Domain and Range of Composite Functions Part One 

Wow, never heard it put like that before- thank you sir!

So glad you enjoyed it- I agree : )

Thanks so much! I wish this video got more love, it's one of my favorites. Glad it helped you 😄

Well, it's a tricky topic, that's why I made the video. Sometimes a new idea doesn't click the first time. You may want to rewatch the video, or look for another video that presents it in a different way.

Glad you found it helpful- and way to go for modeling life-long learning strategies for your son 😀

@@RealMikeDobbs Oh, I sent him the video link, since I found the book's explanation inadequate.

That’s great! Hope the test goes well 😀

Sorry you find it confusing. I don't believe I ever refer to the composition as g(x); rather, the point of the entire video is that compositions are made up of individual functions (in this case g(x) and f(x)). If you think of the behavior of those individual functions separately you will be able to better understand the behavior of the composition. I have found this incredibly useful for my own understanding of compositions, but not everything works as well for everyone. When you compose functions, the individual functions are still there, you can't lose sight of that. The inside function takes in the input, and generates an output, same as always. That output then becomes the input for the next function, and so on until you reach the outermost function. This is precisely why we have to restrict the domains of composed functions in ways that are not obvious if you ONLY look at the composition.

Thanks very much! Glad it helped 😀

Ahhh, I see what you mean. If you watch my video I gave the correct answer verbally, and wrote it down correctly when I said it, but it changes to an incorrect answer at about 13:20 in the video. I must written down the wrong number when I first recorded it, and then "fixed" it when I did the editing. Apparently, I didn't notice that the correct version blinked off the screen at 13:20 and it reverted to the incorrect domain I initially wrote. Excellent catch! To clarify, the correct answer is the one I give initially. Domain is reals, take out -3 and 1, and Range is reals take out -2 and 0. I hope that helps.

@@RealMikeDobbs thankyou for replying , but if we take compositon i.e. h(x)=fog(x) i.e. if we substitute (4/x+3) in f(x) then h(x)= (2x+6)/(1-x) the function formed by composition, domain and range for h(x) is reals take out 1 and reals take out -2 . this domain and range is different than what we find out by finding domain and range for each function and as per given by your method ? SIR is there any mistake in this????? plz help sir

OK, thanks for clarifying. I think you're making an error in the composition. You seem to be multiplying the denominator of g(x) by the numerator of f(x). If we take the composition h(x) = f(g(x)), then we are replacing the input of f(x) with the expression (4/(x+3)). Doing so yields h(x) = 2/((4/(x+3)) -1). In other words, the numerator of the composition is just 2, and the denominator is 4/(x+3) - 1. From that equation you can only see that you need to remove -3 from the domain. That's the reason I consider both functions that make up the composition, so we can see that we also need to eliminate 1 from the domain because g(1) = 1, and f(1) undefined. Hopefully that makes sense? I am not sure why you feel you need to remove -2 from the domain, but if you explain your reasoning I'll try to help. Here's a link to a graph of the composition for reference: www.desmos.com/calculator/acfy8kefgo. (note that there is a hole in the graph at (-3, 0), even though Desmos shows the graph being defined at that point)

@@RealMikeDobbs Thankyou so much sir for taking your time, if we simplify the function h(x) = 2/((4/(x+3)) -1) , we get h(x)=(2x+6)/(1-x) , by looking at this function it seems we can't put x=1 so the domain must be all real except 1 but actually it is all real except (-3,1). Is this happening due to simplifying of the function. SIR What is meant by hole in the graph? Also when we look at the graph we see there is (x,y) coordinate corresponding to (-3,0) but as we discussed we can't put x=-3 how the graph is plotting this point. Same in the range we can't get (-2) . BTW thankyou so much for sending the link to graph it helps me understanding it better.

OK yes, I'm sorry I didn't realize the function simplifies to what you wrote, now I see what you are doing. The simplification is the issue. Let's call the simplified version q(x), and the unsimplified version h(x). So, q(x) = (2x+6)/(1-x), and h(x). = 2/((4/(x+3))-1. Here's the thing about simplification, anytime you divide by something with a variable you are loosing solutions. Before you simplify, it is clear that in h(x) you cannot let x equal -3. When you "simplify" the complex fraction, what you get is a fraction that is equivalent to the original, ONLY IF x does not equal -3. That information is lost when you simplify algebraically, and that is the reason I look at the functions individually when doing compositions. So it is in fact not true that h(x) = q(x). What is true is that h(x) = q(x) for all values of x EXCEPT -3. At x = -3, they are both undefined, although q(x) APPEARS to be fine at (-3,0). That is what is mean by a hole in the graph. h(x) and q(x) are identical for all values except x=-3, so at the point (-3,0) q(x) is actually undefined. Since the "size" of the undefined portion of the graph is only a single point, it is infinitely small and takes up no space. That is why the graph seems to be complete at that point. As to why Desmos plots the point if you click on it, this is a common limitation of graphing programs due to the way they actually calculate the function values. A hole in the graph is a single point where an otherwise perfectly fine function in undefined. Does that makes sense?

Thanks so much! I’m really glad you found it helpful 👍

Thanks so much! 😀

I agree some notations can be confusing. The input for the composite function is x. Let's say we are looking at g(f(x)). You can clearly see that the input of this function is "x". Where it gets confusing is that the composite function is really two different functions, with one "following" the other. So, f "gets" the original input (in this case x), and transforms it to a (probably) different value f(x). Then, f "hands this value" over to g, which transforms it again. It's this handing over of the output from one function to another that is the key to understanding compositions. I hope that helps some, it can be a tricky topic.

You're very welcome!

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@@RealMikeDobbs 5:40