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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.

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    $\begingroup$ 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. $\endgroup$ Dec 24, 2023 at 6:25
  • $\begingroup$ but g(x) comes only when i further solve f(x) $\endgroup$ Dec 24, 2023 at 6:30

1 Answer 1

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$f(x)$ 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.

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  • $\begingroup$ now what is the domain for g(x)?? $\endgroup$ Dec 24, 2023 at 6:34
  • $\begingroup$ 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., $x\ge1$ $\endgroup$
    – zxen
    Dec 24, 2023 at 6:36
  • $\begingroup$ So you mean you cannot multiply them and do it first. Ok, now I understand. Thank you sir $\endgroup$ Dec 24, 2023 at 6:39

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