all right so we're going to make a quick video on finding domain of some more specific functions and then we'll start playing around with um shifts of these functions um so let's go ahead and graph we're going to start with the rational function in this case this is the reciprocal function very specific rational function and we know that X cannot be zero here so we remember that we have a vertical asymptote at x equals zero which is the equation of the line that is the y-axis we know that we can never divide one by any number and get the result of zero so we're going to have a horizontal asymptote of y equals zero because we'll never be able to cross the line Y equals zero all right and then any positive X values will be 1 over a positive so a positive again and I've got to make sure that I'm approaching both of these asymptotes this is the behavior we're going to get and if I plug in Negative X values I will result in negative y values and I've got to approach both asymptotes down here as well and so that is the graph of the function 1 over X and you'll notice that the domain is all X values except x equals zero so the domain of this function in interval notation is everything from negative Infinity to a zero not including zero or any value on the interval from zero to Infinity again not including 0. so with that I want to look at the function one over X plus 2 which is basically a shift of this function to the left two units and so one way that we could look at this is to say well if the denominator couldn't be zero in the original function that is still the case here so we know that X plus 2 cannot be zero which means that X cannot be negative 2. so the domain for this function is all real numbers except for negative 2. and just a reminder this is the union symbol it is a big cup it is not a letter u okay so this is our domain and interval notation all right let's take a look at the root function next so now we have the root function we know that the square root of zero is zero but we also know that we cannot take the square root of any negative numbers the square root of positive numbers will also be positive numbers and as you're taking square roots you're getting smaller numbers than what you started with and get this shape here all right and we see that our domain is any real number from zero all the way to positive Infinity so any positive number and zero we could say all non-negative numbers so in this case we will include zero since we can plot a point at zero so the domain is all real numbers from 0 to infinity and we are including zero all right so another way that you could look at this is we basically are saying that X is non-negative greater than or equal to zero so we need whatever is underneath the root to be greater than or equal to zero so down here the thing that we have under the root is x minus five so this is basically a shift of this graph to the right five units we need x minus 5 to be greater than or equal to zero if I add 5 to both sides we know that X must be greater than or equal to positive 5. and so the domain of this G is going to be all real numbers from 5 to Infinity again we are including 5 in that domain all right so the next one we're going to go off to logarithm land and we are not afraid of this it's not as bad as we think it is the graph of the natural log function one of the most important things to remember is that X must be greater than zero when we are looking at the natural log of x we have a vertical asymptote on the graph of the logarithm we will go through this again later on in the semester as well in a little bit more detail all right if we take the natural log of 1 we get 0 which we've talked a little bit about in review so far this semester and we have an increasing behavior for this graph so if we're trying to approach this asymptote but we're still trying to increase over time or over the x-axis this is the shape we are going to get all right so again X must be greater than zero that means our domain is all real numbers greater than zero anything from zero to Infinity but we cannot include zero in that so with this example down here we have 1 minus X inside the logarithm I can't take the log of a negative number and I cannot take the log of 0 I have to take the log of a strictly positive number so that means this 1 minus X quantity must be strictly positive if I add X over I get 1 is greater than x don't love the way that's written I'm just going to reflect that which means X is less than 1. for X to be less than one that means I'm going from negative Infinity to one but again I cannot include one in my domain all right exponential functions these are quite a bit easier you can't raise anything to a power and get a negative number all right nothing like this so if I have e as an example or any real number a I can't raise that real number to a power and get a negative result so what I'm going to have I'm sorry real positive number excuse me so what I'm going to end up with is asymptotic behavior which we have a horizontal asymptote here and I want you to remember these guys are inverses of each other so where I have a vertical asymptote for logarithm I have a horizontal asymptote for exponential if I take e to the zeroth power anything to the zeroth power is equal to one notice again this is going to be an inverse situation one zero becomes 0 1. exponential functions are also increasing functions but I have to approach that asymptote and I do that in the negative Direction and that is the graph of e to the X while that being said that means my domain is all real numbers I have zero restrictions on the domain of this function the domain is all real numbers so that means no matter what I do to e in its