Let's solve each problem step by step.
---
Problem 1:
If \( f(x) = 4x + 1 \) and \( g(x) = 4x - 3 \), find \( f(x) \cdot g(x) \).
#### Solution:
We need to multiply the two functions:
\[
f(x) \cdot g(x) = (4x + 1)(4x - 3)
\]
Use the distributive property (FOIL method):
\[
(4x + 1)(4x - 3) = 4x \cdot 4x + 4x \cdot (-3) + 1 \cdot 4x + 1 \cdot (-3)
\]
\[
= 16x^2 - 12x + 4x - 3
\]
Combine like terms:
\[
= 16x^2 - 8x - 3
\]
Thus, the answer is:
\[
\boxed{C}
\]
---
Problem 2:
If \( f(n) = n^2 + 1 \) and \( g(n) = 2n - 5 \), find \( f(n) - g(n) \).
#### Solution:
We need to subtract the two functions:
\[
f(n) - g(n) = (n^2 + 1) - (2n - 5)
\]
Distribute the negative sign:
\[
= n^2 + 1 - 2n + 5
\]
Combine like terms:
\[
= n^2 - 2n + 6
\]
Thus, the answer is:
\[
\boxed{A}
\]
---
Problem 3:
If \( f(x) = 3x^2 + 7x \) and \( g(x) = 2x^2 - x - 1 \), find \( (f + g)(x) \).
#### Solution:
We need to add the two functions:
\[
(f + g)(x) = (3x^2 + 7x) + (2x^2 - x - 1)
\]
Combine like terms:
\[
= 3x^2 + 2x^2 + 7x - x - 1
\]
\[
= 5x^2 + 6x - 1
\]
Thus, the answer is:
\[
\boxed{D}
\]
---
Problem 4:
All of the \( x \)-values or inputs are called what?
#### Solution:
In a function, the set of all possible input values (\( x \)-values) is called the
domain.
Thus, the answer is:
\[
\boxed{D}
\]
---
Problem 5:
When \( f(x) = x^2 - 9 \) and \( g(x) = 3x \), find \( \frac{f(x)}{g(x)} \).
#### Solution:
We need to divide the two functions:
\[
\frac{f(x)}{g(x)} = \frac{x^2 - 9}{3x}
\]
Factor the numerator \( x^2 - 9 \):
\[
x^2 - 9 = (x + 3)(x - 3)
\]
So,
\[
\frac{f(x)}{g(x)} = \frac{(x + 3)(x - 3)}{3x}
\]
This expression cannot be simplified further without additional context (e.g., restrictions on \( x \)).
Thus, the answer is:
\[
\boxed{D}
\]
---
Final Answers:
1. \(\boxed{C}\)
2. \(\boxed{A}\)
3. \(\boxed{D}\)
4. \(\boxed{D}\)
5. \(\boxed{D}\)
Parent Tip: Review the logic above to help your child master the concept of 10th grade printable worksheet.
