one we're going to go back into functions we're going to talk about a composite function composite function you have done this before just never given the official title if you were doing orders of operation you sometimes ran into a case where you had inner parentheses outer parentheses or brackets and you were always told you had to simplify the inner one first then work yourself out to the outer grouping symbol and that's exactly what's Happening Here a composite function is denoted by f with not an O it's just a circle if you will G which reads F composed with G or F of G is defined by F composed with G of X or I prefer this notation F of G of X the first function letter given is always the outer function the last letter is always the inner function name now it's important to understand that the domain of f of G or F composed with G is the set of all numbers X in the domain of the inner function G such that g of X which is your y-coordinates is the domain of F and that's what makes it seem weird because you're saying y-coordinates are going to be the domain well the y-coordinates of the G are the do range of the G function but in terms of the F function they're the domain so a flow chart looks like this I start with my X put it into my inner function I simplify it and the output is the y-coordinate of the G function but then that value is going to become the input or the x value going into the F function which gives us the output of the y coordinate of Y of the F function excuse me so if we look at a numerical example here we have H ofx is 2x^2 minus 3 J of X is 4X find H of J of one so this is better written as H parentheses J parentheses of one so one goes into the inner function J first that's 4 * 1 but it stays in the parentheses of the H function so 4 * 1 is 4 and now four goes into the H function so now we take this four which again was the y coordinate of the J function now that four becomes the x coordinate of the H function 4^ 2ar is 16 * 2 is 32 - 3 is 29 so h of J of one is equal to 29 and it's important to realize that if you switch the numbers or excuse me the letters around but keep the same number you do not always get the same answer so J of H of one one goes into the H function first so we'll square the one that's a one time the two that's a two minus the three is a negative 1 so the y coordinate of the H function is1 when X X was a positive one then we take that negative 1 we put it into the J function we get four * 1 is4 so J of H of 1 is equal to4 as you can see we don't get the same values we can also have a function of itself h of H of -2 -2 goes into the H function for that X -22 is 4 4 * 2 is 8 - 3 is 5 now that five has to go back into the H function one more time so 2 * 5^ 2 25 * 2 is 50 minus the 3 is 47 so h of H of -2 is 47 J of J of1 it's going to look like this nea1 goes into the J function 4 * 1 is -4 then take4 put it back in the J function 4 * 4 -16 so J of J of1 is-16 now what happens when they don't give us numbers don't panic we saw this before if they give you a numerical value you come back with a numerical value if they give you a variable expression you come back with a variable expression using the same value so h of J of X written like this means go find the J of X function 4X and substitute it into the H function so this 4X is going to go into that X of the H function so 4X all that squared is 16 x^2 * 2 is 32 x^2 minus 3 that's as far as I can simplify it therefore I am done h of J of X is 32 x^2 - 3 J of H ofx now we're going to take the H function that's the inner function 2x^2 minus 3 put that in for the X of the J function so 4 * the 2x^2 -3 we'll do a distributive property 4 * 2x^2 is 8 x^2 4 * -3 is - - 12 you are now done there's nothing to combine so we are done another problem that you'll look at is they'll say find the individual functions f and g such that F of G is equal to Capital H and they give you Capital H and it's let's say it's equal to the square root of x Cub + 5 so by notation f of G of X is equal to Capital H so looking at this the first thing I ask myself is who's the inner function oh that would be the g ofx g of X is inside the parentheses of the F function then I say okay let's go to Capital H who is inside of a grouping symbol here well that'd be X Cub + 5 okay so G of X is equal to X Cub + 5 I go up to where Capital H is I take out the X Cub + 5 replace it with an X and that's my f ofx function so if I take g ofx x Cub + 5 and I substitute it into that X I get Capital H another way that you can find f and g is to go after the variables with powers so I look at Capital H again I say okay where is my X with the exponent there it is that's my inner function G of X take that out put an X back in that's my f ofx the S otk of x + 5 so again if you take the G function and you put it into the x of the F function I get the square OT of X Cub + 5 which is capital H now you only have to find one set you don't have to find both down below same concept now my Capital H is x^2 + 1 all to the 5th power G of X again is my inner function so who has ever inside my grouping symbol parentheses x^2 + 1 that's my G function take out all of x^2 + 1 replace it with an X you get X in the parentheses to the 5th power which simplifies to x to the 5th in second method find the X to the power that's your G function take it out put an X back in x + one to the 5th power that's your F ofx function you can use either technique which best suits you now our composite function is going to lead us to what we call one: one functions and inverse functions a function is one to one if any two different inputs in the domain that means two different X coord coordinates correspond to two different