Let's solve each part step-by-step.
### Given:
[tex]\[ f(x) = 6x + 9 \][/tex]
[tex]\[ g(x) = 5 + x \][/tex]
### (a) [tex]\((f + g)(x)\)[/tex]
To find [tex]\((f + g)(x)\)[/tex], we add the functions [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]:
[tex]\[ (f + g)(x) = f(x) + g(x) \][/tex]
[tex]\[ = (6x + 9) + (5 + x) \][/tex]
Combine like terms:
[tex]\[ = 6x + x + 9 + 5 \][/tex]
[tex]\[ = 7x + 14 \][/tex]
Thus,
[tex]\[ (f + g)(x) = 7x + 14 \][/tex]
### (b) [tex]\((f - g)(x)\)[/tex]
To find [tex]\((f - g)(x)\)[/tex], we subtract [tex]\(g(x)\)[/tex] from [tex]\(f(x)\)[/tex]:
[tex]\[ (f - g)(x) = f(x) - g(x) \][/tex]
[tex]\[ = (6x + 9) - (5 + x) \][/tex]
Distribute the negative sign and combine like terms:
[tex]\[ = 6x + 9 - 5 - x \][/tex]
[tex]\[ = 6x - x + 9 - 5 \][/tex]
[tex]\[ = 5x + 4 \][/tex]
Thus,
[tex]\[ (f - g)(x) = 5x + 4 \][/tex]
### (c) [tex]\((f \cdot g)(x)\)[/tex]
To find [tex]\((f \cdot g)(x)\)[/tex], we multiply the functions [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]:
[tex]\[ (f \cdot g)(x) = f(x) \cdot g(x) \][/tex]
[tex]\[ = (6x + 9)(5 + x) \][/tex]
Distribute to simplify:
[tex]\[ = 6x \cdot 5 + 6x \cdot x + 9 \cdot 5 + 9 \cdot x \][/tex]
[tex]\[ = 30x + 6x^2 + 45 + 9x \][/tex]
Combine like terms:
[tex]\[ = 6x^2 + 39x + 45 \][/tex]
Thus,
[tex]\[ (f \cdot g)(x) = (x + 5)(6x + 9) \][/tex] (Alternatively, [tex]\(6x^2 + 39x + 45\)[/tex], after distribution)
### (d) [tex]\(\frac{f}{g}(x)\)[/tex]
To find [tex]\(\frac{f}{g}(x)\)[/tex], we divide [tex]\(f(x)\)[/tex] by [tex]\(g(x)\)[/tex]:
[tex]\[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} \][/tex]
[tex]\[ = \frac{6x + 9}{5 + x} \][/tex]
The fraction is already in its simplified form, so:
[tex]\[ \left(\frac{f}{g}\right)(x) = \frac{6x + 9}{x + 5} \][/tex]
### (e) The domain of [tex]\(\frac{f}{g}(x)\)[/tex]
The domain of [tex]\(\frac{f}{g}(x)\)[/tex] is all [tex]\(x\)[/tex] such that [tex]\(g(x) \neq 0\)[/tex]. First, we need to determine where [tex]\(g(x) = 0\)[/tex]:
[tex]\[ g(x) = 5 + x = 0 \][/tex]
[tex]\[ x = -5 \][/tex]
So, the function [tex]\(\frac{f}{g}(x)\)[/tex] is not defined when [tex]\(x = -5\)[/tex].
Therefore, the domain of [tex]\(\frac{f}{g}(x)\)[/tex] in interval notation is:
[tex]\[ (-\infty, -5) \cup (-5, \infty) \][/tex]
Putting it all together:
(a) [tex]\((f + g)(x) = 7x + 14\)[/tex]
(b) [tex]\((f - g)(x) = 5x + 4\)[/tex]
(c) [tex]\((f \cdot g)(x) = (x + 5)(6x + 9)\)[/tex]
(d) [tex]\(\left(\frac{f}{g}\right)(x) = \frac{6x + 9}{x + 5}\)[/tex]
(e) The domain of [tex]\(\frac{f}{g}(x)\)[/tex] is [tex]\((-\infty, -5) \cup (-5, \infty)\)[/tex].
