in this video we're going to talk about how to find the domain of an inverse function so we have the regular function 3x minus 4. if we were to draw a rough sketch of the graph of this function it will look something like this it has a y intercept of negative 4 and a slope of positive three so this is a generic shape of that graph the domain ranges from negative Infinity to Infinity the domain represents the X values of the function that it can have and X could be anything the range has to do with the Y values y can be anything in this graph there's no restrictions now in order for this function to have an inverse function it must be a one-to-one function which means that it must pass the horizontal line test it can only touch the horizontal line at one point so because this function is one to one the inverse function exists now let's go ahead and find the inverse function of F so replace f with y and then switch X and Y and then solve for y if we add 4 to both sides we get that X plus 4 is equal to 3y divide both sides by 3 we get this so now we're going to replace y with the inverse function symbol so the inverse function is X plus four over three now what you need to know is that the domain of the inverse function is the range and the range of the inverse function is the domain so for this particular video we're going to focus on finding the domain of the inverse function and the answer for this example is all real numbers it's negative Infinity to Infinity now let's consider another problem so let's say we have the function x squared minus 3. find the domain of the inverse function now let's draw a rough sketch of the graph of that function x squared is a parabola that opens upward but it's been shifted three units down so it's going to look like this and it has a y-intercept of negative 3. now this particular function does not have an inverse function because it does not pass the horizontal line test it touches it at two points so we can't find an inverse function for this unless we restrict the domain so what we're going to do is we're only going to focus we're going to create a new f of x we're only going to use the right side of that function so we're going to restrict the domain of f of x to X is equal to or greater than zero so we still have a y-intercept of negative 3 but we don't have anything on the left side now this new restricted f has an inverse function because it passes the horizontal line test so we could find the domain of the inverse function for this particular function but before we do that let's find the domain and range so we restrict the domain to X being equal to or greater than zero so that means from 0 to Infinity and the range is also restricted as well focusing on the Y values the lowest y value is negative three the highest is infinity now keep in mind the range of f of x is the domain of the inverse function so the answer for this problem is what we see here but let's confirm that let's find the inverse function let's replace f of x with y and then let's switch X and Y and then let's solve for y I'm going to add 3 to both sides and to get y by itself I'm going to take the square root of both sides and then let's replace y with the inverse function symbol so the inverse function is the square root of X Plus 3. if we were to graph that the square root of x looks like this but this has been shifted three units to left so it's going to start at negative three and it's going to go towards the right so the blue line is f the red line is the inverse function when you graph F and the inverse of f they reflects about the line Y equals X so that's how you know if you did it correctly now when dealing with radical functions the inside of the radical has to be equal to or greater than zero it can't be negative otherwise you'll get an imaginary number so subtracting both sides by three we get that X is equal to or greater than negative three for the range I need to correct this because I put positive 3 for the range but it's actually it should be negative 3. I missed that negative sign but nevertheless this is the answer so the domain for the inverse function is from negative 3 to Infinity I just forgot my negative 3 here it's very easy to forget a negative sign now if we want to find a range of the inverse function it's going to be the domain of the regular function so that's a 0 to Infinity and we can see that's the case with a square root function focusing on the red line the lowest y value is zero and this will continue to increase to Infinity so it's from 0 to Infinity the red line never goes below the x-axis so it doesn't have any negative y values therefore this is the domain for the inverse of F and that's the range for it now let's consider another example so let's say f of x is equal to the square root of 8 minus 2X go ahead and find the domain of the inverse function but let's focus on a regular function first we know that the inside part has to be equal to or greater than zero subtracting both sides by eight we get this and then if we divide by negative two negative eight divided by negative two is positive four and since we're dividing by a negative number we need to change the direction of the inequality so the domain for f is negative Infinity to 4 with 4 being included if you were to sketch this graph it would have an X intercept of four if you have a positive sign in front of it it's going to be above the x-axis and it's going to go this way so we can see why the domain will be from negative Infinity to 4. now looking at the range the lowest y value is zero the highest it can continue going up to Infinity so the range is 0 to Infinity which means this is our answer for the problem that's going to be the domain of the inverse function let's find the inverse function by the way this function is one to one it passes the horizontal line test so the inverse function exists let's replace f of x with y and then let's switch X and Y now let's Square both sides if we take the square of both sides the square root will disappear on the right side now I'm going to add 2y to both sides and I'm going to subtract both sides by x squared when