outputs y-coordinates in the range that means your X's can't repeat and your y's can't repeat so if we wrote this out in function notation F ofx sub1 cannot equal f of x sub 2 if this happens you have what is known as a one toone function all right now when we look at one to one functions as a graph we remember there is the vertical line test okay here's X2 it's a parabola it passes the vertical line test here's X cubed passes the vertical line test these are both functions but in order to be a one to one function now they also have to pass a horizontal line test so if I put horizontal lines on my graphs okay here we got a problem now the blue lines touch the graph in two locations this is not one: one it's still a function just not a one to one function over an X cubed all the blue lines touch only once therefore that's a onetoone function so remember our rule of what's supposed to take place here in order to be a one toone function the X has to go into it's a composite function X coordinate goes in for the Y so let's say x is 2 you get a four X2 I take -2 put it in for X that's positive4 2^ 2ar 4 -2 being squared is four so my problem is if I start at four how do I know where to get back to that's the key and that's the concept of inverse functions this little NE one that's looks like it's in the exponent position it's not an exponent you read this as the inverse of f ofx f with that negative 1 that's inverse of F ofx this definition says it's a relation such that every x coordinate in the domain corresponds to exactly one y-coordinate okay stop that's the definition of a function a relation such that every x coordinate in the domain corresponds to exactly one y-coordinate in the range and every y-coordinate in the range corresponds to exactly one x coordinate in the domain remember with our original function we said X's can't repeat but the Y's could now if I switch x coordinate y-coordinate y-coordinate x coordinate now nobody gets to repeat X's can't repeat y's can't repeat so to be an inverse function the number one thing I'm I want you to remember is switch all X and Y Concepts I don't care if we're talking points we're talking on a graph we're talking variable expressions and equations we're going to switch X and y's so the domain of F this is the X's they will be the Y's of the inverse function the range of f that's the Y's of the original function is the domain of the inverse function those are your X's so chart looks like this if I have a function my X goes into the F ofx function and out comes a y now you take the y-coordinate you put it into the inverse function I have to come back to where I started X this concept gives us the inverse identity the inverse of f of f ofx has to equal x f of the inverse of X has to equal x so this one here F of the inverse of X that was this path that I have with the arrows I come back to X if I do this oh excuse me that this one here that function first to the Y then the Y through the inverse that's this one if I look here on the right starting with the inverse function starting with y or again your X I would go here to here then into the F function I'm right back to where I started so that's the key now if you were asked to verify functions are inverses suppose you have 2X + 3 and G ofx = 12 * x - 3 well let's color code this let the F function be in blue G function in red and you have to show one of those two identities it doesn't have to be both you pick the one you want to work with so if I opted to do F of G of x g is the in function f is the outer function you take all of G put it into the blue X that's what I have right here and now you start simplifying 2 * the halftimes the binomial the two gets multiplied to the half that cancels each other out x - 3 + 3 - 3 + they cancel out F of G ofx is equal to X that's good enough for me they are inverse functions or if you decide to go g of F ofx and again you only have to do one one half is the G function xus 3 you take the blue 2X + 3 put it in for the Red X okay start to simplify nothing's being multiplied to the blue parentheses so therefore plus three minus three cancel out 12 * 2 cancels out um back to X let's verify if these are inverse functions f ofx is 2 * the Cub x - 5 G of X is in parentheses x over2 Cub + 5 you pick which one you want to do all right so pause the video either do F of G of X or G of F ofx and see if you can get it to equal x if you do they're inverses so pause it and when you got your answer turn it back on all right welcome back here we go we're going to take F of G of X we're going to take all of this put it in for that X inside the square root there it is really don't need those parentheses because nothing's being multiplied to it so plus 5 - 5 they cancel then we have 2 * the cube root of something cubed those cancel it's 2 * X over2 the twos cancel I'm left with X if you tried G of f ofx then we take all of the F ofx put it in the numerator of this fraction here well since I'm in this fraction two and the two on the top and the bottom they cancel then I'm left with the cubot of x - 5 cubed that cancels then I'm left with x - 5 + 5 those cancel out leaving me the X they're inverse functions if you're given a function that is a set of points switch all your X and Y Concepts if they ask you to find the inverse the inverse of f ofx means the x coordinate becomes the y-coordinate the y-coordinate becomes the x coordinate so this first point 27 comma -3 -2 comma 8 becomes 8 comma -2 negative 1 negative 1 I switched them did you see that still negative 1 negative 1 0 0 switched them one one switched them 28 switch them 82 327 273 close off the set with the brace there is the inverse function we can do the same thing