which functions have removable discontinuities (holes)? check all of the boxes that apply.\n$f(x)=\frac{x…

which functions have removable discontinuities (holes)? check all of the boxes that apply.\n$f(x)=\frac{x - 1}{x^{2}-1}$\n$f(x)=\frac{x^{2}-9}{x^{2}+7x + 12}$\n$f(x)=\frac{x^{2}+4x + 4}{x^{2}+2x - 8}$\n$f(x)=\frac{x + 7}{x^{2}+5x - 14}$\ndone

which functions have removable discontinuities (holes)? check all of the boxes that apply.\n$f(x)=\frac{x - 1}{x^{2}-1}$\n$f(x)=\frac{x^{2}-9}{x^{2}+7x + 12}$\n$f(x)=\frac{x^{2}+4x + 4}{x^{2}+2x - 8}$\n$f(x)=\frac{x + 7}{x^{2}+5x - 14}$\ndone

Answer

Explanation:

Step1: Factor the functions

For (f(x)=\frac{x - 1}{x^{2}-1}), factor (x^{2}-1=(x + 1)(x - 1)), so (f(x)=\frac{x - 1}{(x + 1)(x - 1)}). There is a removable discontinuity at (x = 1) since we can cancel out the common factor ((x - 1)) (when (x\neq1)). For (f(x)=\frac{x^{2}-9}{x^{2}+7x + 12}), factor (x^{2}-9=(x + 3)(x - 3)) and (x^{2}+7x + 12=(x+3)(x + 4)), so (f(x)=\frac{(x + 3)(x - 3)}{(x + 3)(x + 4)}). There is a removable discontinuity at (x=-3) since we can cancel out the common factor ((x + 3)) (when (x\neq - 3)). For (f(x)=\frac{x^{2}+4x + 4}{x^{2}+2x-8}), factor (x^{2}+4x + 4=(x + 2)^{2}) and (x^{2}+2x-8=(x + 4)(x - 2)). There are no common factors, so no removable discontinuity. For (f(x)=\frac{x + 7}{x^{2}+5x-14}), factor (x^{2}+5x-14=(x + 7)(x-2)), so (f(x)=\frac{x + 7}{(x + 7)(x - 2)}). There is a removable discontinuity at (x=-7) since we can cancel out the common factor ((x + 7)) (when (x\neq-7)).

Answer:

(f(x)=\frac{x - 1}{x^{2}-1}), (f(x)=\frac{x^{2}-9}{x^{2}+7x + 12}), (f(x)=\frac{x + 7}{x^{2}+5x-14})