exponent the domain will remain all real numbers so there's really nothing more to say than that the domain is from negative Infinity to Infinity all right we're moving into trigonometric functions next we have six of these so the sine function the sine of zero is zero we know that the sine of pi over 2 is 1. the sine of Pi is also zero the sine of 3 pi over 2 is negative one try to make that about the same distance and then the sine of 2 pi which I ran out of space here apologize for that is also going to be zero and so you get this Behavior like this and it's a wave function and this is a periodic function which means this Behavior continues on in this way in both directions so what that means is I can take the sign of any real number X and get a result between negative 1 and 1. so the domain is all real numbers again so what does that mean about any shifting horizontally means that the domain stays the same all real numbers shifting it horizontally will not change that cosine we're going to go ahead and shift the sine function to the left pi over two units to get cosine as you think about the fact that sine and cosine are co-functions of each other and this intersection happens at pi over negative pi over two pi over 2 pi 3 pi over 2 and 2 pi and we're up at 1 and down to negative one I'll go ahead and put this guy in here negative pi as well all right so the cosine function is just a shift of the sine function to the left pi over two units which means again the domain is all real numbers shifting something horizontally does not change the domain if the domain is all real numbers stays all real numbers no matter what I do to the inside of the function so next we're going to look at tangent so tangent I want to remind you that is the same thing as saying sine over cosine and so we cannot have the cosine be equal to zero because we cannot divide by zero so let's come back over here when is the cosine equal to zero at negative pi over 2 positive pi over 2. 3 pi over 2 and if I keep going in that pattern 5 pi over 2 7 pi over 2. so basically all odd multiples of pi over 2 will make cosine zero which means tangent will be undefined at all of those odd multiples of pi over 2 and we will get vertical asymptotes and I'm trying to be as symmetric as I can I didn't have any graph paper I'm just going to draw a couple of them all right and we're going to go ahead and label these all right so we have pi over 2 here so x equals pi over 2. x equals 3 pi over 2 this one would be x equals negative pi over 2 and then this one would be negative 3 pi over 2 I'm out of space but you get the idea so we're going to go ahead and bring in our increasing behavior that is tangent and if you want to go through why the behavior of tangent goes in the increasing Direction make sure you go see your professor in office hours because we are happy to go through that with you and this is just going to keep going on forever and ever and ever so the domain of tangent is all real numbers besides odd multiples of pi over 2. so there's a way that we can say this very easily we can say that x minus PI what's inside here cannot be odd and I'm just going to use n for now multiples of pi over 2 and I'll say n is odd foreign so that means that X cannot be pi plus n pi over 2. so this is one way to write it but now if you want to write it in interval notation think about the fact that if you're going to shift this to the right Pi units which is What's Happening Here all of these negative pi over 2 and pi over 2 are just going to shift to the next one and the next one so if I add pi to negative pi over 2 I get positive pi over 2. if I add pi to pi over 2 I get 3 pi over 2. so the domain for tangent of x minus Pi is actually the same as the domain for tangent of x so here we're just going to say because this can go on infinitely anything from negative 3 pi over 2 to negative pi over 2 negative pi over 2 to pi over 2. pi over 2 to 3 pi over 2 and so on so these three dots mean that the pattern continues on in that fashion all right so now cotangent we'll take a look at that cotangent is cosine over sine and so we are looking for the values of sine to not be equal to zero here so let's go ahead and bring back our sine graph which says that sine is 0 at 0 Pi 2 pi 3 Pi so on so sine is equal to 0 at integer multiples of Pi so we will have vertical asymptotes at integer multiples of Pi zero oops sorry moving the paper too much x equals pi 2 pi and then I'll just put one more over here at negative pi and cotangent is going to be a decreasing function so it will look something like this basically the reciprocal of tangents so whatever tangent was doing cotangent of the multiplicative inverse doing my best with my graphs here hopefully they're clean enough for you so in this case the domain for cotangent is all real numbers except integer multiples of Pi all right and so in this case what's inside here X plus pi cannot be an integer multiple of Pi so if I subtract PI from both sides I'm taking an integer multiple of Pi minus Pi well for example if I take 2 pi minus Pi I just get to Pi same discontinuity if I go Pi minus Pi get to 0 same issue if I go from PI minus Pi I get to negative pi same issue so in this case the domain is all real numbers except for integer multiples of Pi so I'll start with negative pi to zero zero to Pi and then pi to 2 pi and again I'll just use my Ellipsis at the ends of either side there to identify that this pattern continues on forever all right and then you can use this same reasoning for secant and cosecant if you write secant as 1 over cosine or if you write cosecant as 1 over sine and then you can use the same domain restrictions as we did here with Tangent and cotangent