Range of $\sqrt{x-1}$ Problem: 

Find the range of $f(x)=\sqrt{x-1}$

Now the problem I face is this: is the range $[0,\infty),$ or is it $(-\infty,\infty)$? $$$$I had learnt that $\sqrt{x^2} = \pm x$. However, on the Net, I read that $\sqrt{x^2}=|x|$ ie the output of a square root function is positive. If the output of a square root is always positive then the Range of  $\sqrt{x-1}$ is obviously $[0, \infty)$ $$$$
I would be very grateful for any help in clearing this doubt. Thanks very, very much in advance!
 A: $\sqrt{z}$ is a function(Maps a point to a single point). So $\sqrt{z^2}$can only take a single value. By convention, $\sqrt{z}\ge0$. So,$\sqrt{z^2}=|z|$. So to answer your question, range will be $[0,\infty)$
A: We think of $\sqrt{}$ as the positive square root. This is for convenience, but it is the consensus of all mathematicians. It is of course true that $(-x)^2 = (x)^2$, so you might even say something like $x$ and $-x$ are both square roots of $x^2$. In solving an equation like $(x+4)^2 = 25$, you can only "take the square root" of both sides if you remember that there can always be up to two square roots, and you would want to write $(x+4) = \pm\sqrt{25}$.
But mathematicians agree that $\sqrt{}$ should always refer to one number, as functions must do. Confusion would ensue if when you wrote $\sqrt{}$ and when I wrote $\sqrt{}$, we might be referring to two separate numbers. So we want $\sqrt{}$ to be a function. Your question emphasizes that $\sqrt{}$ is a function by referring to it as $f(x)$ and asking about its range. Functions must output a $single$ value for each input. This is why we define $\sqrt{ x^2} = |x|$.  The range is $[0,\infty)$.
A: Here are a few fallacy i have observed in your answer.


*

*The expression $f(x)=\sqrt{x-1}$ ,first of all requires a domain, normally you'd assume that $x\geq1$ i.e., $f(x)\mid \forall x\in [1,\infty) \to f(x)\in [0,\infty)$

*Your expression, $\sqrt{x^2}=\pm x$ is invalid, it should be $\color{red}{\pm}\sqrt{x^2}=\pm x.$

*Normally, $\sqrt{x^2}=|x|$ which is defined as, $$\begin{cases}x& x\geq0 \\ -x & x<0 \end{cases}$$
A: The definition of the square root includes that it is the positive square root:
$$\sqrt{x^2}=|x|$$ (which means that the solutions to $x^2=a$ are $x=\sqrt a$ and $x=-\sqrt a$).
As the modulus function $f(x)\ge0$ has a range $[0,\infty)$, this means that the range of $f(x)=\sqrt{x-1}$ is $[0,\infty)$ just like the modulus function, and the domain is $x\ge1$.
A: for square root function to be defined x-1>=0,or x>=1 there for the range of function is [0,+inf)
