Question

Find
domain & Range of
1/1-x2​

Answer 1

Answer:

Domain is  R - { - 1 , 1 }

Range is  ( - ∞ , 0 ) U [ 1 , ∞ )

Step-by-step explanation:

Given :

$\displaystyle{f(x)=\dfrac{1}{1-x^2}}$

For Domain

F ( x ) to be real , we must have

$\displaystyle{1-x^2\neq 0}\\\\\displaystyle{x^2\neq 1}\\\\\displaystyle{x\neq 1 \ or \ -1}$

Domain of  f ( x ) = R - { - 1 , 1 }

For Range

$\displaystyle{Let \ f(x) \ is \ to \ equal \ y}\\\\\displaystyle{y=\dfrac{1}{1-x^2}}\\\\\displaystyle{y(1-x^2)=1}\\\\\displaystyle{x^2=\dfrac{y-1}{y}}\\\\\displaystyle{x=\sqrt{\dfrac{y-1}{y}}}$

x can take other value other than - 1 and 1 if

$\displaystyle{\sqrt{\dfrac{y-1}{y}}>0}\\\\\\\displaystyle{\dfrac{y-1}{y}>0}\\\\\\\displaystyle{y-1\geq 0}\\\\\displaystyle{y\geq 1}$

y ∈  [ 1 , ∞ )

y > 0

y ∈  ( - ∞ , 0 )

Intersection of both :

y ∈  ( - ∞ , 0 ) U [ 1 , ∞ )

Thus we get answer .

Answer 2

SOLUTION:-

Given:

$f(x) = \frac{1}{1 - {x}^{2} }$

Here,

$1 - {x}^{2} = 0 \\ = > 1 = {x}^{2} \\ = > x = ± x \ \\ = > Df = R- ( - 1,1)$

Now,

Let y= f(x)

Then:

$y = \frac{1}{1 - {x}^{2} } \\ \\ = > y - y {x}^{2} = 1 \\ \\ = > - y {x}^{2} = 1 - y \\ \\ = > y {x}^{2} = y - 1 \\ \\ = > {x}^{2} = \frac{y - 1}{y} \\ \\ = > x = \sqrt{ \frac{y - 1}{y} }$

Hence, range of f(x) is all real number.

=) y<0 & y≥1

Hope it helps ☺️