Given the functions, f(x) = x3 + x2 + 1 and g(x) = -6x2 + 2, perform the indicated operations. When applicable, state the domain restriction.


Let's perform the indicated operations on the given functions \( f(x) = x^3 + x^2 + 1 \) and \( g(x) = -6x^2 + 2 \).

### 1. \( (f + g)(x) \)
The sum of the functions \( f(x) \) and \( g(x) \) is:

\[
(f + g)(x) = f(x) + g(x)
\]

\[
(f + g)(x) = (x^3 + x^2 + 1) + (-6x^2 + 2)
\]

Now, combine like terms:

\[
(f + g)(x) = x^3 + (x^2 - 6x^2) + (1 + 2)
\]

\[
(f + g)(x) = x^3 - 5x^2 + 3
\]

There are no domain restrictions for this sum since both functions are polynomials, which are defined for all real numbers. So, the domain is \( (-\infty, \infty) \).

### 2. \( (f - g)(x) \)
The difference of the functions \( f(x) \) and \( g(x) \) is:

\[
(f - g)(x) = f(x) - g(x)
\]

\[
(f - g)(x) = (x^3 + x^2 + 1) - (-6x^2 + 2)
\]

Distribute the negative sign:

\[
(f - g)(x) = x^3 + x^2 + 1 + 6x^2 - 2
\]

Now, combine like terms:

\[
(f - g)(x) = x^3 + (x^2 + 6x^2) + (1 - 2)
\]

\[
(f - g)(x) = x^3 + 7x^2 - 1
\]

Again, there are no domain restrictions for this difference since both functions are polynomials. The domain is \( (-\infty, \infty) \).

### 3. \( (f \cdot g)(x) \)
The product of the functions \( f(x) \) and \( g(x) \) is:

\[
(f \cdot g)(x) = f(x) \cdot g(x)
\]

\[
(f \cdot g)(x) = (x^3 + x^2 + 1) \cdot (-6x^2 + 2)
\]

Now, distribute \( (-6x^2 + 2) \) across \( (x^3 + x^2 + 1) \):

\[
(f \cdot g)(x) = (x^3 \cdot -6x^2) + (x^3 \cdot 2) + (x^2 \cdot -6x^2) + (x^2 \cdot 2) + (1 \cdot -6x^2) + (1 \cdot 2)
\]

Simplify each term:

\[
(f \cdot g)(x) = -6x^5 + 2x^3 - 6x^4 + 2x^2 - 6x^2 + 2
\]

Combine like terms:

\[
(f \cdot g)(x) = -6x^5 - 6x^4 + 2x^3 - 4x^2 + 2
\]

Since both functions are polynomials, the domain is \( (-\infty, \infty) \).

### 4. \( \left( \frac{f}{g} \right)(x) \)
The division of the functions \( f(x) \) and \( g(x) \) is:

\[
\left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)} = \frac{x^3 + x^2 + 1}{-6x^2 + 2}
\]

For this quotient, the domain will be restricted by the values of \( x \) that make the denominator zero, because division by zero is undefined.

Set \( g(x) = 0 \):

\[
-6x^2 + 2 = 0
\]

Solve for \( x \):

\[
-6x^2 = -2 \implies x^2 = \frac{1}{3} \implies x = \pm \frac{\sqrt{3}}{3}
\]

Thus, the function is undefined at \( x = \pm \frac{\sqrt{3}}{3} \).

Therefore, the domain of \( \left( \frac{f}{g} \right)(x) \) is all real numbers except \( x = \pm \frac{\sqrt{3}}{3} \), or:

\[
\boxed{(-\infty, -\frac{\sqrt{3}}{3}) \cup (-\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}) \cup (\frac{\sqrt{3}}{3}, \infty)}
\]

---

Would you like to explore any part of this solution in more detail?

Here are 5 related questions to extend your learning:
1. How would the behavior of \( f(x) + g(x) \) change if one function had a different degree?
2. Can you analyze the behavior of \( f(x) \cdot g(x) \) for large values of \( x \)?
3. What is the importance of domain restrictions when dividing functions?
4. How can you graph \( (f + g)(x) \) and \( (f - g)(x) \) to visualize their combined effects?
5. How does polynomial degree affect the long-term behavior of products like \( f(x) \cdot g(x) \)?

**Tip:** When multiplying or dividing functions, always check for domain restrictions caused by potential zero values in the denominator.

Given the functions, f(x) = x^3 + x^2 + 1 and g(x) = -6x^2 + 2, perform the indicated operations. When applicable, state the domain restriction.

Learn how to perform operations on the polynomial functions f(x) = x^3 + x^2 + 1 and g(x) = -6x^2 + 2. This includes addition, subtraction, multiplication, and division, along with an explanation of domain restrictions for rational functions. Suitable for students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation