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Find $ f \circ g \circ h $. $ f(x) = 3x -2 $ …

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Problem 38 Hard Difficulty

Find the function (a) $ f \circ g $, (b) $ g \circ f $, (c) $ f \circ f $, and (d) $ g \circ g $ and their domains.

$ f(x) = \dfrac{x}{1 + x} $ , $ g(x) = \sin 2x $

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Video Transcript

All right, let's find F of G. So G is the inside function. So we're putting sign of two X in place of X in the outside function. So that's going to look like sign of two X over one plus sign of two X, and we'll just leave it like that. We won't simplify it. Okay, now let's find Gff, so F is the inside function. So we're going to put the F function inside the G function. So that means we're putting X over one plus X inside the sine function where the exits. So that's going to look like the sign of two times the quantity X over one plus X. We could write that as the sign of two X over one plus X. Now let's think about the domains first of all, the domains of the individual functions. F N G. When you look at F, you notice that it's rational and it has a denominator. We can't let the denominator B zero so X cannot be negative. One g is just a sign function, and its domain will be all real numbers. Okay, For part A. The domain is going to be pretty tricky because we can't let the denominator b zero so we can't let sign of two x b negative one. So we're going to have to work that out. So we want one plus sign of two X to not be equal to zero. So what would make the sign of two X equal to negative one? Well, let's take the inverse sign of both sides. And then let's think about our unit circle. Where is the sign value equal to negative one. That would be at the point right down here on the unit circle with coordinates. Zero negative one. And that would be three pi over two radiance. Okay, so two X would be three pi over two radiance. So if we divide that by two, we get X is three pi over four radiance. However, that's not the only answer. Because we could go all the way around the circle another time and come back to that same point on. We could do that infinitely many times. So there are infinitely many points where the sign of two X is equal to negative one. So we have the three X are the three pi over two, and then once around the circle again would add two pi to that. So that would be seven pi over two. And rather than a plus sign, let me put a comma there. Seven pi over two. Once around the circle again would add another two pi, so that would be 11 pi over two, etcetera. So the way to describe that more generally would be that we have three pi over two plus two pi n where n is an integer 1234 Negative one negative to etcetera. Okay, so now let's divide this by two and we get X equals three pi over four plus pi times in Those are all the numbers that cannot be in the domain. So that was pretty complicated. Just to find the domain for one part of the function the domain is X is not equal to see if we can find it again. Three pi over four plus Pyin. Okay. What about the domain of G of F? Well, this one's a little bit simpler. We just need to look at that denominator. The denominator cannot be zero so x cannot be negative one. Other than that, you can take the sign of anything, so that's a relief. Okay, let's move on to part C. So here we want to find FF. Okay, so we have a rational function that's going to be put inside a rational function, So this is going to get pretty complicated, too. We have to take this whole thing and put it in for X there and put it in for X there. So we're going to have X over one plus X over one plus X over one plus X. You see how we put the X over one plus X in both spots? Now that's going to take some work to simplify. So here we go. By the way, I don't know if this is your experience in calculus, but this has been my experience with all the students I've taught over the years that it's usually the algebra that we get hung up on. Not so much the calculus. So here's a perfect example. What we're about to do is just a bunch of algebra, and there is so much room for error here. So here's what I would do just what I would do To simplify this, I would multiply the numerator and the denominator by a special form of one, and that would be one plus X. And I'm going to write it as one plus X over one. Because that way my enumerators lined up nicely and my denominators line up nicely on the top and the bottom. Okay, when we multiply one plus X over one by this top fraction, the one plus X is there going to cancel and what we have left is just X Now on the bottom. We're going to multiply one plus X over one by one. So nothing cancels and we have one plus X. Then we're going to multiply one plus x over one by X over one plus X and the X plus. The one plus X is on the top and bottom are going to cancel. And all we have left is X. Okay, so all of that led us to X over one plus two X. That whole messy fraction was equal to X over one plus two x. So let's go back and write that as our answer, we can find where that was X over one plus two x. Okay, Now we still want to find the domain of that. So before we get too far along, let's go ahead and find the domain. We can't have zero on the bottom, so X cannot be negative 1/2. But also remember that before that we couldn't have negative one on the bottom, the native one for X, either because we can't have zero on the bottom of our original expression. Okay, let's move on to G of G. So G is going inside of G. So sign of two X is going in for X, so that gives us the sign of two times the quantity. The sign of two X and perhaps will put an extra set of parentheses around that. And there's really nothing we can do to simplify that now, because Sign is a function with the domain of all real numbers and two exes as well. The domain of this will just be all real numbers. What a relief

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James Stewart

Calculus: Early Transcendentals

Catherine Ross

Caleb Elmore

Michael Jacobsen

Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Catherine Ross

Caleb Elmore

Michael Jacobsen

Joseph Lentino

James StewartCalculus: Early Transcendentals

Catherine Ross

Caleb Elmore

Michael Jacobsen

Joseph Lentino

James Stewart

Calculus: Early Transcendentals