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onegirl
Determine the intervals on which f(x) is continuous. f(x) = sqrt( x + 3)
so should i just plug in the number like 6 then add it with 3 and take the square root of the answer?
@zepdrix can u help?? me
Are you familiar with the graph of \(y=\sqrt x\)? It's a good one to memorize, the basic shape.
|dw:1359680242637:dw|So 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.
|dw:1359680353027:dw|So our function would look like this, shifted over to the left 3 units.
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.
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?
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.
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.
So our interval would start from -3 and go up to positive infinity, those are all of the x values we can use.
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.