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Finding inverse functions: quadratic (example 2)

Video transcript

we have the function f of X is equal to X minus 1 squared minus 2 and they've constrained the domain to X being less than or equal to 1 so we have the left half of a parabola right here and they constraint it so that it's not a full u parabola and I'll let you think about why it would make that would make finding the inverse difficult but let's try to find the inverse here and a good place to start let's just set Y being equal to f of X so we could say Y is equal to f of X or we could just write that Y is equal to X minus 1 squared minus 2 we know it's 4 X is less than or equal to 1 but right now we have Y solved for in terms of X or we've solved for y to find the inverse we're going to want to solve for X in terms of Y and we're going to constrain Y some early and we can look at the graph and we can say well in this in this graph right here this is defined for y being greater than or equal to negative 2 so we could maybe put in parentheses Y being greater than or equal to negative 2 because this is going to be right now this is our range but when we swap the X's and Y's this is going to be our domain so let's just keep that in parentheses right there so let's solve for X so that's all you have to do to find the inverse solve for X and make sure you keep track of the domains and the ranges so let's see if we could add 2 to both sides of this equation we get y plus 2 is equal to X minus 1 squared right minus 2 plus 2 so those cancel out and then I'm just going to switch to the Y constraint because now it's not clear what we're salt you know whether Y is the domain in the range or the X is the domain in the range but we know by the end of this problem y is going to be the domain so let's just swap this here so 4 for y is greater than or equal to negative 2 and we could also say in parentheses X is less than 1 this is we haven't solved explicitly for either one so we'll keep both of them around right now now to solve for y or solve for X you might be tempted to just take the square root of both sides here and you wouldn't be completely wrong we have to be very very very careful here and this is this might not be something that you've ever seen before so this is is an interesting point here we want to we want the right side to just be to just to be X minus 1 that's our goal here and taking the square root of both sides we want to just have an X minus 1 over there now is X minus 1 a positive or a negative number well we've constrained our X is 2 being less than 1 so we're dealing only the situation where X is less than or equal to 1 so if X is less than or equal to 1 this is negative this is negative so we want to take the negative square root let me just be very clear here if I take negative 3 I mean if I take negative 3 and I were to square that is equal to 9 now if we want to if we take the square root of 9 if we take the square root let's say we take the square root of both sides of this equation and our goal is to get back to negative 3 if we take the positive square root if we just take the principal root of both sides of that we would get 3 is equal to 3 but that's not our goal we want to get back to negative 3 so we want to take the negative square root of that of our of our squared so because this expression is negative and we want to get back to this expression we want to get back to this X minus 1 we need to take the negative square root of both sides you can always every perfect square has a has a positive or a negative route the printt root the principal root is a positive root but here we want take the negative root because this expression right here is going to be negative and that's what we want to solve for so let's take the negative root of both sides so you get the negative square root of y plus 2 is equal to and I'll just write this extra step here just so you realize what we're doing is equal to the negative square root the negative square root of x minus 1 squared for y is greater than or equal to negative 2 and X is less than or equal to 1 that's why the whole reason we're going to take the negative square root there and then this expression right here so let me just write the left again negative square root of y plus two is equal to the negative square root of x minus one squared is just going to be X minus one is just going to be X minus one X minus 1 squared is some positive quantity the negative route is root is the negative number that you have to square to get it to get X minus one squared so that just becomes X minus one hopefully that doesn't confuse you too much we just want to get rid of this squared sign we want make sure we want to get the negative version we don't want the positive version which would have been 1 minus X don't want to confuse you so here we now I just have to solve for x add 1 and let me write for y is greater than or equal to negative 2 add 1 to both sides you get negative square root of y plus 2 plus 1 is equal to X for 4 y is greater than or equal to negative 2 or if we want to rewrite it we could say that X is equal to the negative square root of y plus 2 plus 1 for y is greater than equal to negative 2 if we want to write it in terms as an inverse function of Y we could say so we could say F inverse of Y is equal to this or F inverse of Y is equal to the negative square root of y plus 2 plus 1 for y is greater than equal to negative 2 and now if we wanted this in terms of X if we just want to rename Y is X we just replace the Y's with X's so we could write f inverse of X and I'm just renaming the Y here is equal to the negative square root of x plus 2 plus 1/4 I'm just renaming the Y for X is greater than or equal to negative 2 and if we were to graph this let's see if we started X is equal to negative 2 this is zero so the point negative 2 1 is going to be on our graph so negative 2 1 is going to be on our graph let's see if we go to if we go to negative 1 negative 1 this will become a negative 1 negative 1 0 is on our graph negative one zero is on our graph and then let's see if we were to do if we were to put X is equal to two here so X is equal to two is four four square root principle root is two but it becomes a negative two so it becomes to negative one so that's on our graph right there so the graph is going to look something like this of F inverse it's going to look something like something like that right there as you can see it is a reflection of our original f of X along the line y is equal to X along the line along the line y is equal to X because we've essentially just swapped the X and the y this is about as hard of an inverse problem that I expect you to see especially in a precalculus class because it really is tricky to realize that you have to take the negative square root here because the way our domain was constrained this value right here is going to be negative so to solve for it you want to have the negative square root