# Domain of $f(x) = \sqrt{x+1}\cdot\sqrt{x-1}$ and domain of $g(x)=\sqrt{x^2-1}$

Recently I came across this question in my homework. The problem is as above. While doing the problem I converted the $$f(x)$$ into $$g(x)$$ since they are the same and I got domain as $$(x \leq -1$$ and $$x \geq 1)$$ but the answer showed the domain is $$x\geq 1$$. If the two functions are the same then why do their domains differ? Or are the functions not the same? Please clarify my doubt.

• The functions are same only in the intersection of the domains. You cannot say that the functions are the same and conclude that they have the same domain. Dec 24, 2023 at 6:25
• but g(x) comes only when i further solve f(x) Dec 24, 2023 at 6:30

$$f(x)$$ is not the same as $$g(x)$$.
The square root function is only defined for a positive expression inside it. Function $$f(x)$$ has a term $$\sqrt{x-1}$$, and in order for this term to be defined, $$x-1\ge0$$. This gives the domain for $$f(x)$$ as $$x\ge1$$. When you convert $$f(x)$$ into $$g(x)$$, you obtain $$\sqrt{x^2-1}$$, this transformation is only possible if you specify that $$x\ge1$$. Essentially, if you do not use this condition, you could be multiplying the square root of a negative number with another expression, and this is not defined.
• The domain for $g(x)$ is as you said, $x\le-1$ and $x\ge1$. But the point is that functions $f(x)$ and $g(x)$ are only equivalent under certain conditions, i.e., $x\ge1$