 Let's look at a couple of piece function examples of how to determine continuity at a point. We will look at both of these problems from both a graphical and an algebraic point of view, this letter using the three tests for continuity that we studied in this lesson. So here we have a piece function defined as the linear equation 10-2x provided that x does not equal 3 but when x equals 3 the y value is given by x plus 1. So let's take a look at the graph of it first and see what we can conclude from that before we walk through the three tests for continuity. So the first piece of this piece function is the linear equation 10-2x which we know is going to have a y intercept up at 10 and then the slope simply is negative 2 so we'll come down to over 1, down to over 1. Now when we do this one more time down to an over 1 that puts us at the x value of 3 which the value for which this particular piece is not defined so we're going to have a hole there and then we'll continue on down to an over 1 so that piece then looks like this. When x equals 3 though notice the y value is given by the equation x plus 1 which then fills in this hole and that is at a y value of 4 so in the end the two pieces together create simply a line. Now let's walk through the actual three tests for continuity. Recall that the first test requires that the function value exists at the x value in question so in this case we would be concerned with the value of f of 3. Well that's given by the second piece and as we confirm from the graph that y value simply is 4. For the second test we need to check the limit of the function as x approaches 3. Now remember you would have to do this by looking at both the left and the right hand limits so let's go ahead and take a look at that. Let's take a look at the limit from the left. If we wanted to do this algebraically we would substitute into the piece that applies for values of x just to the left of 3 so we're talking about values such as 2.7, 2.8, 2.9 and that would be the first piece. Remember that we simply substitute in the number itself because we're approaching from the left does not mean we substitute in something a little less than 3. We actually substitute in the 3 so we get 4 for that. We would do the same in looking at the limit from the right. Now we would be substituting into the piece that applies for values of x just a little bit larger than 3 so for example 3.3, 3.2, 3.1 and we find once again that is the first piece so we can see these two left and right limits equal each other so therefore we know the limit as we approach 3 does exist and it equals 4. The third requirement is that the function value in this case at x equals 3 has to equal that limit which we see that it does. They both equal 4 so therefore we can conclude that this function f of x is continuous at x equals 3. Let's look at another piece function. Here we have something similar we have the first piece defined by the rational expression 1 over x provided r x is are not equal to 0 and the function is defined by 5 when r x does equal 0. So once again let's look at a graph of this. The graph of 1 over x simply is that of a rational expression with two branches one in the first quadrant one in the third quadrant and the second piece when r x is equal to 0 we have the y value of 5 so that simply gives us an isolated point up here on the y-axis at 5. We can see that this is an infinite discontinuity so our hunch would be that this is going to be found to be not continuous at x equals 0 simply from the way the graph looks but let's go ahead and apply the three tests of continuity to prove it. Recall that the first requirement is that the function value has to exist so in this case we would be inquiring about f of 0 and we do know that to equal 5. The second requirement is that the limit of this function as x approaches 0 has to exist. Well we could probably just do this graphically if you'd like. We can see from the right this limit as we approach 0 does not exist but it does approach positive infinity. As we approach 0 from the left the limit once again does not exist but this time approaches negative infinity so definitely an infinite discontinuity here. So this limit simply does not exist and this is where the three tests fall apart is here at the limit requirement so therefore we can conclude without even looking at the third test that f of x is not continuous at x equals 0.