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

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2 Answers

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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.

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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.

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