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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) = \sqrt {x + 1} $ , $ g(x) = 4x - 3 $

$f(x)=\sqrt{x+1}, D=\{x | x \geq-1\} ; g(x)=4 x-3, D=\mathbb{R}$.

(a) $(f \circ g)(x)=f(g(x))=f(4 x-3)=\sqrt{(4 x-3)+1}=\sqrt{4 x-2}$

The domain of $f \circ g$ is $\{x | 4 x-3 \geq-1\}=\{x | 4 x \geq 2\}=\left\{x | x \geq \frac{1}{2}\right\}=\left[\frac{1}{2}, \infty\right)$.

(b) $(g \circ f)(x)=g(f(x))=g(\sqrt{x+1})=4 \sqrt{x+1}-3$

The domain of $g \circ f$ is $\{x | x \text { is in the domain of } f \text { and } f(x) \text { is in the domain of } g\} .$ This is the domain of $f$, that is, $\{x | x+1 \geq 0\}=\{x | x \geq-1\}=[-1, \infty)$.

(c) $(f \circ f)(x)=f(f(x))=f(\sqrt{x+1})=\sqrt{\sqrt{x+1}+1}$

For the domain, we need $x+1 \geq 0,$ which is equivalent to $x \geq-1,$ and $\sqrt{x+1} \geq-1,$ which is true for all real values of $x$, Thus, the domain of $f \circ f$ is $[-1, \infty)$.

(d) $(g \circ g)(x)=g(g(x))=g(4 x-3)=4(4 x-3)-3=16 x-12-3=16 x-15, \quad D=\mathbb{R}$.

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All right, let's start by finding f of G. So we're going to put the G function inside the F function. So for X minus three goes in place of X in the F function and we end up with the square root of four X minus three plus one, and we can simplify that, and that would be the square root of four X minus two. Now let's find G of F. So we're putting the F function inside the G function. So this whole thing, this whole square root goes inside G where the X is, and that gives us four times the square root of X plus one minus three. Nothing we can really do to simplify that. Now let's look at the domains. So first, let's think about the domain of F in the domain of G. So for F, you have to have the square root of a non negative number, so X plus one has to be greater than or equal to zero. So X has to be greater than or equal to negative one, and G is just a polynomial. So it's domain is all real numbers now. Those individual domains for F and G in some cases will help us with the composition domains and in some cases they will not. For example, for Part B. They will help us because we have the same radical that we have to deal with that we had in finding the domain of death. So it's going to be X is greater than or equal to negative one, just like it was for the domain of F, however, for part A, we have a different quantity under the radical, and we're going to have to make sure that that quantity is greater than or equal to zero. So we have four X minus two is greater than or equal to zero. So for X is greater than or equal to two. So X is greater than or equal to 1/2 and we can write it that way. Or we can ride the interval from 1/2 to infinity. If we want to use interval notation on the other domain, then we'll right the interval from negative one to infinity. All right, now we're finding FF, so we're putting the whole F function inside itself where the X is, and that's going to look like the square root of the square root of X plus one plus one. We'll just leave it like that. And now it's fine. G a. G. So we're putting the whole G function inside of G, where the X is so we get four times the quantity for X minus three minus three if we distribute before we get 16 X minus 12 minus three. So that's 16 X minus 15. Now for the domains. Well, let's get the domain of GMG over with. That's the easy one. That's just a polynomial. So it's domain is all real numbers negative. Infinity to infinity. Now let's talk about the domain of F of F. It looks pretty intimidating right now, but if we start on the inside and realize that that's the same thing, we have to consider that we considered for the domain of F. Inside this inside radical, we have to have X be greater than or equal to negative one. And as long as that's true, then that radical will either be zero or positive. And if you take a zero or positive and you add it, toe one, you get something you can take the outer square root of. So we're going to stick with the domain. Is Thean Derval from negative one to infinity that works for the inside function so that works for the whole function.