3
$\begingroup$

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 $dom(f)=[2,\infty)$.

$\endgroup$
  • 1
    $\begingroup$ and $Ran (f)=[0,\infty)$. Where is the problem here? $\endgroup$ – Peter Melech Nov 18 '17 at 3:04
  • $\begingroup$ @PeterMelech, how did you get that range for $f(x)$? please give the procedure $\endgroup$ – pi-π Nov 18 '17 at 3:40
  • 1
    $\begingroup$ @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. $\endgroup$ – rb612 Nov 18 '17 at 10:56
  • $\begingroup$ @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 $\endgroup$ – Peter Melech Nov 19 '17 at 11:54
0
$\begingroup$

Since (x-2) is always positive for all x greater than or equal to 2, √(x-2) is a positive real number. Since (x-2) increases as x increases, and √(x-2) increases as (x-2) increases, (and the least possible value of √(x-2) is zero), the range will contain all real numbers greater than zero. A better mathematical proof would probably require the knowledge of the first derivative of a function.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.