How to find the domain of this function?

Question

f(x)= $\frac{(\sqrt{x}-\sqrt{x-1} )}{( \sqrt{x}+\sqrt{x-1} )}\;$

first off $\sqrt{x}$ is defined for:

$$x > 0 \tag{1}$$

and $\sqrt{x-1}$ is defined for:

$$x \ge 1 \tag{2}$$

from $(1)$ and $(2)$, we get domain of $f(x)$ should be $\{x | x\ge 1\}$

We found the domains of elementary functions contained in $f(x)$ and intersected them to find the domain of $f(x)$.

but putting $x = 0$, we get

$$f(0)= -1$$

why this solution was not contained in the above mentioned solution? I always find the domains of composite functions by this way, How can I be sure that some points, like $(0,-1)$ in above problem, are still not included in the solution?

Answer 1

We have $$f(x) = \frac{(\sqrt{x}-\sqrt{x-1} )}{( \sqrt{x}+\sqrt{x-1} )}$$ The domain of $f$ is: $$D_f = \{ x \in \mathbb{R} : (\sqrt{x}+\sqrt{x-1} \ne 0) \wedge (x \ge 0) \wedge (x-1 \ge 0) \}$$

• Let we consider the first inequality: $\sqrt{x}-\sqrt{x-1} \ne 0$
  To make the explanation clearer let we consider to negation: $$\sqrt{x}+\sqrt{x-1} = 0 \Leftrightarrow \sqrt{x-1} = -\sqrt{x}$$ Because $(\forall x \in \mathbb{R}): \sqrt{x} \ge 0 \Rightarrow \sqrt{x-1} = -\sqrt{x}$ is not solvable($\sqrt{x-1}$ can not be negative)
  The solution is $\emptyset$, because we considered the negation, so we must negate it again what result $\mathbb{R}$ Let $D_1$ denotes the first solution set, so $D_1 = \mathbb{R}$
• Now let consider the second inequality: $x \ge 0$
  This inequality is already solved. In analogue to the first case let $D_2$ denotes the second solution set, so $D_2 = [0,+\infty[$
• Now let consider the last inequality: $ x-1 \ge 0 $
  $ x-1 \ge 0 \Leftrightarrow x \ge 1 \Leftrightarrow D_3 = [1,+\infty[$
  The whole solution $$D_f= D_1 \cap D_2 \cap D_3$$ $$\Leftrightarrow D_f= \mathbb{R} \cap [0,+\infty[ \cap [1,+\infty[$$ $$\Leftrightarrow D_f= [1,+\infty[$$

Answer 2

$f$ is a fraction, and fractions are defined everywhere the denominator is non-zero (which it is). The numerator is defined when $[0,\infty) \cap [1,\infty) = [1,\infty)$, (the square root is defined at $0$), and the denominator is defined for $$ \{x : \sqrt{x} + \sqrt{x-1} \neq 0, \, x\geq 0, x\geq 1\} = [1,\infty) $$ so the domain is $[1,\infty)$.

By this, we should see that $f(0)$ is not defined.

Answer 3

To me, your observation seems correct. I assume this is a real function. We have

$$f(x) = \frac{(\sqrt{x}-\sqrt{x-1} )}{( \sqrt{x}+\sqrt{x-1} )}\ = (\sqrt{x} - \sqrt{x-1})^2$$ For $x\geq 1$ this is defined, of course. For $x\leq0$ substitute $x \to -t$ so that we have $$(i\sqrt{t} - i\sqrt{t+1})^2 = -(\sqrt t - \sqrt {t+1})^2$$ $f$ maps values of $x\leq0$ to $\mathbb{R}$ so this is, in fact, in the domain (though since complex numbers are being used in the process, some may disagree with this).

If $x \in (0,1)$, the above argument fails. In particular, we then have $(\sqrt{x} - i\sqrt{1-x})^2$ which has an imaginary component.

We thus conclude that the domain is $\mathbb{R} \setminus (0,1)$.

Answer 4

Putting $x=0$, we get $$f(x) = \dfrac{\sqrt{0}-\sqrt{-1}}{\sqrt{0}+\sqrt{-1}}$$ So we are getting $\sqrt{-1}$ somewhere in this function which is not real. Therefore $f(0)$ is not defined.