## Calculus, 10th Edition (Anton)

(i) $x=-2, 1$ (ii) $g(x) = x+1$
(i) The holes of the function occur when the denominator of the function equals zero. The denominator is $(x+2)(x-1)$ and is $0$ when $x=-2, 1$. (ii) Note that the holes of the function are created because the denominator equals $0$ are the two points mentioned in part (i). We are able to remove these holes by cancelling the factors in the bottom with the factors in the top as such: $g(x) = \dfrac{(x+2)(x^2-1)}{(x+2)(x-1)} = \dfrac{(x+2)(x-1)(x+1)}{(x+2)(x-1)} = x+1$. Therefore, the desired function identical to $f(x)$ but without holes is $g(x) = x+1$.