Let's tackle part (a) first:
a) Finding the domain of \( f, g, f+g, f-g, fg, ff, \frac{f}{g}, \frac{g}{f} \).
To begin with the basic definitions of the functions:
- \( f(x) = x^3 \)
- \( g(x) = 4x^2 + 21x - 18 \)
#### Domain Definitions:
- The domain of a polynomial function is all real numbers (\( \mathbb{R} \)).
Thus:
- The domain of \( f(x) \) = \( \mathbb{R} \) because it is a polynomial function.
- The domain of \( g(x) \) = \( \mathbb{R} \) because it is a polynomial function.
Now, let's find the domain of the combined functions:
1. \( f+g \):
[tex]\[ (f+g)(x) = f(x) + g(x) = x^3 + 4x^2 + 21x - 18 \][/tex]
The domain of \( f+g \) is \( \mathbb{R} \) since both \( f \) and \( g \) have the domain \( \mathbb{R} \).
2. \( f-g \):
[tex]\[ (f-g)(x) = f(x) - g(x) = x^3 - (4x^2 + 21x - 18) = x^3 - 4x^2 - 21x + 18 \][/tex]
The domain of \( f-g \) is \( \mathbb{R} \).
3. \( fg \):
[tex]\[ (fg)(x) = f(x) \cdot g(x) = x^3 \cdot (4x^2 + 21x - 18) \][/tex]
The domain of \( fg \) is \( \mathbb{R} \).
4. \( ff \):
[tex]\[ (ff)(x) = f(f(x)) = f(x^3) = (x^3)^3 = x^9 \][/tex]
The domain of \( ff \) is \( \mathbb{R} \).
5. \( \frac{f}{g} \):
[tex]\[ \left( \frac{f}{g} \right)(x) = \frac{x^3}{4x^2 + 21x - 18} \][/tex]
We need to exclude values of \( x \) for which the denominator is zero, i.e., the solutions to \( 4x^2 + 21x - 18 = 0 \).
To find these zeros, we solve the quadratic equation \( 4x^2 + 21x - 18 = 0 \). Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
[tex]\[ a = 4, \, b = 21, \, c = -18 \][/tex]
[tex]\[ x = \frac{-21 \pm \sqrt{21^2 - 4 \cdot 4 \cdot (-18)}}{2 \cdot 4} = \frac{-21 \pm \sqrt{441 + 288}}{8} = \frac{-21 \pm \sqrt{729}}{8} = \frac{-21 \pm 27}{8} \][/tex]
[tex]\[ x = \frac{6}{8} = \frac{3}{4} \quad \text{or} \quad x = \frac{-48}{8} = -6 \][/tex]
Therefore, the domain of \( \frac{f}{g} \) is \( \mathbb{R} \setminus \left\{ -6, \frac{3}{4} \right\} \).
6. \( \frac{g}{f} \):
[tex]\[ \left( \frac{g}{f} \right)(x) = \frac{4x^2 + 21x - 18}{x^3} \][/tex]
We need to exclude values of \( x \) where the denominator is zero, i.e., \( x = 0 \).
Thus, the domain of \( \frac{g}{f} \) is \( \mathbb{R} \setminus \{ 0 \} \).
To summarize:
- The domain of \( f \) is \( \mathbb{R} \).
- The domain of \( g \) is \( \mathbb{R} \).
- The domain of \( f+g \) is \( \mathbb{R} \).
- The domain of \( f-g \) is \( \mathbb{R} \).
- The domain of \( fg \) is \( \mathbb{R} \).
- The domain of \( ff \) is \( \mathbb{R} \).
- The domain of \( \frac{f}{g} \) is \( \mathbb{R} \setminus \left\{ -6, \frac{3}{4} \right\} \).
- The domain of \( \frac{g}{f} \) is \( \mathbb{R} \setminus \{ 0 \} \).
b) Finding the expressions for \( (f+g)(x), (f-g)(x), (fg)(x), (ff)(x), \left( \frac{f}{g} \right)(x), \left( \frac{g}{f} \right)(x) \):
1. \( (f+g)(x) \):
[tex]\[ (f+g)(x) = f(x) + g(x) = x^3 + 4x^2 + 21x - 18 \][/tex]
2. \( (f-g)(x) \):
[tex]\[ (f-g)(x) = f(x) - g(x) = x^3 - (4x^2 + 21x - 18) = x^3 - 4x^2 - 21x + 18 \][/tex]
3. \( (fg)(x) \):
[tex]\[ (fg)(x) = f(x) \cdot g(x) = x^3 \cdot (4x^2 + 21x - 18) = 4x^5 + 21x^4 - 18x^3 \][/tex]
4. \( (ff)(x) \):
[tex]\[ (ff)(x) = f(f(x)) = f(x^3) = (x^3)^3 = x^9 \][/tex]
5. \( \left( \frac{f}{g} \right)(x) \):
[tex]\[ \left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)} = \frac{x^3}{4x^2 + 21x - 18} \][/tex]
6. \( \left( \frac{g}{f} \right)(x) \):
[tex]\[ \left( \frac{g}{f} \right)(x) = \frac{g(x)}{f(x)} = \frac{4x^2 + 21x - 18}{x^3} \][/tex]
So, here are the final expressions and their domains accordingly:
- \( (f+g)(x) = x^3 + 4x^2 + 21x - 18 \) with domain \( \mathbb{R} \).
- \( (f-g)(x) = x^3 - 4x^2 - 21x + 18 \) with domain \( \mathbb{R} \).
- \( (fg)(x) = 4x^5 + 21x^4 - 18x^3 \) with domain \( \mathbb{R} \).
- \( (ff)(x) = x^9 \) with domain \( \mathbb{R} \).
- \( \left( \frac{f}{g} \right)(x) = \frac{x^3}{4x^2 + 21x - 18} \) with domain \( \mathbb{R} \setminus \left\{ -6, \frac{3}{4} \right\} \).
- [tex]\( \left( \frac{g}{f} \right)(x) = \frac{4x^2 + 21x - 18}{x^3} \)[/tex] with domain [tex]\( \mathbb{R} \setminus \{ 0 \} \)[/tex].