with a graph find your starting points find any turning points write them down 1 2 3 four to the left4 -3 -20 and 42 write them in order as you read it from left to right then just like we did with these points switch them4 neg3 is now3 ne4 -2 0 0 -2 42 now two 4 plot the points in order -34 0 -2 and then finally 2 4 now you saw that line appear and disappear this is the line Y equals X some authors refer it to the inverse line you definitely can see it right here negative 1 negative 1 0 0 one one all those points are on that green line if you take your blue function and you flip it over that green line yals X you get the inverse graph if I have a function that's an equation F ofx = 3x + 5 again I have to switch X and Y Concepts so F ofx I take it out I rewrite it with a Y that's worth a point rewrite it with the Y to get a point step two switch X and Y switched that's worth a point step three solve for y get y by itself everything that you UND to do worth a point subtract over the five that's a point divide over the three that's a point one excuse me one two 3 4 and the last step take out the Y replace it with your inverse notation that's a point now folks be very careful remember it's got to match the letter here you notice there's a lot of f of XS going on here don't get brainwashed into everything is f ofx next example oh here we go g of X is 34 xus one okay follow the steps take out the G of X put in a y That's a point physically switch the X and Y around x = 34 Yus 1 solve it for y add over the one to get rid of the multiplication of three fors multiply by its reciprocal 4/3 put parentheses around the left side 4/3 time the binomial x + one you do not have to distribute it just leave it you got y by itself now put in the inverse notation for the G function the inverse of G ofx is 4/3 times the parentheses x + one all right let's do H ofx it's the cube root of x + 6 - 4 follow your steps take out the H of X put in a y physically switch X and Y two of the easiest points you'll ever earn now we need to get the cube root by itself to solve for y so we add over the four get rid of the cube root symbol by cubing it now I have y + 6 get y by itself subtract the six leave it on the outside did you notice there's no arithmetic being done whatsoever there's my y replace it now with h inverse of X there's my inverse function now delicate matter let me pause this for a second go back to this graph right here remember how we said x s was not one: one it has to be one: one we're going to figure out a way to get it to pass the horizontal line test that's our goal all right so here we go find the inverse of f ofx which is the square root of x + 5 I put in the Y for f ofx steps are good switch the X and Y good add over the three even better Square both sides x + 3^ 2 square root gone y + 5 subtract over the five good put in my inverse notation good I do all that and then all of a sudden it says minus two I lost two points here's the problem you wrote down a parabola which we have not proven it to be a one: one function yet so I went to my calculator I typed in y1 the of x + 5us 3 I typed in my inverse function and I typed in yals X here's what my graph gave me so I blew this up so you can see it here that is the S < TK of x + 5us 3 here's my yals X line now remember we said a true inverse graph flips over that line this point lands right at the vertex of my Parabola but my y = the binomial x + 3^2 - 5 gives me the whole binomial I just want half of it so what I got to do is find my vertex oh yeah that's right this is in vertex form my vertex is at -35 so at this x coordinate I want my x's to be -3 or bigger once you attach this domain restriction you don't lose the two points now I forced that red graph on your screen is the binomial x + 3^ 2us 5 with a domain restriction of bracket -3 comma infinity or X is greater than or equal to -3 that means this part of the graph here is gone and now it passes the vertical and horizontal line test so now this is one to one you have to add the domain restriction it's a must so here's the rule if you're given y equals a positive square root you want the right side of the problem your domain will be H comma Infinity I like using inequalities because it says X greater than or equal to H which which is the x coordinate of your vertex I want to go to the right of that x coordinate right side of the parabola if y equals a negative square root you want the left side of the parabola which is the domain negative Infinity comma H bracket but what I have found when people do interval notation they sometimes get the H and the negative Infinity switched around so safest thing to do write it as an inequality I want the X's that are the left of the x coordinate of my vertex then I get the left side all right memorize that rule find the inverse of f ofx = -2 * x + 1^2 - 7 wait a minute that's a parabola it's not one: one oh yes it is I attached a domain restriction x + one H is ne1 that's where this comes from so then this is perfectly legal to find the inverse now because I want the right side of the parabola it's going to the right then I'm going to be looking for a positive square root symbol somewhere in my work so I put in my y for f ofx I switch my X and Y again the two easiest points you'll ever get add over the seven divide by -2 just leave it as a fraction don't do any work with it get rid of the Power of Two by the square root aha Here Comes Your plus or minus we want it to be positive so leave this positive subtract over the one there is my inverse function so again one point one. one23 three four rewrite it with your inverse notation five oh that's about seven points right there all right and that wraps up 12.1