Composition of Functions
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Composition of Functions

October 8, 2019


HELLO!!! Mr. Tarrou again. Same shirt, same
day. Multiple recordings. I am going to talk about composition of functions. Not just combining
them, not just adding subtracting, multiplying, or dividing but actually a composition of
functions. Composition notation is f open dot g of x, or… This is the notation my
textbook uses a lot but you may also see it as f of g of x. I like this notation better
because really more clearly explains what is happening. We are going to take function
g and put in into f. With this notation you are forced to read right to left which is
not how normally read. So, let’s talk about domain. Basically this is fancy language to
say that when you talk about composition functions you have to check the domain of the individual
pieces and the final answer. If x is not in the domain of g, if x makes g undefined it
must not be in the domain of your final answer f(g(x)), even if your final simplified answer
makes it look like it will be ok. Any x for which g of x is not in the domain of f, so
you take x and plug into the function of g and it works just fine, but then you plug
into the domain of f and that comes out to be undefined then again that must be thrown
out for the domain values of f(g(x)). This is a pretty simple idea as long as you are
comfortable with function notation. Let’s see what it looks like. f of x is going to
be two x minus three and g of x is going to be five x squared minus one. So, let’s first
find out what f of g is, or f of g of x. We want to start off with the outside function,
the overall pattern that we are following is function f. What is function f? Function
f is not 2x minus three. No, no, no. X is a place where plug something in, so I like
to say that f or the outside function is two times something minus three. I don’t care
what I am plugging in. X is some empty space where you are going to plug in something else,
that something else is g. See it is f of g of x and g of x is five x squared minus one.
So, five x squared minus one. Clean it all up, work it out, simplify, blah blah blah.
Two and five is ten x squared plus, two and negative one, is negative two minus three
is negative five. What about your domain and range? No fractions, no radicals, no fractions
no radicals, actually not radical but even roots, and no fractions or even roots so domain
is all real numbers. Lets try this the other way and see what happens. What is g of f,
well that is g of f of x. We are just going to substitute in the opposite direction. What
is g? Well I am glad you asked! It is five times something squared minus one. So it is
five times something squared minus one. What is that something you ask? It is f of x and
f of x is two x minus three. Now I am going to run out of of space so I am going to do
this off the top of my head and hopefully I am right, or I can put this off to the side
and distribute this out and make sure you get all three terms not two. This is going
to be five times four x squared minus twelve x plus nine minus one. If you you did not
get that, go off to the side of your paper, write (2x-3) twice and distribute that together
and you will go OHHH I know where that came from. Now distribute the five through and
you get 20 x squared minus 60x plus 45 minus 1. Combine like terms, clearly that is going
to be 44. And BAM!!! I got the composition working both ways. Again F has a domain of
all real numbers and g(f(x)) is defined for all real numbers. What if, I am going to take
this away so make sure you have a copy, we take a look at…or I ask you, “What is f
of g of 2?” I don’t know why I wrote that in parenthesis, I don’t need to. I just worked
out what f(g(x)) was, it was ten x squared minus five. So if I take x out and plug in
2, what am i going to do? I am going to take out the x and plug in two. Ten times two squared
minus five. That is four so ten times four is forty, and forty minus five is thirty-five.
So there you go, just take out the x and plug in the number. Badda Bing, badda boom. Ok,
any questions…didn’t think so:P Let’s take a look at…. If you had any questions, I
wouldn’t know on this video. Let’s take a look at the next example. A little bit of
fraction work. We have f of x is equal to x over x plus one and g of x is equal to three
over x. I am sure you have some questions, this stuff can get a little bit confusing.
Just ask your teacher about the function notations if you need some extra help. We want to find
f of g of x, or f(g(x)). Well again, this is your framework function f, the outside
function, or the first notation. F is something over something plus one. That something you
are plugging in is three over x. Ok. I am going to need a little bit of space so let
me take out these parenthesis. They were not doing anything except highlighting where I
am going to do my substitution. Again like the last video, if you just watched it, we
are going to have a fraction over a fraction. We need that denominator to be one fraction
and then we are going to flip it up. So we are going to get a common denominator and
that common denominator is going to be x, so this is going to be x over x. You know
multiply the top and bottom by x. So three plus x over x. Voila, three over x over three
plus x over x again. And a fraction divided by a fraction, you cannot really divide fractions
you have to multiply by the reciprocal. Do your little pig tail and flip that denominator
up. Look the x’s are going to cancel out. So our final answer is going to be three over
three plus x. Now let’s talk about domain for a second. This is going to be undefined
when x equals negative three, so x cannot be equal to negative three. But is that the
only problem with our domain? I don’t think so. G of x is undefined at zero, and the rules
say “If x is not in the domain of g, it must not be in the domain of f of g. So x can also
not be equal to zero. And negative one? Any x for which g of x is not in the domain of
f, must not be in the domain of f(g(x)). So can x be negative one…well let’s see. This
becomes undefined but negative one does not make g of x undefined which is what we are
plugging in and the final equation is not undefined at negative one, so that is going
to be ok. It does not say anything about x making f, the outside function, undefined.
It is about x making g (your inside function) undefined and x making g of x making f undefined.
So maybe some of this stuff isn’t quite so simple. But we have a domain which is restricted
at negative three and zero. Let’s look at the other direction. What is
g of f of x, or g(f(x)). That is going to
be, the framework is g so it is going to be three over something. What is that something
that we are plugging in? The something we are plugging in is x over x plus one. And
fraction over fraction again, so that is going to get flipped up. If you want to write this
because three does not look like a fraction, it is three over one times x plus one over
x. So our final answer is going to be three x plus three over x. X cannot be equal to
zero and x cannot be equal to negative on. Check the domain of the inside function, then
check the domain of the final answer. BAM!!!

