What is the domain of the function y SQR root X?

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Find the Domain and Range y=- square root of x

Step 1

Set the radicand in greater than or equal to to find where the expression is defined.

Step 2

The domain is all values of that make the expression defined.

Interval Notation:

Set-Builder Notation:

Step 3

The range is the set of all valid values. Use the graph to find the range.

Interval Notation:

Set-Builder Notation:

Step 4

Determine the domain and range.

Domain:

Range:

Step 5

Domain is #x>=1#. Range is all real numbers.

Explanation:

Note that #(x-1)# cannot take negative values of #y# is real. Assuming that we are working in real number domain, it is obvious x cannot take values less than one. Hence, domain is #x>=1#.

However, as #sqrt(x-1)#, #y# can take any value. Hencr, range is all real numbers.

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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