To determine the domain of the product function [tex]\((cd)(x)\)[/tex], we need to consider the domains of the individual functions [tex]\(c(x)\)[/tex] and [tex]\(d(x)\)[/tex] and identify any values for [tex]\(x\)[/tex] that would make the product function undefined.
Given functions:
[tex]\[ c(x) = \frac{5}{x-2} \][/tex]
[tex]\[ d(x) = x+3 \][/tex]
1. Domain of [tex]\(c(x)\)[/tex]:
The function [tex]\(c(x)\)[/tex] is defined as [tex]\( \frac{5}{x-2} \)[/tex]. This function is undefined when the denominator is zero. Thus, we need to find when [tex]\( x - 2 = 0 \)[/tex]:
[tex]\[ x - 2 = 0 \][/tex]
[tex]\[ x = 2 \][/tex]
So, [tex]\(c(x)\)[/tex] is undefined at [tex]\(x = 2\)[/tex]. The domain of [tex]\(c(x)\)[/tex] is all real values of [tex]\(x\)[/tex] except [tex]\(x = 2\)[/tex].
2. Domain of [tex]\(d(x)\)[/tex]:
The function [tex]\(d(x)\)[/tex] is defined as [tex]\( x + 3 \)[/tex]. This function is defined for all real values of [tex]\(x\)[/tex]. There are no restrictions on [tex]\(x\)[/tex] for [tex]\(d(x)\)[/tex].
3. Domain of [tex]\((cd)(x)\)[/tex]:
The product function [tex]\((cd)(x)\)[/tex] is the product of [tex]\(c(x)\)[/tex] and [tex]\(d(x)\)[/tex]:
[tex]\[
(cd)(x) = c(x) \cdot d(x) = \left(\frac{5}{x-2}\right) \cdot (x+3)
\][/tex]
To find the domain of [tex]\((cd)(x)\)[/tex], we need to consider the domains of both [tex]\(c(x)\)[/tex] and [tex]\(d(x)\)[/tex]. Since [tex]\(d(x)\)[/tex] is defined for all real values of [tex]\(x\)[/tex], the only restriction comes from [tex]\(c(x)\)[/tex]. As we established, [tex]\(c(x)\)[/tex] is undefined when [tex]\(x = 2\)[/tex]. Therefore, the product function [tex]\((cd)(x)\)[/tex] will also be undefined at [tex]\(x = 2\)[/tex].
Thus, the domain of [tex]\((cd)(x)\)[/tex] is all real values of [tex]\(x\)[/tex] except [tex]\(x = 2\)[/tex].
[tex]\[
\boxed{\text{all real values of } x \text{ except } x = 2}
\][/tex]
So, the correct answer is:
[tex]\[
\boxed{\text{all real values of } x \text{ except } x = 2}
\][/tex]
