In Problems 1-4, determine whether the function is continuous at the given points. 1. \( f(x)=\frac{x+4}{x-2} ; \quad-2,0 \). 2. \( f(x)=\frac{x^{2}-4 x+4}{6} \quad 2,-2 \) 3. \( g(x)=\frac{x-3}{x^{2}-9} ; 3,-3 \) 4. \( f(x)=\frac{3}{x-2} \) at \( \mathrm{c}=2 \)

For the first function \( f(x) = \frac{x+4}{x-2} \), it is continuous at -2 and 0. At \( x = -2 \), the function evaluates to a defined value, and at \( x = 0 \), it again has a defined output \( f(0) = -2 \). In the second function \( f(x) = \frac{x^{2}-4x+4}{6} \), which simplifies to a constant \( \frac{(x-2)^{2}}{6} \), it is continuous everywhere, including at \( x = 2 \) and \( x = -2 \), since both points yield defined values without any restrictions. In the third function \( g(x) = \frac{x-3}{x^{2}-9} \), we have a point of discontinuity at \( x = 3 \) because it results in division by zero. However, for \( x = -3 \), the function is defined. Finally, for \( f(x) = \frac{3}{x-2} \) at \( c = 2 \), the function is not continuous as it results in division by zero when substituting \( x = 2 \). The function does not exist at that point. Thus, the functions are evaluated for continuity by checking defined values and points that cause division by zero or other discontinuities.