What is the domain of the function $f(x)=\sin^{-1}\left(\frac{8(3)^{x-2}}{1-3^{2(x-1)}}\right)$?
I started using the fact $-\frac{\pi}{2}\le f(x) \le \frac{\pi}{2}\implies-1 \le\frac{8(3)^{x-2}}{1-3^{2(x-1)}}\le1$.Now,on dissecting it into two cases.
$CASE (1): -1 \le\frac{8(3)^{x-2}}{1-3^{2(x-1)}}$
$CASE (2): \frac{8(3)^{x-2}}{1-3^{2(x-1)}}\le1$
The calculations is inboth cases are bewidering,that's why i'm not showing it.
I need someone who can help me in solving this.
On
Examine that,
$$\lim_{x\to-\infty}\frac{8(3)^{x-2}}{1-3^{2(x-1)}}=0$$
$$\lim_{x\to\infty}\frac{8(3)^{x-2}}{1-3^{2(x-1)}}=0$$
Also the function is discontinuous at $x=1$.
Now check where it assumes the values $1$ and $-1$.
$$\frac{8(3)^{x-2}}{1-3^{2(x-1)}}=1$$
$$8(3)^{x-2}=1-3^{2(x-1)}$$
$$8(3)^{x-2}+3^{2(x-1)}=1$$
$$3^{x-1}(\frac{8}{3}+3^{x-1})=1$$
Let $3^{x-1}=t$
$$t(8+3t)=3$$
$$t=\frac{1}{3} \implies x=0 $$
$$\frac{8(3)^{x-2}}{1-3^{2(x-1)}}=-1$$
$$8(3)^{x-2}=-1+3^{2(x-1)}$$
$$8(3)^{x-2}-3^{2(x-1)}=-1$$
$$3^{x-1}(\frac{8}{3}-3^{x-1})=-1$$
Let $3^{x-1}=t$
$$t(8-3t)=-3$$
$$t=3 \implies x=2 $$
Therefore, Domain is, $$(-\infty,0]\cup[2,\infty)$$
On
Hint:
Let $3^{x-2}=a\implies a>0$ for real $x$
$$-1\le\dfrac{8a}{1-9a^2}\le1$$
$$\dfrac{8a}{1-9a^2}\le1\iff0\le\dfrac{9a^2+8a-1}{9a^2-1}=\dfrac{(9a-1)(a+1)}{(3a-1)(3a+1)}$$
As $a.0,$ we need $\dfrac{9a-1}{3a-1}\ge0$
$\implies$ either $9a-1=0\iff a=?$ or $ a>$max$\left(\dfrac19,\dfrac13\right)$ or $a<$min$\left(\dfrac19,\dfrac13\right)$
$\implies x<-2$ or $x\ge-1$
Similarly for $$-1\le\dfrac{8a}{1-9a^2}\iff\dfrac{8a+1-9a^2}{1-9a^2}\ge0$$
$$\iff\dfrac{(9a+1)(a-1)}{(3a+1)(3a-1)}\le0$$
As $a.0,$ we need $\dfrac{a-1}{3a-1}\ge0$
Can you take it from here?
Hint:
Step1 : Write the first case as $$-1\le \frac {\frac 83 \cdot t}{1-t^2}$$
Step 2: And second case as $$ \frac {\frac 83 \cdot t}{1-t^2}\le 1$$
Where $t=3^{x-1}$ and hence
Step 3: $t\ge 0$
Now you can take intersection of the intervals obtained from 3 above steps to get domain of $t$ and Resubstitute it in form of $x$ to get original domain