Answer :
To find the domain and range of the function [tex]\( f(x) = \frac{3x - 1}{x + 2} \)[/tex], we need to consider the points where the function is undefined and the y-values that the function cannot take.
### Domain
The function [tex]\( f(x) \)[/tex] will be undefined at points where the denominator is zero. So we need to solve for [tex]\( x \)[/tex] in the equation:
[tex]\[ x + 2 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = -2 \][/tex]
Thus, the domain of [tex]\( f \)[/tex] is all real numbers except [tex]\( x = -2 \)[/tex]. Therefore, we have:
[tex]\[ a = -2 \][/tex]
### Range
Next, we need to determine the range of [tex]\( f(x) \)[/tex]. This requires considering the y-values that the function cannot take. To find such [tex]\( y \)[/tex]-values, we set [tex]\( y = \frac{3x - 1}{x + 2} \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[ y = \frac{3x - 1}{x + 2} \][/tex]
We want to identify the y-values for which no [tex]\( x \)[/tex] in the domain will produce that particular [tex]\( y \)[/tex]. Rearranging the equation to solve for [tex]\( x \)[/tex]:
[tex]\[ y(x + 2) = 3x - 1 \][/tex]
[tex]\[ yx + 2y = 3x - 1 \][/tex]
[tex]\[ yx - 3x = -2y - 1 \][/tex]
[tex]\[ x(y - 3) = -2y - 1 \][/tex]
[tex]\[ x = \frac{-2y - 1}{y - 3} \][/tex]
To find the [tex]\( y \)[/tex]-value that makes this expression invalid, we need to check when the denominator is zero:
[tex]\[ y - 3 = 0 \][/tex]
[tex]\[ y = 3 \][/tex]
Thus, the function [tex]\( f(x) \)[/tex] cannot take the value [tex]\( y = 3 \)[/tex] because, at this point, the denominator becomes zero, making the expression undefined.
Hence, the range of [tex]\( f \)[/tex] is all real numbers except [tex]\( y = 3 \)[/tex], which means:
[tex]\[ b = 3 \][/tex]
### Final Answer
The ordered pair [tex]\((a, b)\)[/tex] is:
[tex]\[ (-2, 3) \][/tex]
### Domain
The function [tex]\( f(x) \)[/tex] will be undefined at points where the denominator is zero. So we need to solve for [tex]\( x \)[/tex] in the equation:
[tex]\[ x + 2 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = -2 \][/tex]
Thus, the domain of [tex]\( f \)[/tex] is all real numbers except [tex]\( x = -2 \)[/tex]. Therefore, we have:
[tex]\[ a = -2 \][/tex]
### Range
Next, we need to determine the range of [tex]\( f(x) \)[/tex]. This requires considering the y-values that the function cannot take. To find such [tex]\( y \)[/tex]-values, we set [tex]\( y = \frac{3x - 1}{x + 2} \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[ y = \frac{3x - 1}{x + 2} \][/tex]
We want to identify the y-values for which no [tex]\( x \)[/tex] in the domain will produce that particular [tex]\( y \)[/tex]. Rearranging the equation to solve for [tex]\( x \)[/tex]:
[tex]\[ y(x + 2) = 3x - 1 \][/tex]
[tex]\[ yx + 2y = 3x - 1 \][/tex]
[tex]\[ yx - 3x = -2y - 1 \][/tex]
[tex]\[ x(y - 3) = -2y - 1 \][/tex]
[tex]\[ x = \frac{-2y - 1}{y - 3} \][/tex]
To find the [tex]\( y \)[/tex]-value that makes this expression invalid, we need to check when the denominator is zero:
[tex]\[ y - 3 = 0 \][/tex]
[tex]\[ y = 3 \][/tex]
Thus, the function [tex]\( f(x) \)[/tex] cannot take the value [tex]\( y = 3 \)[/tex] because, at this point, the denominator becomes zero, making the expression undefined.
Hence, the range of [tex]\( f \)[/tex] is all real numbers except [tex]\( y = 3 \)[/tex], which means:
[tex]\[ b = 3 \][/tex]
### Final Answer
The ordered pair [tex]\((a, b)\)[/tex] is:
[tex]\[ (-2, 3) \][/tex]
