#### Answer

(i) The rational function has factors $x+2$ and $x-1$ which can be cancelled.
(ii) $g(x) = x + 1$

#### Work Step by Step

(i)
Restrictions on a rational function include any $x$ that makes its denominator equal to zero. The function's graph will have a hole in its graph at that $x$ if that factor can be cancelled.
$f(x) = \frac{(x + 2)(x^2 - 1)}{(x + 2)(x - 1)}$ $= \frac{(x + 2)(x - 1)(x + 1)}{(x + 2)(x - 1)} = x+1, x \ne -2, 1$
with holes at $x = -2$ and $x = 1$ since both factors can be cancelled.
(ii) A function whose graph is identical to that of $f$ but without the holes will therefore be $g(x) = x + 1$ for all values of $x$, since this is the reduced form of $f$ without the restricted/cancelled factors