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Problem 33 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) = 3x + 5 $ , $ g(x) = x^2 + x $


$f(x)=3 x+5 ; g(x)=x^{2}+x . \quad D=\mathbb{R}$ for both $f$ and $g,$ and hence for their composites.
(a) $(f \circ g)(x)=f(g(x))=f\left(x^{2}+x\right)=3\left(x^{2}+x\right)+5=3 x^{2}+3 x+5, \quad D=\mathbb{R}$.
(b) $(g \circ f)(x)=g(f(x))=g(3 x+5)=(3 x+5)^{2}+(3 x+5)$
=9 x^{2}+30 x+25+3 x+5=9 x^{2}+33 x+30, D=\mathbb{R}.
(c) $(f \circ f)=f(f(x))=f(3 x+5)=3(3 x+5)+5=9 x+15+5=9 x+20, D=\mathbb{R}$.
$(d)(g \circ g)(x)=g(g(x))=g\left(x^{2}+x\right)=\left(x^{2}+x\right)^{2}+\left(x^{2}+x\right)$
=x^{4}+2 x^{3}+x^{2}+x^{2}+x=x^{4}+2 x^{3}+2 x^{2}+x, D=\mathbb{R}.

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Video Transcript

all right. This is a function composition problem, and we're first going to find f of G. So G is the inside function. So we're going to take X squared plus X, and we're going to substitute it in to the F function, and that's going to look like three times a quantity X squared plus X Plus five. And when we simplify that, we get three x squared plus three x plus five. Now it's fine. GF so f will be the inside function. So we'll take F and we'll substitute it in two G in both places that we see an X so that will give us three X plus five quantity squared plus three X plus five. And then we'll go ahead and simplify by squaring that quantity using the foil method. And I gives us nine X squared plus 30 X Plus 25 and then we'll add the other terms at the end and combine like terms and we get nine x squared plus 33 X plus 30. Now let's talk about domain. So the domain of F since it's a polynomial, is all real numbers, and the no domain of G is also all real numbers. It's a polynomial to, and what we got for effigy is a polynomial. So it's domain is all real numbers. We can also write that as negative Infinity to infinity and G of F is also a polynomial. So it's domain is also all real numbers negative. Infinity to infinity and now, for part C. Let's find f of F. So we're going to take function F and substituted into itself where the exits. So that's going to be three times a quantity three X plus five plus five. And we can simplify that and we get nine X plus 15 plus five. Add the like terms and we have nine X plus 20. Now it's fine G of G. So we're going to take G and substituted into itself for acts in both places. So we're going to have X squared plus X quantity squared plus X squared plus X, and we can simplify that. First will square the quantity X squared plus X using the foil method. And we get extra the fourth plus two x cubed plus X squared. And now we can add the other terms by combining like terms we end up with X to the fourth power plus two x cubed plus two X squared plus X. Now let's look at the domains So we know that the domain of F was all real numbers. It's just a polynomial, and the domain of G was all real numbers. Just a polynomial look at FF. It's also a polynomial, so it's domain is all real numbers negative. Infinity to infinity. Same thing with G. It's a polynomial. Its domain is also all real numbers.

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