# Find the domain and range of $y=\sqrt {x-2}$

Find the domain and range of $y=\sqrt {x-2}$

My Attempt: $$y=\sqrt {x-2}$$ For $y$ to be defined, $$(x-2)\geq 0$$ $$x\geq 2$$ So $dom(f)=[2,\infty)$.

• and $Ran (f)=[0,\infty)$. Where is the problem here? – Peter Melech Nov 18 '17 at 3:04
• @PeterMelech, how did you get that range for $f(x)$? please give the procedure – pi-π Nov 18 '17 at 3:40
• @blue_eyed_... the range is the possible outputs for the function. What's the smallest value of the function? 0. The largest value is unbounded, because I can plug in any really large number and get the square root. There is no limit to how big this number can be, so the range goes to infinity. – rb612 Nov 18 '17 at 10:56
• @blue_eyed Consider the comment by rb612 or just use the domain You found and the monotonicity of the square root, it´s rarely a procedure but quite simple to see – Peter Melech Nov 19 '17 at 11:54

That's absolutely correct. The domain of a function is the set of all input values that you're "legally" allowed to plug into the function. For the function $y=\sqrt {x-2}$, that's going to be $x\geq 2$ because if you were to plug in, say, -1, you would end up with a negative number under the square root and, as you probably know, the square root function is not defined for negatives in the real number system.
The range of a function is usually a bit tougher to find. In your case, the range is the same as that of $f(x)=\sqrt{x}$: $[0,\infty)$. How do we know that? Well, because the square root function is one of those well-known functions whose behavior we all should be familiar with. And we also know, that when you add or subtract a constant before the prevailing operation takes place (for $2(x+1)^2-1$, the prevailing operation would be squaring, for $5\sqrt{x+2}+1$—taking the square root), you are shifting the graph left or right. So, in our case here, the graph is shifted 2 units to the right. And that's the only transformation that's happening. No shifts up or down. So, the range does not change.