Answer :
To determine which function meets the criteria of having a domain where [tex]\( x \neq 3 \)[/tex] and a range where [tex]\( y \neq 2 \)[/tex], let's analyze each given function step by step.
### Function 1: [tex]\( f(x) = \frac{x-5}{x+3} \)[/tex]
- Domain: The domain is all [tex]\( x \)[/tex] except where the denominator is zero. Setting the denominator to zero:
[tex]\[ x + 3 = 0 \rightarrow x = -3 \][/tex]
So, the domain is [tex]\( x \neq -3 \)[/tex].
- Range: [tex]\(\frac{x-5}{x+3}\)[/tex] does not exactly target a specific discontinuity at [tex]\( y = 2 \)[/tex], but let's solve for [tex]\( y = 2 \)[/tex] and see if there is a restriction.
[tex]\[ 2 = \frac{x-5}{x+3} \][/tex]
[tex]\[ 2(x+3) = x-5 \][/tex]
[tex]\[ 2x + 6 = x - 5 \][/tex]
[tex]\[ x = -11 \][/tex]
A specific input [tex]\( x = -11 \)[/tex] may create the value [tex]\( y = 2 \)[/tex]. So, this function's range does include [tex]\( y = 2 \)[/tex].
### Function 2: [tex]\( f(x) = \frac{2(x+5)}{x+3} \)[/tex]
- Domain: Similar to Function 1:
[tex]\[ x + 3 = 0 \rightarrow x = -3 \][/tex]
The domain is [tex]\( x \neq -3 \)[/tex].
- Range: To see if [tex]\( y = 2 \)[/tex] is part of the range, solve:
[tex]\[ 2 = \frac{2(x+5)}{x+3} \][/tex]
[tex]\[ 2(x+3) = 2(x+5) \][/tex]
[tex]\[ 2x + 6 = 2x + 10 \][/tex]
[tex]\[ 6 \neq 10 \][/tex]
A contradiction is obtained, hence [tex]\( y = 2 \)[/tex] is not part of the range.
### Function 3: [tex]\( f(x) = \frac{2(x+5)}{x-3} \)[/tex]
- Domain: Set the denominator:
[tex]\[ x - 3 = 0 \rightarrow x = 3 \][/tex]
Hence, the domain is [tex]\( x \neq 3 \)[/tex].
- Range: Test for [tex]\( y = 2 \)[/tex]:
[tex]\[ 2 = \frac{2(x+5)}{x-3} \][/tex]
[tex]\[ 2(x-3) = 2(x+5) \][/tex]
[tex]\[ 2x - 6 = 2x + 10 \][/tex]
[tex]\[ -6 \neq 10 \][/tex]
So, [tex]\( y = 2 \)[/tex] is not in the range due to a contradiction.
### Function 4: [tex]\( f(x) = \frac{x+5}{x-3} \)[/tex]
- Domain: Setting the denominator to zero:
[tex]\[ x - 3 = 0 \rightarrow x = 3 \][/tex]
Hence, the domain is [tex]\( x \neq 3 \)[/tex].
- Range: Next, test for [tex]\( y = 2 \)[/tex]:
[tex]\[ 2 = \frac{x+5}{x-3} \][/tex]
[tex]\[ 2(x-3) = x+5 \][/tex]
[tex]\[ 2x - 6 = x + 5 \][/tex]
[tex]\[ x = 11 \][/tex]
Specific value [tex]\( x = 11 \)[/tex] makes [tex]\( y = 2 \)[/tex], showing [tex]\( y = 2 \)[/tex] is indeed a part.
