Problem

Find the function (a) $ f \circ g $, (b) $ g \cir…

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Problem 37 Hard Difficulty

Find the functions (a) $f \circ g,$ (b) $g \circ f,$ (c) $f \circ f,$ and (d) $g \circ g$ and their domains.
$$
f(x)=x+\frac{1}{x}, \quad g(x)=\frac{x+1}{x+2}
$$

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Video Transcript

All right. Time to have fun with functions. Okay, first thing we're gonna do, let's take a look at F F G. Uh or you can write an F of G of X like this. Okay, so that means wherever I normally would have an X for F. It's now G F X. So that gives me G of X plus one over G of X. Now let's substitute in G. Of X. So that's X plus one over X plus two. And then I take the reciprocal. So I get plus X plus two over X plus one. You could do a common denominator and turn it into a different form. But this is the solution. So we could stop here. The main thing we're looking at is the domain values of acts that we don't want. We don't want the denominator to go to zero. So far. Domain We do not want it's all real except X cannot be -2 And X cannot be -1. Right? Because those are the two values that make the denominator go to zero. So those are not good. Alright, excellent. So that was our first one. Let's do our second one. Our 2nd 1 is G up F of X. So G of F of X. That means wherever jihad and access. Now ffx. So that's the same as F of X plus one Over FFX Plus two. So we'll go ahead and substitute an X plus one over X plus one and X plus one over X plus two. A nice way to clean up when you have fraction within a fraction is to go ahead and multiply everything by the denominator, which is X. So that will give us X squared plus one plus X over X squared plus one plus two X. Now I can only do that trick of multiplying everything by X. If X is not equal to zero. Which by the way, we don't want anyway because we see it in the denominator. So for our domain will take note already, X cannot be zero. Let's keep going. We also now have a more complicated expression. We also don't want bad expression to equal zero. So let's see what that is. If I have expert plus two x Plus one, we don't want that to be zero. Well, I can factor that. That's X plus one. Uh quantity squared cannot be zero. Therefore X should not be minus one. So the domain also includes acts should not be -1. Okay, so that's our domain name for our part. B Let's see if I have room for the 3rd 1. If not, I will I'll have to raise the boards. Let's try let's try at least one more in here. Okay, 3rd 1 is F of F. So that's F F F F X. So I replace every X in the original ffx by guess what itself. So that means I'm gonna have F of X Plus one over FFX. Well, what is F of X? F of X is X plus one over X. Uh So then that all that goes here on the bottom. So like before I want to clean up that fraction, I'm going to multiply top and bottom in the second fraction. Bye. By acts. So that gives me plus X over X squared plus one. We already know if our domain, I'll kind of keep track here and I think I will have to raise the board before last one. Woo. Okay, so our domain for this one X cannot be zero. We know that already. Um And then uh I accidentally wrote an S instead of an X. Here. Uh and then okay, that the new denominator cannot go to um zero ever because exports positive. And then I add once, it's always going to be positive. I think we've covered it. Also, our only limit on the domain is X cannot be zero. So give me a second. I'll clear the board and we'll finish up with our last one which is G. F. G. Okay, so our last one is find G. G. G. Okay, let's fix that. G. F. G. So that is G of G of X. So again, that means wherever we have X, it's now G of X. So we have G of X plus one over G. Of X plus two. But G of X. Is X plus one Over X-plus two. So we have all this a lot in there. A lot to put in and we'll see what happens. Okay, so if I want to clean this up I'm going to multiply everything. Uh Top and bottom by X plus two to take that denominator. It cleans up fraction within a fraction. Domain wise we already know X cannot be minus two. We don't want that denominated here to be zero. If I clean it up by multiplying top and bottom by X plus two I get X plus one Plus X-plus two over X plus one plus twice X plus two which is two X plus four. So we look to make sure this bottom cannot go to zero by the way. Just clean up a little bit. It's two X plus three Over three X Plus five. Alright so we do not want this to nominate her to go to zero. So we say three X plus five cannot be allowed to be zero. So three x cannot be -5 x. Cannot be minus five thirds. So that adds to our bad access in the domain. Otherwise we're good for all other values and we've done it. Yay. Alright. Have a great rest of your day

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James Stewart

Calculus: Early Transcendentals

Caleb Elmore

Kristen Karbon

Michael Jacobsen

Joseph Lentino

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

Calculus: Early Transcendentals

Caleb Elmore

Kristen Karbon

Michael Jacobsen

Joseph Lentino

James StewartCalculus: Early Transcendentals

Caleb Elmore

Kristen Karbon

Michael Jacobsen

Joseph Lentino

James Stewart

Calculus: Early Transcendentals