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 whole here's another type of discontinuity and this one is called a jump discontinuity because the left side and 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 Asm toote on the right side it approaches positive Infinity on the left negative Infinity now you need to know that the hole 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/x^2 if you're given a rational function you could find the vertical ASM toote which is an infinite discontinuity by setting the denominator equal to zero so in this case the vertical ASM toote is x equal Z 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 / by x + 2 X+ 2 cannot be zero because that's going to be the vertical ASM toote so therefore X cannot be -2 that is the point of discontinuity if we graph this function we're going to have a vertical ASM toote at x = -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 * x + 2 / x + 2 * x - 5 the x - 5 Factor on the bottom leads to the vertical asto so X cannot equal Al POS so that is an infinite discontinuity anytime It's associated with a vertical ASM toote it's an infinite discontinuity now notice that we can cancel x + 2 however this is still a point of discontinuity so X cannot equal -2 but this type of discontinuity is known as a whole if it can be cancelled here's another problem f X is equal to the absolute value of x / 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 zero so X cannot equal zero but notice that we have an absolute value function so this could be positive X which means F ofx can be 1 or it could be Nega X which means F ofx can be equal to 1 this is true when X is greater than zero and this is true when X is less than zero so the graph looks something like this when X is greater than zero it's going to be positive 1 and when it's less than zero 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 peie wise 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 pie wise function 5x + 3 x^2 + 4 and X Cub so it's going to be 5x + 3 when X is less than 1 and when X is between 1 and 2 f ofx is going to equal X2 + 4 and then it's going to be X Cub when X is equal to or greater than two so determine the points of discontinuity now looking at this function linear function funs are always continuous everywhere quadratic functions are also continuous even cubic functions so any polom function is continuous everywhere rational functions like this one will be discontinuous when the denominator is zero but we don't have a rational function here so for peie wise function the only possible locations where it can be discontinuous is at an x value of one or at an x value of two and a quick way to tell is to see if the yv value is the same at those points so let's start with one if we plug in one into 5x + 3 we're going to get a value of eight and if we plug it into the second part x^2 + 4 we're going to get a value of five therefore it's discontinuous at x = 1 now what about x = 2 so we need to use these two functions if we plug in 2 into x^2 + 4 it's going to be 2^ 2 + 4 which is 8 and 2 to the 3 power is also 8 now because these two have the same y value you know that it's continuous at x = 2 but is discontinuous at x = 1 consider the function f ofx which is equal to CX + 3 when X is less than 2 and that's equal to 3x + 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 xal 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 xal 2 these two functions have to have the same yvalue 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 two so the second step is to replace x with two and then find the value of C so we're going to have 2 C + 3 is equal to 6 + C so let's subtract both sides by C and let's subtract both sides by 3 2 c - c is C 6 - 3 is 3 so C is equal to three now what about this problem let's say that f ofx is equal to ax - 2 and also x^2 - 5 when X is less than 3 and when X is equal to or greater than three 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 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 * 3 - 2 is = to 3^ 2 - 5 so this is going to be 3 a - 2 which is equal to 9 - 5 and 9 - 5 is 4 now if we add two to both sides 4 + 2 is six and then we have to divide both sides by three so 6 / 3 is 2 so the value of the constant is two let's try another problem so let's say f ofx is ax + 5 when X is less than 1 and that it's x^2 - BX + 9 when X is between 1 and 4 4 and it's ax^2 - BX - 7 when X is equal to or greater than four go ahead and find the values of A and B so first let's make the function continuous when X is one so therefore we need to set these two equations equal to each other so ax + 5 is equal to x^2 - BX + + 9 now let's replace x with 1 so this is going to be 1 a or just a 1^ 2 is 1 - B * 1 or B + 9 1 + 9 is 10 so a + 5 is equal to 10 minus B now let's add B to both sides and let's subtract both sides by five so on the left what we have is A + B on the right we have 10 - 5 so a plus b is equal to 5 now let's use the next number four so we need to set these two functions equal to each other so x^2 - BX + 9 is equal to ax^2 - BX - 7 and X is 4 so 4^ 2 - 4 b + 9 is equal to 4 or a * 4^ 2 - B * 4 - 7 so 4^ 2 is 16 and this is going to be 16 A minus 4 b - 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 16 a - 7 and let's not forget about the plus 9 on that side 16 + 9 is 25 and that's equal to 16 a minus 7 but now since we're run out of space let's clear away a few things so now let's add seven to both sides 25 + 7 is 32 so 32 is equal to 16 a and if we divide both sides by 16 a is 32 / 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 + B is equal to 5 therefore B is 5 - 2 which is 3 so now you have the value of a and b