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

$f(x)=x+\frac{1}{x}, \quad D=\{x | x \neq 0\} ; \quad g(x)=\frac{x+1}{x+2}, \quad D=\{x | x \neq-2\}$(a) $(f \circ g)(x)=f(g(x))=f\left(\frac{x+1}{x+2}\right)=\frac{x+1}{x+2}+\frac{1}{\frac{x+1}{x+2}}=\frac{x+1}{x+2}+\frac{x+2}{x+1}$$=\frac{(x+1)(x+1)+(x+2)(x+2)}{(x+2)(x+1)}=\frac{\left(x^{2}+2 x+1\right)+\left(x^{2}+4 x+4\right)}{(x+2)(x+1)}=\frac{2 x^{2}+6 x+5}{(x+2)(x+1)}$since $g(x)$ is not defined for $x=-2$ and $f(g(x))$ is not defined for $x=-2$ and $x=-1$,the domain of $(f \circ g)(x)$ is $D=\{x | x \neq-2,-1\}$.(b) $(g \circ f)(x)=g(f(x))=g\left(x+\frac{1}{x}\right)=\frac{\left(x+\frac{1}{x}\right)+1}{\left(x+\frac{1}{x}\right)+2}=\frac{\frac{x^{2}+1+x}{x}}{\frac{x^{2}+1+2 x}{x}}=\frac{x^{2}+x+1}{x^{2}+2 x+1}=\frac{x^{2}+x+1}{(x+1)^{2}}$since $f(x)$ is not defined for $x=0$ and $g(f(x))$ is not defined for $x=-1$, the domain of $(g \circ f)(x)$ is $D=\{x | x \neq-1,0\}$.(c) $(f \circ f)(x)=f(f(x))=f\left(x+\frac{1}{x}\right)=\left(x+\frac{1}{x}\right)+\frac{1}{x+\frac{1}{x}}=x+\frac{1}{x}+\frac{1}{\frac{x^{2}+1}{x}}=x+\frac{1}{x}+\frac{x}{x^{2}+1}$$=\frac{x(x)\left(x^{2}+1\right)+1\left(x^{2}+1\right)+x(x)}{x\left(x^{2}+1\right)}=\frac{x^{4}+x^{2}+x^{2}+1+x^{2}}{x\left(x^{2}+1\right)}$$=\frac{x^{4}+3 x^{2}+1}{x\left(x^{2}+1\right)}, \quad D=\{x | x \neq 0\}$$(d)(g \circ g)(x)=g(g(x))=g\left(\frac{x+1}{x+2}\right)=\frac{\frac{x+1}{x+2}+1}{\frac{x+1}{x+2}+2}=\frac{\frac{x+1+1(x+2)}{x+2}}{\frac{x+1+2(x+2)}{x+2}}=\frac{x+1+x+2}{x+1+2 x+4}=\frac{2 x+3}{3 x+5}$since $g(x)$ is not defined for $x=-2$ and $g(g(x))$ is not defined for $x=-\frac{n}{3}$, the domain of $(g \circ g)(x)$ is $D=\left\{x | x \neq-2,-\frac{5}{3}\right\}$.

10:30

Jeffrey P.

