To determine the domain of the function [tex]\((c \cdot d)(x)\)[/tex], where [tex]\(c(x) = \frac{5}{x-2}\)[/tex] and [tex]\(d(x) = x + 3\)[/tex], we need to identify the values of [tex]\(x\)[/tex] for which this combined function is defined.
First, let's understand the individual domains of the functions [tex]\(c(x)\)[/tex] and [tex]\(d(x)\)[/tex]:
1. Domain of [tex]\(c(x) = \frac{5}{x-2}\)[/tex]:
- The function [tex]\(c(x)\)[/tex] will be undefined when the denominator is zero.
- Therefore, solve [tex]\(x-2=0\)[/tex] to find the value that makes the denominator zero:
[tex]\[
x - 2 = 0 \implies x = 2
\][/tex]
- Hence, [tex]\(c(x)\)[/tex] is undefined at [tex]\(x = 2\)[/tex].
- The domain of [tex]\(c(x)\)[/tex] is all real numbers except [tex]\(x = 2\)[/tex].
2. Domain of [tex]\(d(x) = x + 3\)[/tex]:
- The function [tex]\(d(x)\)[/tex] is a linear function, and it is defined for all real numbers [tex]\(x\)[/tex].
- Therefore, the domain of [tex]\(d(x)\)[/tex] is all real numbers.
Now, to find the domain of the product function [tex]\((c \cdot d)(x)\)[/tex]:
[tex]\[
(c \cdot d)(x) = \left( \frac{5}{x-2} \right) \cdot (x + 3)
\][/tex]
For the product [tex]\((c \cdot d)(x)\)[/tex] to be defined, both [tex]\(c(x)\)[/tex] and [tex]\(d(x)\)[/tex] must be defined. The only restriction comes from [tex]\(c(x)\)[/tex], which is undefined at [tex]\(x = 2\)[/tex].
Thus, the domain of [tex]\((c \cdot d)(x)\)[/tex] is all real values of [tex]\(x\)[/tex] except [tex]\(x = 2\)[/tex].
Therefore, the correct answer is:
[tex]\[
\boxed{\text{all real values of } x \text{ except } x = 2}
\][/tex]
