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Prove It $\displaystyle f(x) = \frac{2x - 1}{x - 1}$
$\displaystyle = \frac{2x - 2 + 1}{x - 1}$
$\displaystyle = \frac{2x - 2}{x - 1} + \frac{1}{x - 1}$
$\displaystyle = 2 + \frac{1}{x -th 1}$.
This is a hyperbola with a horizontal and a vertical translation. Can you go from here?
Erm, where did you get the -2 and +1 from. And no, I don't think so. I already have the answer to the problem: (-infinity, 1)U(1, +infinity), so how do you go from $\displaystyle = 2 + \frac{1}{x -th 1}$ to (-infinity, 1)U(1, +infinity)?
$\displaystyle f(x) = \frac{|x-3|}{2}$.
Remember that absolute values gobble up all negatives and turn them positive. Can you go from here?
I think so. So the range would be (-infinity, 3)U(3, +infinity)?