Question

Find the domain and the range of the real function, $f\left(x\right)=\frac{x^2+1}{x^2-1}$

Answer 1

We have, $f\left(x\right)=\frac{x^2+1}{x^2-1}$

Clearly, $f\left(x\right)$ is defined for all real values of $x$ except those for which $x^2-1=0$, i.e., $x=\pm 1$.

$\therefore \text{dom (f)}=R-\{-1,1\}$

Let $y=f\left(x\right)$. Then,

$y=\frac{x^2+1}{x^2-1}\Rightarrow x^{2}y-y=x^2+1\Rightarrow x^2(y-1)=(y+1)$

$\Rightarrow x^2=\frac{y+1}{y-1}\Rightarrow x=\pm \sqrt{\frac{y+1}{y-1}}$ ..... (i)

It is clear from (i) that $x$ is not defined when y - 1 = 0 or when $\frac{y+1}{y-1}<0$.

Now, $y-1=0\Rightarrow y=1$ ..... (ii)

And $\frac{y+1}{y-1}<0\Rightarrow (y+1>0)\text{and}y-1<0$ or $y+1<0\text{and}y-1>0$

$\Rightarrow (y>-1\text{and}y<1)$ or $(y<-1\text{and}y>1)$

$\Rightarrow -1<y<1$ ...... (iii)

$[\because y<-1\text{and}y>1\text{is not possible}]$

Thus, $x$ is not defined when $-1<y\le 1$. [using (ii) and (iii)]

$\therefore$ range (f) = R - (-1, 1]

Hence, $\text{dom (f)}=R-\{-1,1\}$ and range (f) = R - (-1, 1].