Determining continuity of a piecewise function

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OK, I must have been daydreaming in class when the teacher went over how to determine continuity of a piecewise function. I am hoping someone can help. Here is what the problem is:
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f(x)={ x^2-x+3 if x<=2
{ 11-3x if x>2

Determine whether the piecewise defined function is continuous at x=0 and at x=2.
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How would I determine if they are continuous or not? I would appreciate any help. Thanks.
 
What is the limit when you approach 2 from the left?

What is the limit when you approach 2 form the right?

try that.
 
I put 2 in from the left and right and they both came out to 5. I am guessing that means they are continuous? When I put in zero from the left and the right, it came out to 3 and 11. Does this mean they are discontinuous at zero? I hope Im right.
 
One of your answers is correct, the other is not.

Here is a hint. Check the domain (values of x) over which each function exists.
 
Hello, Bobalino!

f(x) = { x<sup>2</sup> - x + 3 . if x <u><</u> 2
. . . . . { 11 - 3x . . . if x > 2

Determine whether the function is continuous at x=0 and at x=2.
Click to expand...
Recall the three requirements for continuity of f(x) at x = a.

(1) .f(a) must exist . . . . (some finite value; there is a point on the graph)

(2) .lim f(x) must exist . . (same value from "both sides")
. . .x→a

(3) .lim f(x) .= .f(a) . . . . .(the limit must lead to that point)
. . .x→a
 
When I look at a value of x>2, and try to apply zero to it, this does not work because 0<2. So I will say continuous at x=2, Does not exist at x=0?
 
Why are you looking at x = 0, when that is not where the function is broken into two pieces?

Eliz.
 
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