Answer
(a) $\frac{(3 − x)}{x}$
(b) No
Work Step by Step
(a) $f(x) = \frac{(2x − 1)}{(x + 1)}$, $g(x) = \frac{1}{(x − 1)}$
$f(g(x)) = \frac{2(\frac{1}{(x - 1)}) - 1}{(\frac{1}{x - 1}) + 1} = \frac{\frac{2 - (x - 1)}{x -1}}{\frac{1 + (x - 1)}{x -1}}$
$f(g(x)) = \frac{2 - x + 1}{1 + x -1} = \frac{3 - x}{x}$
(b) No; the definition of $f(g(x))$ requires $g(x)$ to be defined, so $x \ne 1$, and $f(g(x))$ requires $g(x) \ne −1$, so we must have $g(x) \ne −1$, i.e. $x \ne 0$; whereas $h(x)$ only requires $x \ne 0$
