where they often appear in trigonometric and geometric contexts.
In our exercise, \( \pi \) is input into the function \( g(x) \), producing numerical values in expressions, linking the geometrical concept of circumference or area to calculus operations.

• \( \pi \) represents the ratio of a circle's circumference to its diameter.
• It's an irrational number, meaning it can't be expressed as a simple fraction.

Using \( \pi \) in calculus can often lead to more accurate and meaningful results, especially in real-world geometric problems.