Now let's talk about continuity. This graph is continuous everywhere. There's no jumps or breaks in the graph. Now in this example, we have a point of discontinuity at some x value, which we'll call c.
Now this particular type of discontinuity is known as a hole. Here's another type of discontinuity. And this one is called a jump discontinuity. Because the left side and the right side do not connect. And then there's another type.
This type of discontinuity is known as an infinite discontinuity. As you approach the vertical asymptote, on the right side it approaches positive infinity, on the left, negative infinity. Now, you need to know that the whole is a removable discontinuity.
But the other two, the jump discontinuity and the infinite discontinuity, are non-removable discontinuities. Now, let's talk about how to identify points of discontinuity in a graph. So here's the first example, 1 over x squared. If you're given a rational function, you could find the vertical asymptote. which is an infinite discontinuity by setting the denominator equal to zero.
So in this case the vertical asymptote is x equals zero. So therefore that's a point of discontinuity because in this graph there is no x value of zero, it's undefined at that point. And as you can see we have an infinite discontinuity. So here's another example. Let's say if we have 5 divided by x plus 2. x plus 2 cannot be 0 because that's going to be the vertical asymptote.
So therefore, x cannot be negative 2. That is the point of discontinuity. If we graph this function, we're going to have a vertical asymptote at x equals negative 2. And so the graph is going to look something like this. So this is another infinite discontinuity. Now what about a function that looks like this?
Let's say 3 times x plus 2 divided by x plus 2 times x minus 5. The x minus 5 factor on the bottom leads to the vertical asymptote, so x cannot equal positive 5. So that is an infinite discontinuity. Anytime it's associated with a vertical asymptote... out it's an infinite discontinuity.
Now notice that we can cancel x plus 2. However, this is still a point of discontinuity, so x cannot equal negative 2, but this type of discontinuity is known as a whole if it can be canceled. Here's another problem. f of x is equal to the absolute value of x divided by x. So what type of discontinuity do we have in this example?
Is it a whole, a jump discontinuity, or an infinite discontinuity? Also, determine if it's removable or non-removable. Now we know that the denominator can be 0, so...
x cannot equal 0. But notice that we have an absolute value function. So this could be positive x, which means f of x can be 1, or it could be negative x, which means f of x can be equal to negative 1. This is true when x is greater than 0, and this is true when x is less than 0. So the graph looks something like this. When x is greater than 0, it's going to be positive 1, and when it's less than 0, negative 1. So this is the jump discontinuity.
And it usually occurs whenever you have an absolute value function like this, or it could be due to a piecewise function as well. Now a jump discontinuity is also a non-removable discontinuity. The only one that's removable is the whole.
Now let's say if we have a piecewise function. 5x plus 3, x squared plus 4, and x cubed. So it's going to be 5x plus 3 when x is less than 1, and when x is between 1 and 2, f of x is going to equal x squared plus 4. And then it's going to be x cubed when x is equal to or greater than 2. So, determine the points of discontinuity.
Now, looking at this function, linear functions are always continuous everywhere. Quadratic functions are also continuous, even cubic functions. polynomial function is continuous everywhere.
Rational functions like this one will be discontinuous when the denominator is 0, but we don't have a rational function here. So for a piecewise function, the only possible locations where it can be discontinuous is at an x value of 1 or at an x value of 2. And a quick way to tell is to see if the y value is the same at those points. So let's start with 1. If we plug in 1 into 5x plus 3, we're going to get a value of 8. And if we plug it into the second part, x squared plus 4, we're going to get a value of 5. Therefore, it's discontinuous at x equals 1. Now, what about at x equals 2? So we need to use these two functions. If we plug in 2 into x squared plus 4, it's going to be 2 squared plus 4, which is 8. And 2 to the third power is also 8. Now because these two have the same y value, you know that it's continuous at x equals 2, but it's discontinuous at x equals 1. Consider the function f of x which is equal to cx plus 3 when x is less than 2 and that's equal to 3x plus c when x is equal to or greater than 2. What is the value of the constant c that will make the function continuous at x equals 2?
So what do we need to do in order to find the value of that constant C? Well, if it's going to be continuous at x equals 2, these two functions have to have the same y value, which means that they must be equal to each other. So the first thing is to set them equal to each other. And they have to be equal to each other at an x value of 2. So the second step is to replace x with 2, and then find the value of C.
So we're going to have... have 2C plus 3 is equal to 6 plus C. So let's subtract both sides by C and let's subtract both sides by 3. 2C minus C is C, 6 minus 3 is 3. So C is equal to 3. Now what about this problem? Let's say that f of x is equal to...
ax minus 2 and also x squared minus 5 when x is less than 3 and when x is equal to a gradient 3. Find the value of the constant a that will make the function continuous. So, go ahead, pause the video, and try this problem. So, the first thing we need to do is set the two portions of the function equal to each other.
And then, in the next step, we need to replace x with 3. so this is going to be a times 3 minus 2 is equal to 3 squared minus 5 so this is going to be 3a minus 2 which is equal to 9 minus 5 and 9 minus 5 is Now if we add 2 to both sides, 4 plus 2 is 6. And then we have to divide both sides by 3, so 6 divided by 3 is 2. So the value of the constant is 2. Let's try another problem. Let's try another problem. Let's try another problem. So let's say f of x is ax plus 5 when x is less than 1 and that it's x squared minus bx plus 9 when x is between 1 and 4. It's ax squared minus bx minus 7 when x is equal to or greater than 4 You go ahead and find the values of a and b so first let's Make the function continuous when x is 1 so therefore we need to set these two equations equal to each other So ax plus 5 is equal to x squared minus bx plus 9. Now, let's replace x with 1. So, this is going to be 1a, or just a. 1 squared is 1, minus b times 1, or b, plus 9. 1 plus 9 is 10. So, a plus 5 is equal to 10 minus b.
Now, let's add b to both sides, and let's subtract both sides by 5. So on the left, what we have is a plus b. On the right, we have 10 minus 5. So a plus b is equal to 5. Now, let's use the next number, 4. So we need to set these two functions equal to each other. So x squared minus bx plus 9 is equal to ax squared minus bx minus 7. And x is 4, so 4 squared minus 4b plus 9 is equal to 4, or a times 4 squared minus b times 4 minus 7. So 4 squared is 16. and this is going to be 16a minus 4b minus 7. So if we add 4b to both sides, notice that we can cancel the b expressions since they're the same.
So therefore 16 is equal to 16a. Minus 7 and let's not forget about the plus 9 on that side 16 plus 9 is 25 and That's equal to 16 a minus 7 but now since we're running out of space let's clear away a few things So now let's add 7 to both sides. 25 plus 7 is 32. So 32 is equal to 16a.
And if we divide both sides by 16, a is 32 divided by 16, which is 2. Now that we have the value of a we can plug it in here to get the value of B So 2 plus B is equal to 5 Therefore B is 5 minus 2 which is 3 So now you have the value of a and B