Question

need help on attachment problem

Answer 1

\[f(x) = \frac{x^2-2}{x+2}\]
Where x not -2:
\[f(x) = \frac{(x+2)(x-2)}{x+2}\]
\[f(x) = x-2\]
Thus the discontinuity at x = -2 is removable as the resulting function is continuous at x=-2