To find the range of the function [tex]\( f(x) = \frac{2x - 1}{x + 3} \)[/tex], we need to understand how the [tex]\( y \)[/tex]-values behave in different scenarios and identify any restrictions on [tex]\( y \)[/tex].
Let's start by analyzing the domain of the function. The function is defined for all real numbers [tex]\( x \)[/tex] except where the denominator equals zero. Setting the denominator to zero:
[tex]\[ x + 3 = 0 \][/tex]
[tex]\[ x = -3 \][/tex]
Thus, [tex]\( x \)[/tex] cannot be [tex]\(-3\)[/tex], but this pertains to the domain, not the range.
Now, we solve for [tex]\( y \)[/tex], the output of the function:
[tex]\[ y = \frac{2x - 1}{x + 3} \][/tex]
We need to rearrange this equation to express [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ y(x + 3) = 2x - 1 \][/tex]
[tex]\[ yx + 3y = 2x - 1 \][/tex]
[tex]\[ yx - 2x = -1 - 3y \][/tex]
[tex]\[ x(y - 2) = -1 - 3y \][/tex]
[tex]\[ x = \frac{-1 - 3y}{y - 2} \][/tex]
From this rearranged equation, we see that [tex]\( x \)[/tex] exists as long as the denominator is not zero. So, we set the denominator equal to zero and solve for [tex]\( y \)[/tex]:
[tex]\[ y - 2 = 0 \][/tex]
[tex]\[ y = 2 \][/tex]
Therefore, [tex]\( y = 2 \)[/tex] is a value that makes the denominator zero, which means there is no corresponding value of [tex]\( x \)[/tex] that could make the function output [tex]\( y = 2 \)[/tex].
Hence, the function [tex]\( f(x) = \frac{2x - 1}{x + 3} \)[/tex] can take any real number value except [tex]\( y = 2 \)[/tex].
The range of the function is all real numbers except [tex]\( y = 2 \)[/tex]. Thus, the correct answer is:
[tex]\[ \boxed{C. \, R \backslash \{2\}} \][/tex]