you do this it has the effect of moving the 2y from the right to left it'll change from negative 2y to positive 2i moving x square to the other side makes the change from positive x squared to negative x squared now dividing both sides by 2 we get that Y is 8 minus x squared over two so the inverse function is this but I'm going to rewrite it as 8 over 2 is 4. and then minus x squared over 2 which is the same as one half x squared we can write that like this negative one-half x squared plus four it makes it easier to graph so x squared is typically a graph that opens upward but since we're dealing with negative x squared we know that opens downward but it's been shifted up four units so it should start somewhere over here and if we didn't restrict this function it will look something like this that's going to be a rough sketch but this function is going to be restricted because if we don't it's not a one-to-one function so what we need is only the right side of this graph if we think about why the range of f of x is 0 to Infinity therefore the domain must be restricted to that it's going to be 0 to Infinity so we only need the right side of the inverse function and we could see if we were to graph y equals x the red line and the blue line they reflect about the line Y equals X so we had to restrict the domain the range of the inverse function is the domain of the original function negative Infinity to 4. so if we follow the blue line this is going to go from negative Infinity and if we follow it up it's going to stop at 4. when you plug in 0 into this function you get 4. so the range is from negative Infinity to 4. but this is the answer we're looking for the domain of the inverse function it's the range of the regular function which is 0 to Infinity now let's work on one more problem this time we're going to focus on a rational function let's say f of x is 3x Plus 2. divided by 5x minus 4. go ahead and find the domain of the inverse function now when dealing with fractional functions you can't have a zero in the denominator of the fraction so we could say that 5x minus 4 cannot be 0. add in 4 to both sides we get this dividing by 5 we get that X can't be four over five it could be anything else but that so for f the domain is going to be from negative Infinity to four over five and then Union four over five to Infinity X can be anything except 4 over 5. now what about the range so in this function we have the numerator which is degree one and the denominator is a linear equation with degree one as well when you have a rational function like this where the degree of the numerator is the same as the degree of the denominator the horizontal asymptote is simply going to be the ratio of the coefficients of the degree of the term with the highest degree so in this case it's going to be 3 over 5. that's the horizontal asymptote what you need to know is that the Y value for this function can be anything except the horizontal asymptote the graph will get close to the horizontal asymptote but it won't touch it so since this is the horizontal asymptote y will never be that value therefore the range is going to be negative Infinity to 3 over 5. Union 3 over 5 to Infinity so in order to write the range of a function a rational function you need to remove the horizontal asymptote so if this is for f then the inverse function has the domain negative Infinity to 3 over 5 Union 3 over 5 to Infinity the range of the regular function becomes the domain of the inverse function so this is the answer that we're looking for now let's find the inverse function so let's replace f of x with y and then let's switch X and Y and then let's solve for y so I'm going to put X over 1 and then we're going to cross multiply so this is going to give us 1 times 3y Plus 2. and that is going to equal X times 5y minus 4. now let's distribute X to 5y minus 4. so it's 3y plus 2 is equal to 5 x y minus 4X I'm going to move this to the other side and this to the left side so we have 3y minus 5xy is equal to negative 2 minus 4X I probably should do it a different way but we can still get the right answer now I'm going to factor out y so this is going to be 3 minus 5X is equal to negative 2 minus 4X and then I'm going to divide by 3 minus 5X so Y is negative 2 minus 4X over 3 minus 5X I'm going to multiply the top and bottom by negative 1. just a make it look better so Y is going to be 4X plus 2 . so the negative 5x will become positive 5x and the positive 3 will become negative 3. so now it looks a lot better and so we can write this as the inverse of f now for the inverse function we can have a zero in the bottom so if we set that not equal to zero and then if we add three then divide by 5 we get that X can't be three over five and so we can see that in the domain for the inverse function three over five was removed which is the horizontal asymptote so that confirms that we have the right answer by the way as a side note if you're wondering why we need to remove the horizontal asymptote from the range of the original function besides the fact that it's been removed in the domain of the inverse function and the fact that the graph never touches it another way to see why we need to remove it is by plugging in this value into F and seeing what happens if we were to replace y with 3 over 5 and try to solve for x notice that we won't get a value for x Let's cross multiply so we have 5 times 3x Plus 2. equal 3 times 5x minus 4. so Distributing the 5 we get 15x plus 10 and distributing the 3 we get 15x minus 12. now if we subtract both sides by 15x completely disappears and we're left with 10 on the left negative 12 on the right and these two do not equal in value so it doesn't work therefore there is no x value that will produce the Y value of three over five and that's why it's the horizontal asymptote and that's why it must be removed from the range of f and that's also why it doesn't exist in the domain of f i mean the domain of the inverse function of F so that's basically it for this video now you know how to find the domain of an inverse function and the key is by finding the range of the original function