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Current time:0:00Total duration:7:12

Video transcript

we've got the function f of X is equal to X plus 2 squared plus 1 and we've constrained our domain that X has to be greater than or equal to negative 2 that's where we've defined our function and we want to find its inverse and I'll leave you to think about why we had to constrain it to X being greater than or equal to negative 2 why it wouldn't have been possible to find the inverse if we had just left it as the full parabola I'll leave you maybe I'll make a future video about that but let's just figure out the inverse here so like we've said in the first video in the introduction to inverses we're trying to find a mapping or if you if we were to say that Y if we were to say that Y is equal to X plus 2 squared plus 1 this is the function you give me an X and it maps to Y we want to go the other way we want to take I'll give you a Y and then map it to an X so what we do is we essentially just solve for X in terms of Y so let's do that one step at a time so the first thing to do we could subtract one from both sides of this equation y minus one is equal to X plus two squared and now to solve here you might want to take the square root and that actually will be the correct thing to do but it's it's very important to think about whether you're going to take the positive or the negative square root at this step so we've constrained our domain to X is greater than or equal to negative two so this value right here X plus two if X is always greater than or equal to negative two x plus two will always be greater than or equal to zero so this expression right here this right here is positive this is positive so we have a positive squared so if we really want to get to the X plus two and the appropriate in the appropriate domain we want to take the positive square root and in the next video or the video after that we'll so an example where you want to take the negative square root so we're gonna take the positive square root or just the principal root which is just the square root sign of both sides so you get the square root the square root of Y minus one is equal to X plus two and one thing I should have remembered to do is from the beginning we have we had a constraint on X we had 4x is greater than or equal to negative two but what constraint could we have on why if you look at the graph right here X is greater than or equal to negative two but what's Y what what what what is the range of Y values that we can get here well if you just look at the graph Y will always be greater than or equal to one and that just comes from the fact that this term right here that this term right here is always going to be greater than or equal to zero so the minimum value that that the function can take on is one so we could say for X is greater than or equal to negative two and we could add we could add that Y is always going to be greater than or equal to one right y is always greater than or equal to 1 the function is always greater than or equal to that right there to one and the reason why I want to write it at this stage is because you know later on we're going to swap the X's and Y's so let's just leave that there so here we haven't explicitly solved for x and y but we can write for Y is greater than or equal to one this is going to be the domain for our inverse so to speak and so here we'll we can keep it for y is greater than or equal to one this Y constraints going to matter more because over here the function the domain is X but for the inverse the domain is going to be the Y value and then let's see we have the square root of Y minus 1 is equal to X plus 2 now we can subtract 2 from both sides we get the square root of y minus 1 minus 2 is equal to X for y is greater than or equal to 1 and so we've solved for X in terms of Y or we could say F let me just write it the other way because say X is equal to I'm just swapping this X is equal to the square root of Y minus 1 minus 2 for y is greater than or equal to 1 so you see now the way we've written it now Y is the input into the function which is going to be the inverse of that function x is the output X is now the range so we could even rewrite this as F inverse of Y that's what X is is equal to the square root of Y minus 1 minus 2 or y is greater than or equal to one and this is the inverse function we could say this is our answer but many times people want the answer in terms of X and we know we could put anything in here if we put an A here we can take f inverse of a and we'll become the square root of a minus 1 minus 2 4 well assuming a is greater than or equal to 1 but we could put an X in here so we can just rename the the Y for X so we could just do a renaming here so we can just rename Y for X and then we would get let me scroll down a little bit we would get F inverse of X and I'll highlight it here just show you we're renaming Y with X you could rename it with anything really is equal to the square root of x minus 1 of X minus 1 minus 2 4 we have to rename this to 4 X for X being greater than or equal to 1 and so we now have our inverse function in 2 as a function of X and if we were to graph it let's try our best to graph it maybe the easiest thing to do is to draw some points here so the smallest value X can take on is 1 if you put a 1 here you get a 0 here so the point 1 negative 2 is on our inverse graph so 1 negative 2 is right there and then if we go to 2 let's see 2 minus 1 is 1 this principal root of that is 1 minus 2 so it's negative 1 so the point 2 negative 1 is right there and let's think about let's see if we did 5 I'm trying to take perfect squares 5 minus 1 is 4 minus 2 so the point 5 2 is let me make sure Oh 5 minus 1 is 4 square root is 2 minus 2 is 0 so the point 5 0 is here and so the inverse graph it's only defined for X greater than or equal to negative 1 so the inverse graph is going to look something like this it's going to look something like I started off well and it got messy so let's see it's going to look something like that just like that and just like we saw in the first the introduction to function inverses these are mirror images around the line y let me let me graph y equal X y equals x y equals x is is that line right there notice they're mirror images around that line over here we map the value 0 to negative 2 5 right if x is 0 we get it gets mapped to 5 here we go the other way we're mapping 5 to we're mapping 5 to the value 0 so that's why they're mirror images we essentially swapped the x and y