Question

using the appropriate theorems, explain why the function is continuous at every number in its domain.
q(x) = \frac{\sqrt3{x - 2}}{x^3 - 2}
\bigcirc q(x) is a polynomial, so it is continuous at every number in its domain.
\bigcirc q(x) is a rational function, so it is continuous at every number in its domain.
\bigcirc q(x) is built up from functions that are continuous for all real numbers, so it is continuous at every number in its domain.
\bigcirc q(x) is not continuous at every number in its domain.
\bigcirc none of these
find the domain. (enter your answer using interval notation.)

Explanation:

Part 1: Explaining Continuity

1. Analyze the function \( Q(x)=\frac{\sqrt[3]{x - 2}}{x^{3}-2} \). The numerator \( \sqrt[3]{x - 2} \) is a cube - root function. Cube - root functions \( \sqrt[3]{u} \) are continuous for all real numbers \( u \) (since the domain of \( y = \sqrt[3]{u} \) is \( (-\infty,\infty) \) and it is a smooth function everywhere). The denominator \( x^{3}-2 \) is a polynomial function. Polynomial functions are continuous for all real numbers. A rational function (a function of the form \( \frac{f(x)}{g(x)} \) where \( f(x) \) and \( g(x) \) are polynomials and \( g(x)

eq0 \)) is continuous at every number in its domain. The domain of \( Q(x) \) is all real numbers except where the denominator \( x^{3}-2 = 0 \), i.e., \( x=\sqrt[3]{2} \). So \( Q(x) \) is a rational function, and rational functions are continuous on their domain.

1. Now let's analyze the other options:

• Option 1: \( Q(x) \) is not a polynomial. A polynomial is of the form \( a_{n}x^{n}+a_{n - 1}x^{n - 1}+\cdots+a_{1}x + a_{0} \), and \( Q(x) \) has a denominator, so it is not a polynomial.
• Option 3: While the cube - root and the polynomial in the denominator are continuous on \( (-\infty,\infty) \), the key here is that \( Q(x) \) is a rational function. The description in option 3 is a more general statement, but the most accurate description is that it is a rational function (since it is a quotient of two polynomials - the numerator \( \sqrt[3]{x - 2}=(x - 2)^{\frac{1}{3}} \) is not a polynomial, wait, correction: The numerator \( \sqrt[3]{x-2}=(x - 2)^{\frac{1}{3}} \) is a power function with a fractional exponent, and the denominator \( x^{3}-2 \) is a polynomial. The function \( Q(x) \) is built from the cube - root function (continuous everywhere) and the polynomial function (continuous everywhere) and the operation of division (which is continuous except where the denominator is zero). But the option that says "Q(x) is built up from functions that are continuous for all real numbers, so it is continuous at every number in its domain" is also correct in a sense, but let's re - evaluate. Wait, the cube - root function \( y=\sqrt[3]{u} \) is continuous for all real \( u \), and the function \( u=x - 2 \) is a polynomial (hence continuous everywhere), so the composition \( \sqrt[3]{x - 2} \) is continuous everywhere. The denominator \( y=x^{3}-2 \) is a polynomial, hence continuous everywhere. The function \( Q(x) \) is the quotient of two functions that are continuous everywhere, except where the denominator is zero. So the function \( Q(x) \) is built up from functions that are continuous for all real numbers (the cube - root function and the polynomial function) and the operation of division (which preserves continuity on the domain of the quotient). So the option "Q(x) is built up from functions that are continuous for all real numbers, so it is continuous at every number in its domain" is correct. Wait, earlier mistake: The numerator \( \sqrt[3]{x - 2} \) is not a polynomial, but it is a function that is continuous everywhere. The denominator \( x^{3}-2 \) is a polynomial (continuous everywhere). So when we take the quotient, the function \( Q(x) \) is continuous on its domain (all real numbers except \( x = \sqrt[3]{2} \)) because it is a quotient of two functions that are continuous everywhere, and we exclude the point where the denominator is zero.
• Option 4: \( Q(x) \) is continuous at every number in its domain. The domain is all real numbers except \( x=\sqrt[3]{2} \), and at all other points, it is continuous. So option 4 is wrong.
• Option 5:…

Step 1: Analyze the denominator

To find the domain of \( Q(x)=\frac{\sqrt[3]{x - 2}}{x^{3}-2} \), we need to find where the denominator \( x^{3}-2
eq0 \).
Solve the equation \( x^{3}-2 = 0 \). We can rewrite it as \( x^{3}=2 \), and then \( x=\sqrt[3]{2} \) (since the cube - root function is one - to - one, and for the equation \( y^{3}=a \), the solution is \( y = \sqrt[3]{a} \)).

Step 2: Analyze the numerator

The numerator is \( \sqrt[3]{x - 2} \). The cube - root function \( \sqrt[3]{u} \) is defined for all real numbers \( u \). So the expression \( \sqrt[3]{x - 2} \) is defined for all real numbers \( x \) (because for any real number \( x \), \( x - 2 \) is a real number, and the cube - root of a real number is defined).

Step 3: Determine the domain

The domain of \( Q(x) \) is all real numbers except \( x=\sqrt[3]{2} \). In interval notation, this is \( (-\infty,\sqrt[3]{2})\cup(\sqrt[3]{2},\infty) \).

Answer:

Q(x) is built up from functions that are continuous for all real numbers, so it is continuous at every number in its domain.

Part 2: Finding the Domain