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Determine the intervals on which f(x) is continuous. f(x) = sqrt( x + 3)
 one year ago
 one year ago
Determine the intervals on which f(x) is continuous. f(x) = sqrt( x + 3)
 one year ago
 one year ago

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onegirlBest ResponseYou've already chosen the best response.0
so should i just plug in the number like 6 then add it with 3 and take the square root of the answer?
 one year ago

onegirlBest ResponseYou've already chosen the best response.0
@zepdrix can u help?? me
 one year ago

onegirlBest ResponseYou've already chosen the best response.0
@blondie16 can u help?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Are you familiar with the graph of \(y=\sqrt x\)? It's a good one to memorize, the basic shape.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
dw:1359680242637:dwSo this is what \(y=\sqrt x\) looks like, yes? Our function that they gave us has a +3 under the square root. That represents a horizontal shift to the `left` 3 units.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
dw:1359680353027:dwSo our function would look like this, shifted over to the left 3 units.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
I guess we actually don't need the graph to figure this one out... Let's just think about what values we're allowed to plug into a square root.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
If you plug in x=3, \(\large f(3)=\sqrt{3+3}\) which equals \(0\) yes? So 3 is fine. Let's try plugging in 4, \(\large f(4)=\sqrt{4+3}\) which equals \(\sqrt{1}\). Uh oh! We can't take the square root of a negative value! :O Remember that from math rules?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
When you need to determine intervals of continuity, you just need to look for numbers that would cause a problem. Those are places where your function is `not` continuous. So we would NOT include those values in our interval.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
It turns out that you can plug in larger and larger values of x in the positive direction and you'll get real solutions for f(x). That just means, none of the positive x values are a problem. It seems that if \(x \lt 3\), it creates a negative value under the square root, which is a problem.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
So our interval would start from 3 and go up to positive infinity, those are all of the x values we can use.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
In interval notation we would write it like this,\[\huge [3,\infty)\] See the square bracket? That's to show that we want to `include` the value 3. If we had done a rounded bracket around the 3, it would mean we start at 3 but don't include the number itself. And also, we always put a rounded bracket on infinity. It's not an actual number so we can't include it.
 one year ago
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