<p>f(x)= 1 + squareroot x</p>
<p>What is the domain and range.
The answer is 1 and there is an explanation in the BBpg256, but I don't get it!</p>
<p>D: All x >= 0
R: All y >= 1</p>
<p>Nvm, I understand it now. Thanks jamesford!</p>
<p>The domain is the x values for which this function is defined. So basically what x values can you use with the x under the square root sign? —Anything except a negative number [since a negative under a radical is nonreal or “not” defined]</p>
<p>Therefore the domain is x is greater than or equal to 0</p>
<p>You can find the range by using the domain that you found from above. Since x cannot be a negative number and has to be at least 0, then the lowest “range value” [which is the same as f(x)] is 1. </p>
<p>Since x can be anything above 0, similarly the range can be anything above 1.</p>
<p>So the range of this function is y is greater than or equal to 1.</p>
<p>Hope that helped :)</p>
<p>You mean I typed that out for no reason :(, lol</p>
<p>Oh well that helped wayy more! Lol thanks, now I get it. There was this wierd problem in the book and I was kind of lost! Thanks!</p>
<p>^Dis! Are you going to take that Jamesford!?</p>
<p>Senior0991: I meant that Manu101’s explanation was more accurate than the book’s explanation. Jamesford gave the accurate answers which helped me understand the problem! Everyone helped:)</p>