Finding Range from Domain In the book Thomas' Calculus,in exercise section i got one question e.g. Find the domain and Range of G(t) = $\frac{2}{t^2-16}$. 
Ans: Domain is (-∞,-4) U (-4,4) U (4,∞)..I understand this.Let us discuss how they find Range


*

*t<-4  as (-∞,-4)
=> -t>4   Multiply by -1
=> $(-t)^2$ > $4^2$   Squiring both side
=> $t^2$ > 16
=> $t^2$ - 16 >0
So $\frac{2}{t^2-16}$ >0 .Is the derivation correct?

*t>4  as (-∞,-4)
=> t2  > 16 Squiring both side
=> $t^2$ - 16 >0
So $\frac{2}{t^2-16}$ >0 .Is the derivation correct?
The following third one is the one i need some understanding


*-4 

=> -16 $\le$  $t^2$ - 16 < 0 ***
=> $-\frac{2}{16} \le \frac{2}{t^2-16} <0$ 
*** How this line is being derived. If square is done in both side -4 should 16 , negative sign should eliminate such like at no 1( -t is $t^2$)?,How we change < to $\le$? Please let me know
Best regards
sabbir

The solutions for $1$ and $2$ are incomplete and for $3$ is just wrong.
In $1$ and $2$, surely $\frac{2}{x^2-16}\gt 0$. However, it says nowhere that each of those values is reached by substituting some $x$.
In $3$, it should say $-4\lt x\lt 4$, and then $-16\le x^2-16\lt 0$, but then $-\frac{2}{16}\ge\frac{2}{x^2-16}$ (not $\le$, because we are taking reciprocals). Again, there is no proof that each of those values is reached.
Now, having said all that, let's solve it rigorously. One potential solution is to try to solve the equation $\frac{2}{x^2-16}=y$ and see for which $y$ it has solutions. Basically:
$$\frac{2}{x^2-16}=y$$
$$2=y(x^2-16)$$
$$yx^2-16y-2=0$$
which has no solution for $y=0$, and, for $y\ne 0$, has solutions if and only if the discriminant of this is $\ge 0$, i.e $0^2-4y(-16y-2)\ge 0$. This is further equivalent to:
$$64y^2+8y\ge 0$$
Solving $64y^2+8y=0$ gives you $y_1=-\frac{8}{64}=-\frac{2}{16}, y_2=0$, so the solutions are $y\in(-\infty, -\frac{2}{16}]\cup[0,\infty)$. However, we dismissed $y=0$ previously, so the real range is:
$$(-\infty, -\frac{2}{16}]\cup(0,\infty)$$