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  1. I'm in my last year of highschool and i've watched your videos ever since I was in grade 9.Thank you so much for helping me through high school math! You're an awesome teacher… BAM!

  2. You are adorable and you don't look old enough to have students! I am showing this video to the kids I tutor to help them, they understood well! Thanks

  3. I'm in sophomore year in one of the toughest schools in our country, and this video still did better in making me understand the subject than my own professor. Thanks Prof RobBob!

  4. Thank god for YouTube! This was so much help, and I'm a college student. Your a lifesaver, I subscribed.

  5. Thanks for the video. I am currently a junior in HS and I want to do engineering but I am a little behind on math. I have algebra 1 and 2 and geometry finished. My algebra skills I know I still need to work on. What does composition of functions exactly benefit you for in pre calc and calculus? And what stuff in calculus is used the most from algebra?

  6. price for algebra class =$2,500. listen to professor babble = 2 hrs sat morning, 3 minutes with your video = priceless. Thanks

  7. You have taught me what my college math teacher could not accomplish in 3 Hours! I'm going to clone you and replace all math teachers with your clone. My math teacher will be shipped off to Pakistan and used as a camel for the remainder of her life as punishment for tainting the education system. Thanks for the great channel and for restoring my hope that amazing teachers exist.

  8. hey, i cant seem to get f(g(h(x)))
    or example a question asks,
    For f(x)=x^7 +3, g(x) = x-10, and h (x) = sqrt(x), find f of g of h.

  9. Oh, wow! Did I just understand this lesson???? I've been struggling with this concept. Won't blame my teacher, it's just me having a hard time. Since I finally understood all the steps, now I'll just watch it a few more times, until I can replicate it myself. I have a test next week, so PADIUSH (thank you in Nahuat Pipil) !!!

  10. I've been back in the zone with my favourite maths teacher …..thought I'd go back and check this out after trying to tackle chain rule …think I may have been punching above my weight class haha

  11. I'm 4 hours away from an exam that I was a 100% sure that I was going to bomb. Thank you for being a saving grace! I actually feel like I stand a chance now!

