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

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

Step 1: Substitute π into g(x)

We first need to substitute the value of \(x = \pi\) into the function \(g(x) = 6x - 11\). Doing so, we obtain \(g(\pi) = 6\pi - 11\).

Step 2: Calculate g³(π)

Next, we calculate \(g^3(\pi)\). From the previous step, we have \(g(\pi) = 6\pi - 11\), so \(g^3(\pi) = (6\pi - 11)^3\). Calculate this value by performing polynomial expansion.

Step 3: Calculate g³(π) - g(π)

With \(g^3(\pi) = (6\pi - 11)^3\) and \(g(\pi) = 6\pi - 11\), compute their difference: \(g^3(\pi) - g(\pi) = (6\pi - 11)^3 - (6\pi - 11)\).

Step 4: Compute the Cube Root

Finally, take the cube root of the result from the previous step. This gives us: \(\left[(6\pi - 11)^3 - (6\pi - 11)\right]^{1/3}\). Simplify this expression to get the answer.

Key Concepts

Polynomial Functions

Polynomial functions are a cornerstone of calculus and algebra. They are represented as expressions involving variables and coefficients, where the variable is raised to non-negative integer powers. A simple example would be something like: \( f(x) = ax^n + bx^{n-1} + \, ... \, + k \). Here,

• Each term consists of a coefficient (like \(a\), \(b\), etc.).
• The powers of \(x\) are whole numbers.
• \(n\) represents the degree of the polynomial.

In the exercise, the function \(g(x) = 6x - 11\) is a polynomial function with a degree of 1, commonly known as a linear function. Understanding the structure of polynomials helps in comprehending how they behave under different operations such as function substitution and polynomial expansion.

Function Substitution

Function substitution is a process where you replace the variable in a function with a specified value. This is a fundamental technique used to evaluate functions at particular points. In our exercise, - The function is \(g(x) = 6x - 11\).- We substitute \(x\) with \(\pi\) to get \(g(\pi) = 6\pi - 11\). Substitution helps in determining the value of the function at a specific point. This straightforward process is essential in problem solving, especially when followed by other operations like polynomial expansion or simplification.

Cube Root Calculation

Calculating the cube root can be challenging if you're not familiar with radical expressions. The cube root of a number \(a\) is a value that, when raised to the power of three, gives \(a\). It's denoted by \( \sqrt[3]{a} \). For example, - Cube root of 8 is 2 because \(2^3 = 8\).In this exercise, the cube root calculation is the final step. You compute the cube root of the expression \( (6\pi - 11)^3 - (6\pi - 11) \). Accurately finding the cube root simplifies the solution significantly. Recognizing patterns or using computational tools often helps with these complex calculations, especially for non-standard constants like \(\pi\).

Polynomial Expansion

Polynomial expansion is a method used to express a polynomial that has been raised to a power in an expanded form. This is achieved through the binomial theorem or by repeated distribution. For example,- Expanding \((a + b)^2\) gives \(a^2 + 2ab + b^2\).In our exercise, we had to expand \((6\pi - 11)^3\). The expansion allows you to multiply the polynomial by itself multiple times, ultimately finding each term involved. This step is crucial for simplifying larger expressions to perform further operations such as subtraction and cube root calculations. Mastering expansion techniques can greatly enhance your ability to tackle complex equations in calculus and algebra.