Find the Domain and range of each function.
1) $$y=\sqrt{x-1}+2$$
2) $$y=-\sqrt{x+3}-1$$
3) $$y=1/2\sqrt{x}+3$$
4) $$y=-\sqrt{x+4}-1$$
Also, how would I graph these?

Question

Find the Domain and range of each function.
1) $$y=\sqrt{x-1}+2$$
2) $$y=-\sqrt{x+3}-1$$
3) $$y=1/2\sqrt{x}+3$$
4) $$y=-\sqrt{x+4}-1$$
Also, how would I graph these?

anonymous · Answer

The domain of the first function is [1, infinity), and its range is [2, infinity).

anonymous · Answer

infinity as in greater than or equal to?

anonymous · Answer

What do you mean? [1, infinity) means all the numbers from 1 to infinity, inclusive.

mertsj · Answer

Do you understand that the radicand cannot be negative?

anonymous · Answer

Listen to Mertsj: He knows what he is talking about. :P

mertsj · Answer

And answer him.

anonymous · Answer

The inifinity thing doesnt seem familiar, i haven't heard about that in class. 
the radicand is the number under the radical, yes

mertsj · Answer

$$x-1\ge0$$
$$x \ge1$$

mertsj · Answer

So since the radicand, which is x-1, must be greater than or equal to 1, x can be any real number that is greater than or equal to 1.  And that is the domain.  Lots of times instead of saying greater than or equal to 1, mathematicians use interval notation which would be:

mertsj · Answer

$$[1,\infty)$$

mertsj · Answer

The bracket shows that x could be 1 but not infinity, sisnce there is no largest number.

mertsj · Answer

Or you could use set builder notation and say that the domain is:

mertsj · Answer

$${x|x \ge 1}$$

mertsj · Answer

Do we need to discuss the range of the first one?

mertsj · Answer

The smallest value of sqrt(x-1) is 0 so the smallest y value is 2.  So the range is all real numbers greater than or equal to 2

anonymous · Answer

so is x−1≥0 okay for the domain? I think my teacher's okay with the simply one.

anonymous · Answer

I'll try.... what do you need help with?

anonymous · Answer

@dpaInc finding the domain and range

anonymous · Answer

@dumbcow can you help

anonymous · Answer

@Luis Rivera @.Sam. @saifoo.khan @KingGeorge

kinggeorge · Answer

Which one are we working on right now?

anonymous · Answer

the first one since i wasnt even sure of the answer

kinggeorge · Answer

First, lets start with the domain. Since the domain of a function $g(x)=\sqrt{x}$ is $x \geq 0$, we know that the domain of a function $f(x)=\sqrt{x-1}$ is $x-1 \geq 0$. In other words, $x\geq1$ is the domain. Since your original function is $$f(x) = \sqrt{x-1} +2$$Your domain is $x\geq1$.

Does this make sense so far?

anonymous · Answer

so you pay attention to what's under the radical?

kinggeorge · Answer

For the domain, yes. 

Since a function $f(x) =x+2$ is defined everywhere, we can ignore the $+2$ in your function if we're finding the domain.

anonymous · Answer

what about the range?

kinggeorge · Answer

For the range, we know that the function is always increasing, so we only have to find the smallest value possible, and extend it to infinity. Since $\sqrt{x}\geq 0$ we know that $\sqrt{x-1}\geq0$ so if we add 2, the range of $y=\sqrt{x-1}+2$, is $y\geq2$

anonymous · Answer

how would the graph of that function look like? like a sketch

kinggeorge · Answer

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