 Welcome back to our lecture series Math 12-10, Calculus 1 for students at Southern Utah University. As usual, I'll be a professor today, Dr. Andrew Misseldine. In our previous lecture, number 12, we introduced the idea of continuous functions. Those functions whose graph can be drawn with one continuous stroke of your pen, or in other words, those functions for which the limit of the function, say the limit as x approaches a of f of x here, is equal to f of a the limit and the function agree with each other for these numbers x equals a inside the domain of the function. We did pay particular importance to piecewise functions. Those functions built out of pieces and particularly the pieces we've built from were continuous pieces. We were interested in how, you know, how do you compute the limits of these piecewise functions and therefore how can we determine whether a piecewise function is continuous or not. Well, there are times where we might want to build a piecewise function but still want it to be continuous. In order for a piecewise function to be continuous, we have to avoid the following problem. When you jump between one piece to the other, we can't have some jump discontinuity. We need the pieces to come together. We need to connect them. And so the concern you see on a picture like this is that when you're approaching the first point, this is going to be your left-handed limit, the limit as x approaches a from the left of f of x. And then we also have the approach from the right, the limit as x approaches a from the right of f of x here. We need to make sure that the left-sided limit and the right-sided limit are the same numbers that the limit exists and agrees with the function. So the key thing to do here is to look for the left and right-handed limits of this function. So if we were to look at the limit as x approaches two from the left of our function f of x, why do we look at two? Well, two is the switching number. It's at x equals two, where the function switches from one portion to the other. So less than or equal to two, it looks like the parabola Cx squared. But then when you're above two, it'll look like the line x plus c. And so we need those to agree with each other. So if you're approaching two from the left, that is you're a little bit less than two, you're going to use this domain, and thus you're going to use this part. So the limit as x is less than two of f of x, this is going to become the limit of Cx squared as x approaches two from the left. Now the function Cx squared, C is just some unspecified parameter. We're trying to find out what constant C will make this function continuous. What constant C will make the left side and the right side touch each other. So we take the limit as x approaches two here of Cx squared because the function is, because a parabola is a continuous function, we can compute this limit using direct substitution. So we're going to end up with C times two squared. That is we end up with a four C. That's what the left-handed limit wants to be. Now the right-handed limit, if we take the limit as x approaches two from the right of f of x, we have to consult which domain inside of f will be a little bit bigger than two. Well that's going to come from this condition right here where x is greater than two. And so on this sector, the function looks like x plus C. So the limit as x approaches two from the right of f of x will be identical to that limit of x plus C, but as the line is a continuous piece, we can direct substitution to compute the limit. So we're going to end up two plus C. And so what we then have to do is we have to equate these things together. We have the equation four C equals two plus C and we have to solve for this parameter C. You subtract two from both sides. You're going to end up with a three C equals two. Divide both sides by three. You're going to end up with the parameter that C equals two-thirds. So if we set this coefficient C to be two-thirds, we're going to get that the function is in fact continuous. Let's take a look at this on Desmos right here so you can see the graph of our function f on the screen. So again, this is the piecewise function where when you are less than two, you get the parabola Cx squared. And when you're to the right of two, you're going to get x plus C. And notice that our function at the point x equals two is defined to be the parabola. And so you see this jump occurring on the graph where the parameter C is then set up here. So if we start changing the C value, you can see what happens here that depending on the C value, our jump can get bigger or smaller, bigger or smaller. So we basically want to find the value where these things come together and touch each other. So playing around with this thing, that's going to happen somewhere around 0.6 or 0.7. You can see it's not exactly perfect, but around 0.6 or 0.7, they get really close to each other. And why is that 0.7 here? Well, remember that the value we found was two thirds, which is 0.66666, right? Two thirds is the devil's fraction. If we were to turn this stuff off for a second, notice if I set C to be two thirds, the function comes to a perfect touch. Boom. And if we zoom out of this, you have perfect continuity. There is a corner, like a little sharp corner, as it changes from the parabola to the line, but it is continuous. There is no break on the function anymore when we plug in two thirds. Let's consider another example. This time, let's take g of x to be the piecewise function where you take x cubed plus C when x is less than or equal to 3, and take Cx minus 5 when x is greater than 3. So like before, we're interested in what parameter can we set for this constant C to make the function be continuous, right? We have to make sure that the graph is touching each other. Let's switch over to the graph for a second. Again, looking at Desmos right here, we see that part of the graph is this cubic function x cubed plus C, and then we see this other portion is the line, Cx minus 5. And so if we play around with the constant C, we see that if we play around with the constant, there are some values that seem to bring it together. That went off the screen. If we bring it a little bit bigger, bigger, bigger, bigger, bigger, they seem to be coming together up. We went too far. If we come back here, looks to be around 16. Let's turn off these points here so we can give it some of the clutter. When you look at 16, you take those points away. It's like, wow, that does look flawless. That looks pretty close. Now, it could be that we're off by a little bit. Kind of like we saw before, the point six was too hot. The point seven was too cold. This one looks really good, but maybe there's just some approximation problems. It looks right. And so if we go back to the algebra, which is going to be much more reliable, let's try to solve this algebraically. We need to compute the left-handed limit as x approaches three from the left here of g of x. Now, if you are less than three, a little bit less than three, you're going to be on this sector right here. It's going to look like x cubed plus c. So we need to compute the limit as x approaches three from the left of x cubed plus c. But as it's a polynomial, the limit can be computed by direct substitution. So we're going to get a three cubed plus c. That is 27 plus c. That's what the left-handed limit turns out to be. If we look at the right-handed limit, that is, if we're a little bit bigger than three, we want to compute g of x right here. We're going to be on this domain. Thus, we need to look at cx minus five. So we get the limit as x approaches three from above of cx minus five. It's a polynomial. We can compute the limit by direct substitution. We end up with three c minus five right here. And so we need to equate these things together. We need the left-handed limit to equal the right-handed limit for the function to be continuous. So we take 27 plus c. This is equal to three c minus five. So let's subtract c from both sides. Let's add five to both sides like so. We end up with, well, three c minus a c is a two c. And then 27 plus five is 32. Divide both sides by two. And we end up with the c value should be 16. 16 is the one value that'll make the two parts touch each other, like we kind of predicted with the Desmos. It wasn't just an approximation. 16 was the exact value. So this shows you how we can find the constant to make this piecewise function be continuous. But this is only if you have two pieces. What if you have three pieces or four pieces? Well, if you have three pieces, you're going to have to have two unknowns, two constants, maybe call them a and b. If you have four pieces, you're going to need three unknowns to put them together. In these examples we saw here, we had to solve linear equations with respect to these constant c's. Well, you might have to solve a system of equations if there's more than one parameter because there are more than two parts in your piecewise function.