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If [tex]\( c(x) = \frac{5}{x-2} \)[/tex] and [tex]\( d(x) = x+3 \)[/tex], what is the domain of [tex]\( (c \cdot d)(x) \)[/tex]?

A. All real values of [tex]\( x \)[/tex]
B. All real values of [tex]\( x \)[/tex] except [tex]\( x = 2 \)[/tex]
C. All real values of [tex]\( x \)[/tex] except [tex]\( x = -3 \)[/tex]
D. All real values of [tex]\( x \)[/tex] except [tex]\( x = 2 \)[/tex] and [tex]\( x = -3 \)[/tex]

Sagot :

GinnyAnswer
To determine the domain of the product of the functions [tex]\( c(x) \)[/tex] and [tex]\( d(x) \)[/tex], where [tex]\( c(x) = \frac{5}{x-2} \)[/tex] and [tex]\( d(x) = x + 3 \)[/tex], we need to find the values of [tex]\( x \)[/tex] for which both functions are defined and their product is also defined.

1. Determine the domain of [tex]\( c(x) = \frac{5}{x-2} \)[/tex]:
- The function [tex]\( c(x) \)[/tex] is undefined when the denominator is zero.
- Thus, [tex]\( x - 2 \neq 0 \)[/tex].
- Solving for [tex]\( x \)[/tex] gives [tex]\( x \neq 2 \)[/tex].

2. Determine the domain of [tex]\( d(x) = x + 3 \)[/tex]:
- The function [tex]\( d(x) \)[/tex] is a linear function and is defined for all real numbers.
- Therefore, there are no restrictions on [tex]\( x \)[/tex] from [tex]\( d(x) \)[/tex].

3. Determine the domain of the product [tex]\( (c \cdot d)(x) = c(x) \cdot d(x) \)[/tex]:
- The product [tex]\( (c \cdot d)(x) = \frac{5}{x-2} \cdot (x+3) \)[/tex].
- This function combines the restrictions from both [tex]\( c(x) \)[/tex] and [tex]\( d(x) \)[/tex].

4. Identify additional restrictions from the product [tex]\( \frac{5}{x-2} \cdot (x+3) \)[/tex]:
- We already established that [tex]\( x \neq 2 \)[/tex] from [tex]\( c(x) \)[/tex].
- Next, we need to ensure that the product itself does not introduce any new restrictions. Although [tex]\( d(x) \)[/tex] does not have inherent restrictions, its impact must be considered when plugging into [tex]\( c(x) \)[/tex].

5. Check for any further restrictions:
- Notice that the expression [tex]\( c(d(x)) \)[/tex] would be evaluated as [tex]\( c(x+3) = \frac{5}{(x+3)-2} = \frac{5}{x+1} \)[/tex].
- Thus, we must ensure the new denominator [tex]\( x + 1 \neq 0 \)[/tex].
- Solving for [tex]\( x \neq -1 \)[/tex] due to the transformation in the original function, leads us to the specific circumstance [tex]\( c(d(-3)) \)[/tex].

Hence, combining all restrictions, the overall domain of [tex]\( (c \cdot d)(x) \)[/tex] is all real values of [tex]\( x \)[/tex] except [tex]\( x = 2 \)[/tex] and [tex]\( x = -3 \)[/tex].

Therefore, the correct answer is:

All real values of [tex]\( x \)[/tex] except [tex]\( x = 2 \)[/tex] and [tex]\( x = -3 \)[/tex].
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