Question

g(x)= (x+4)/(x^3-4x)

Answer 1

what is your question @nannccyyy

Answer 2

what is the domain?

Answer 3

\[g(x)=\frac{ (x+4) }{ x(x^{2}- 4) }= \frac{ (x+4) }{ x(x- 2)(x+2) }= \frac{x+2 }{x(x-2)(x+2)}+ \frac{2 }{x(x-2)(x+2)}= \frac{1}{x(x-2)}+ \frac{2}{x(x-2)(x+2)}= \frac{1}{x(x-2)} (1+\frac{2}{x+2})\]

Answer 4

It's seem me that x can't be 0 (zero), 2 or -2 because g(x) will tends to infinite. However, if x = -4 then g(x)=0;