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

I typed this function into a graphing program


and it gives me a domain of $\Bbb R\setminus(-1,0]$

I tried to find it myself:

let the denominator = $0$

$ x= 0$ is a V.A , and thus is excluded from the domain

$\frac{x+1}{x} \ge 0$

If $x \lt 0 \Rightarrow x+1 \le 0 $

$x \le -1 $

If $ x \gt 0 \Rightarrow x+1 \ge 0 $

$ x \ge -1$

I know that the last part is wrong, but I don't know how to fix it. Also, this function has an interval , rather than just a point over which it's discontinuous. Do I say that the points of discontinuity lay at $x=0$ and $x =-1$ or is that wrong?

Thanks for the help.


There is nothing wrong, you just misinterpreted what you got. In the last part, you assumed that $x>0$ in the first place, so when you get that $x\geq -1$, you need to take intersection with the assumption, i.e. you get $(0,+\infty)\cap [-1,+\infty) = (0,+\infty)$.

Taking union with the first bit, you get that the natural domain is $$(-\infty,-1]\cup (0,+\infty) = \mathbb R\setminus (-1,0].$$

  • 1
    $\begingroup$ Ohh okay. Alright thanks! $\endgroup$ – Dahen Sep 23 '17 at 12:01

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

You must have that $x \geq -1$ and $x \neq 0$

For $x=-1$ we have that $f(-1)=0$ and also for $x \leq -1$ we have that $f(x)>0$

Also for $x>0$ wwe have $f(x)>0$

$f$ is not defined on the set $(-1,0)$ because in this set $\frac{x+1}{x}<0$

Thus the domain of $f$ is the set $A=(-\infty,-1] \cup(0,+\infty)$

Also $f$ is continuous on its domain.


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