Calculus 1 / AB

Calculus 2 / BC

Calculus 3

Chapter 1

Functions and Models

Section 3

New Functions from Old Functions

Functions

Integration Techniques

Partial Derivatives

Functions of Several Variables

Missouri State University

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All right, so let's find f of G. So we're going to put the G function inside the F function in both places where we see the X now. One thing to help us make this a little simpler is to recognize that when we see one over X, it means reciprocal. Okay, so if we put G inside of X G inside of F, we get X Plus one over X plus two plus and then for one over X. Just think of that as the reciprocal. So that's going to be X plus two over X plus one. Now perhaps we want to simplify that and add those fractions together. If so, it's going to look like this. So we have X Plus one over X plus two plus X plus two over X plus one. We need to get a common denominator, so we'll multiply the first fraction by X plus one over X plus one, and we multiply the second fraction by X plus two over X plus two. So our common denominator is X plus one times X plus two, and then when we simplify and add the enumerators together, we're going to go ahead and add them and do the foil process at the same time. So when we multiply the 1st 2 together we get X squared plus two x plus one. And then when we multiply the 2nd 2 together we get X squared plus four x plus four. So all together, that's going to be two x squared plus six x plus five over X plus one times X plus two. So we can write that down as our answer. Okay, and then we'll move on to part B. So we're finding gff. So for GFF goes on the inside. So we're going to take X Plus one over X and add one to it and put that over X plus one over X and add to to it. So that's going to require some simplifying as well. On what we could do with this is multiply the numerator by X and multiply the denominator by X. And that's going to give us X squared plus one plus X over X squared plus one plus two X. Now we might want to write those terms in descending powers of X, in which case we would have X squared plus X plus one over X squared plus two X plus one. Now let's talk about the domains of these. So first of all, sometimes it helps to find the domain of the original functions. Sometimes that's not going to come into play. But if you look at the original function F notice that you have X in a denominator and you can't have a denominator of zero, so X cannot be zero. And then, if you look at the original G function, notice that you have X Plus two in the denominator and the denominator cannot be zero. So X cannot be negative, too. Now, if we think about that same sort of thinking for part, a look at the denominator, you can't have X equals negative, too. That would make the denominator zero, and you can't have X equals negative one that would make the denominator zero. How about the denominator? For Part B X squared plus two x plus one actually factors into X plus one times X plus one so you can't have negative one in your domain, but also because the original problem had a one over accident and you can't have zero in that you have to also say X cannot be zero. Okay, let's check out part C. So here we're finding f of F so f goes inside of death. So this whole thing goes in for the first X, and it goes in for the second X. But remember, the second x means reciprocal that may or may not help us. Okay, so the whole thing goes in for the first x So x plus one over X. Now the whole thing goes in for the second X who hit one over X plus one over X. Oh, boy. OK, so let's work on simplifying this part of it. One over X plus one over X. What I would do to that is I would multiply the top and bottom by X. That way we won't have a fraction inside a fraction anymore. So if x over x squared plus one, So if we put that in here we have X plus one over X plus X over X squared plus one. I think that's sufficient. I'm not going to worry about combining all those fractions together. I think that just takes it way too far. Now it's fine. G of g. All right, So we're plugging G and a G. So we're going to put the whole fraction and for X here and in for X here. So this is another ah, very complicated fraction. So we have X plus one over X plus two plus one as the top and X plus one over X plus two plus two as the bottom. So we're going to simplify that X plus one over X plus two plus one over X plus one over X plus two plus two. Okay, what do you think you would want to use for your special form of one? To multiply by? I would want to multiply by X plus two over X plus two to clear out the denominators of X plus two. I'm going to put it over one just so that my enumerators and denominators line up nicely as I go through this work and you can think of the number one as 1/1. And you could think of the number two as to over one. So first we're going to multiply this whole fraction by the 1st 1 and the X Plus two cancels on the top in the bottom On what we have left is X plus one. Then we're going to multiply this whole fraction by 1/1 and nothing cancels and we have X plus two now for the bottom. We're going to multiply this whole fraction by the first fraction and the X plus two councils on the top of the bottom and all. We have his X Plus one. And then we're going to multiply that hole fraction by two, and nothing cancels and we have two times the quantity X plus two. So that gives us for the top two X Plus three. And for the bottom, distribute the to and you get three X Plus five two X plus 3/3 X plus five. All right, so that is our G of G. Now let's think about the domains. So looking at FF, you can't have zero on the bottom, so X cannot be zero. You also don't want this denominator to be zero, but there's no way for that to be zero, because X squared has to be either zero or positive. So X squared plus one is going to be already either one or pot or greater. Now, how about the domain of G GMG. Okay, so, looking at the final answer, we can't have zero on the bottom, so X cannot be negative 5/3. But looking at the intermediate steps, we also could not have to. We could not have X plus two b zero so X could not be negative, too.

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