## Conclusion:
Among the functions, the fourth function [tex]\( f(x)=\frac{x+5}{x-3} \)[/tex] meets the criteria of having a domain [tex]\( x \neq 3 \)[/tex] and excluding [tex]\( y = 2 \)[/tex] from its range consistently:
- Domain: [tex]\( x \neq 3 \)[/tex]
- Range: [tex]\( y \neq 2 \)[/tex]
Thus, the correct function is:
[tex]\[ \boxed{f(x) = \frac{x+5}{x-3}} \][/tex]
### Function 1: [tex]\( f(x) = \frac{x-5}{x+3} \)[/tex]
- Domain: The domain is all [tex]\( x \)[/tex] except where the denominator is zero. Setting the denominator to zero:
[tex]\[ x + 3 = 0 \rightarrow x = -3 \][/tex]
So, the domain is [tex]\( x \neq -3 \)[/tex].
- Range: [tex]\(\frac{x-5}{x+3}\)[/tex] does not exactly target a specific discontinuity at [tex]\( y = 2 \)[/tex], but let's solve for [tex]\( y = 2 \)[/tex] and see if there is a restriction.
[tex]\[ 2 = \frac{x-5}{x+3} \][/tex]
[tex]\[ 2(x+3) = x-5 \][/tex]
[tex]\[ 2x + 6 = x - 5 \][/tex]
[tex]\[ x = -11 \][/tex]
A specific input [tex]\( x = -11 \)[/tex] may create the value [tex]\( y = 2 \)[/tex]. So, this function's range does include [tex]\( y = 2 \)[/tex].
### Function 2: [tex]\( f(x) = \frac{2(x+5)}{x+3} \)[/tex]
- Domain: Similar to Function 1:
[tex]\[ x + 3 = 0 \rightarrow x = -3 \][/tex]
The domain is [tex]\( x \neq -3 \)[/tex].
- Range: To see if [tex]\( y = 2 \)[/tex] is part of the range, solve:
[tex]\[ 2 = \frac{2(x+5)}{x+3} \][/tex]
[tex]\[ 2(x+3) = 2(x+5) \][/tex]
[tex]\[ 2x + 6 = 2x + 10 \][/tex]
[tex]\[ 6 \neq 10 \][/tex]
A contradiction is obtained, hence [tex]\( y = 2 \)[/tex] is not part of the range.
### Function 3: [tex]\( f(x) = \frac{2(x+5)}{x-3} \)[/tex]
- Domain: Set the denominator:
[tex]\[ x - 3 = 0 \rightarrow x = 3 \][/tex]
Hence, the domain is [tex]\( x \neq 3 \)[/tex].
- Range: Test for [tex]\( y = 2 \)[/tex]:
[tex]\[ 2 = \frac{2(x+5)}{x-3} \][/tex]
[tex]\[ 2(x-3) = 2(x+5) \][/tex]
[tex]\[ 2x - 6 = 2x + 10 \][/tex]
[tex]\[ -6 \neq 10 \][/tex]
So, [tex]\( y = 2 \)[/tex] is not in the range due to a contradiction.
### Function 4: [tex]\( f(x) = \frac{x+5}{x-3} \)[/tex]
- Domain: Setting the denominator to zero:
[tex]\[ x - 3 = 0 \rightarrow x = 3 \][/tex]
Hence, the domain is [tex]\( x \neq 3 \)[/tex].
- Range: Next, test for [tex]\( y = 2 \)[/tex]:
[tex]\[ 2 = \frac{x+5}{x-3} \][/tex]
[tex]\[ 2(x-3) = x+5 \][/tex]
[tex]\[ 2x - 6 = x + 5 \][/tex]
[tex]\[ x = 11 \][/tex]
Specific value [tex]\( x = 11 \)[/tex] makes [tex]\( y = 2 \)[/tex], showing [tex]\( y = 2 \)[/tex] is indeed a part.
## Conclusion:
Among the functions, the fourth function [tex]\( f(x)=\frac{x+5}{x-3} \)[/tex] meets the criteria of having a domain [tex]\( x \neq 3 \)[/tex] and excluding [tex]\( y = 2 \)[/tex] from its range consistently:
- Domain: [tex]\( x \neq 3 \)[/tex]
- Range: [tex]\( y \neq 2 \)[/tex]
Thus, the correct function is:
[tex]\[ \boxed{f(x) = \frac{x+5}{x-3}} \][/tex]
