At what value(s) of x does the function $\frac { ( x + 2 ) ^ { 2 } } { x ^ { 2 } - 4 }$ have an infinite discontinuity?
A)3
B)-4
C)-2
D)-3
E)-1
F)1
G)2
H)4

Question

Quizplus · Accepted Answer

The function has an infinite discontinuity where the denominator is zero and the numerator is not zero. The denominator $x^2 - 4$ factors to $(x-2)(x+2)$, which is zero at $x = 2$ and $x = -2$. At $x = -2$, the numerator $(x+2)^2$ is zero, so there is a removable discontinuity. At $x = 2$, the numerator is not zero, resulting in an infinite discontinuity.

At What Value(s) of X Does the Function (x+2)2x2−4\frac { ( x + 2 ) ^ { 2 } } { x ^ { 2 } - 4 }x2−4(x+2)2​

At What Value(s) of X Does the Function $\frac { ( x + 2 ) ^ { 2 } } { x ^ { 2 } - 4 }$