  12. I am a year 10 student and im kind of confused on a question that you solved, at 4:25 how did you get 5(4x^2 -12x+9)-1, is it not meant to be 5(4x^2 +12x+9)-1, im saying this because when -3 is squared it becomes a positve therfore it should be a +12x and not -12x

  13. Im in a college algebra class right now.
    I havent had a math class in over 10 years, and the teacher I have, moves too fast for me to keep up.
    Im not doing great in the class, but I started to understand a few things with this channel.
    Without your videos I would have already dropped the class.

  14. I graduated high 16 yrs ago and I'm so lost trying to help my daughter with her homework. She doesn't even have a textbook!! I've spent a ton of time on Google and YouTube trying to figure this out. Your video has been the easiest to understand, unfortunately, I'm still gonna need to watch this about a ten more times.

  15. You're such a amazing teacher! Can you answer this question, please?
    Determine whether the statements are True or False. Justify your answer with an explanation.
    If you are given two functions f(x) and g(x), you can calculate (f β—‹g)(x) if and only if the range of g is a subset of the domain of f.

  16. what's the problem with negetives.does the function always hold positive value?even jf the denominator is negative what's the prob with that I wonder

  17. You are my favourate teacher sir………….. I have recommended many of my friends to watch your videos even they found it simplified and enjoying…… You are fantastic sir…….

  18. Thank you.
    My AS level math exam is less than 24 hours away, and I was panicking over functions till now.
    You just helped inhibit another breakdown. Thank you.

  19. I have a huge math final this week and I just stumbled across this video and it helped me tremendously. I understand functions now! thanks so much!

  20. Im having trouble doing comp of function with functions such as:

    f(x)= 1/6x-1 and g(x) = 1/x^2 for instance. Trying to find fβ—¦g(x) I set it all up ok, but have trouble simplifying? Havent had much luck finding videos going over these types of problems.

  21. My College Algebra teacher attaches your videos to her emails on each chapter as extra help, youre funny and make the math feel easy. Thank you

  22. Just a friendly observation regarding the domain . At about 8:30 you ask "can x be -1?" I believe we should be looking for an x-value that would make g(x)=-1, since this output will be used as an input by f(x). Therefore , the x-value which makes g(x)=-1 is actually x=-3, which you had already excluded.

  23. My teacher plays u all the time. I love your entrance. Starts the lesson right. U made me love math for the first time.

  24. I have a question

    How to solve this question.

    For which of the following functions f(x)f(x) is the relation f(f(x))=f(f(f(f(x))))f(f(x))=f(f(f(f(x)))) NOT true for at least some values of xx not equal to zero?
    f(x)=βˆ’|x|f(x)=βˆ’|x|f(x)=2βˆ’xf(x)=2βˆ’xf(x)=3xf(x)=3xf(x)=4xf(x)=4xf(x)=5f(x)=

  25. Domain restrictions on the input function and the composition.
    (3/x) / ((3/x) + (1/1)) => (3/x) / ((3/x)+(x/x)) => (3/x) / (3+x/x) =>
    (3/x) * (x/3+x) => (3x) / (3x+x^2) = Factor out an x in denominator => 3(x) / (x)(3+x) ==X bound by multiplication okay to cancel ==> 3/(3+x)
    |||Domain restrictions on the input function|||
    [g(x) = 3/x] is the the input function. g(x) is a rational function. rational functions cannot have 0 in the denominator. set the denominator = 0. [[x = 0]] remove 0 from the domain.
    ||The composition||
    3/(3+x)==> same rules apply ==> [[3+x=0]] ===>[[x= -3]] remove -3 from the domain.
    (-oo, -3) U (-3,0) U (0,+oo)

  26. i've been consistently horrible at algebra my entire high school career, at one point failing a semester of it. now i've finally made it to my final exam in college algebra and this just saved my life. thanks dude!!

  27. Love your videos, I always come home and watch the ones pertaining to whatever the lecture was that day to ACTUALLY understand so THANK YOU!! I'm wondering if you've created a video specifically to the decomposition of functions?? I've been through your website and can't seem to find one. for some reason this topic has boggled me.

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