Answer
Domain: $(-\infty,-4)\cup(-4,4)\cup(4,\infty)$
Range: $\left(-\infty,-\frac{1}{8}\right)\cup(0,\infty)$
Work Step by Step
$G(t)=\frac{2}{t^2-16}=\frac{2}{(t-4)(t+4)}$
When $t$ is equal to $4$ or $-4$, the denominator becomes $0$. So $4$ and $-4$ are not in domain.
The domain is: $(-\infty,-4)\cup(-4,4)\cup(4,\infty)$
The function $t^2-16$ has a minimum at $t=0$, the minimum is
$G(0)=\frac{2}{-18}=-\frac{1}{8}$
As $t^2-16$ is in the denominator, the function $G$ will have a relative maximum at $t=0$.
When $t$ approaches Infinity, the function approaches $0$. The function will be symmetric about $y$ axis because it is an even function. Furthermore, when $t$ approaches $4$ from right side, the function approaches positive infinity but when $t$ approaches $4$ from left side, function approaches negative infinity.
As $t^2-16$ is positive for $t<-4$ or $t>4$ and negative for $-4
