Question

f(x)= √x-1/x+1 find domain​ (got the answer)

Answer 1

Given :

$\sf f(x) = \dfrac{\sqrt{x - 1} }{x + 1}$

To find:

• Domain of the given function

Solution :

$\sf \implies f(x) = \dfrac{\sqrt{x - 1} }{x + 1}$

Now , if any number is under root then it is always greater than or equal to zero.

Therefore , x-1 ≥ 0 because it is under root.

$\sf \implies {x - 1} \geqslant 0$

$\sf \implies \boxed{\sf {x } \geqslant 1}.........(1)$

Now , we know that denominator can never be zero.

Therefore , x+1≠0

$\implies \sf x + 1 \ne0$

$\implies \sf \boxed{\sf \: x \ne - 1}.......(2)$

Therefore , from (1) and (2) we get :-

$\sf x \in[1, \infty)$

X can have any value from 1 to infinity.

Answer :

$\sf domain : \\ \sf \large x \in[1, \infty)$