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
 A: In fact $G(t)=\dfrac 2{t^2-16}$ is defined everywhere the denominator is not zero.
So we solve $d(t)=t^2-16=0$ and we find $t=\pm 4$.
Thus the domain of definition is $\mathbb R\setminus \{-4,4\}$
And we can rewrite it as an union of intervals $]-\infty,-4[\ \cup\ ]-4,4[\ \cup\ ]4,+\infty[$

Now let's look at the sign of $G$ (or $d$ which has the same sign):
$d(t)=(t-4)(t+4)$ so


*

*$t<-4$ both factors are negative so $d(t)>0$

*$t>4$ both factors are positive so $d(t)>0$

*$-4<t<4$ one factor is negative and the other is positive so $d(t)<0$



Regarding the range, $G(t)$ is continuous everywhere $d(t)\neq 0$. Around the zeroes of $d$ it has $\pm \infty$ limits and around infinity is has $0$ limits.
So combining with the sign study and continuity we can say:


*

*in $]-\infty,-4[$ range is $]0,+\infty[\quad \begin{cases}\lim\limits_{x\to-\infty}G(t)=0\\
\lim\limits_{x\to-4^-}G(t)=+\infty\end{cases}$

*in $]-4,4[$ range is $]-\infty,-\frac 18[\quad\begin{cases}\lim\limits_{x\to-4^+} G(t)=-\infty\\\lim\limits_{x\to4^-}G(t)=-\infty\end{cases}$ 
We have also to determine the local maximum in this interval, which happen when $t^2-16$ is minimum i.e. when $d'(t)=2t=0\iff t=0$ and $G(0)=-\frac 18$.


*

*in $]4,+\infty[$ range is $]0,+\infty[\quad \begin{cases}\lim\limits_{x\to4^+}G(t)=+\infty\\\lim\limits_{x\to+\infty}G(t)=0\end{cases}$ 


So the overall range is $]-\infty,-\frac 18] \cup ]0,+\infty[$
Note: You studied the signs correctly but you need to add that $G$ is continuous on the three intervals to justify that all values of the range are reached.
All this can advantageously be summarized in a variation array knowing that $G'(t)=-4t/d^2(t)$ which has sign of $t$.
$\begin{array}{|c|cccc|}\hline
x & -\infty && -4 && 0 && 4 && +\infty\\\hline
d(t) &&+& 0 && -16 && 0 &&+\\\hline
G(t) & 0 &\nearrow& +\infty\mid-\infty &\nearrow& -\frac 18 &\searrow& -\infty\mid +\infty &\searrow& 0\\\hline
\end{array}$
A: i would say $$-\infty<f(x)<\infty$$ with $$t\ne \pm 4$$
the local Minimum is $$\frac{-2}{16}=-\frac{1}{8}$$ so we have additionally $$-\infty<y\le -\frac{1}{8}$$
A: 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:
$$(-\infty, -\frac{2}{16}]\cup(0,\infty)$$
