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Determine (a) $f(x)$ ) and the domain of the composite function, (b) $g(f(x))$ and the domain of the composite function.$f(x)=3 x+7 \quad g(x)=\frac{x-7}{3}$

(a) $x \quad-\infty<x<\infty$(b) $x \quad-\infty<x<\infty$

Algebra

Chapter 1

Functions and their Applications

Section 2

Basic Notions of Functions

Functions

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Lectures

01:43

In mathematics, a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. An example is the function that relates each real number x to its square x^2. The output of a function f corresponding to an input x is denoted by f(x).

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Determine (a) $f(x)$ ) and…

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for this problem we've been given to functions F of X equals three x plus seven and G of X equals X minus seven, divided by three. And our goal for this problem is to create two composite functions f of G of X and G of F of X. So how do we put together a composite function? Well, let's take a step back. If I just had a single function, F and I had something like f of three. What does that mean? Well, that means I'm gonna go to my F function and then anywhere where there's an ex I'm going to substitute in 33 is gonna go into place of any ex. Well, in this case, I don't have a three. What am I putting in? What I'm putting in g of X. So I'm going to my f function and in place of X, I'm going to substitute g of X every time that X appears in my function. So let's take a look at the functions we've been given first f of g of X. So f is my outside function. I go toe F, which is three x plus seven. I'm taking out the X And what am I putting in? I'm putting in G of X, which is X minus seven, divided by three. So let me simplify this a little. I have a three divided by three. So they cancel and I left with X minus seven plus seven. Well, those seven, they're going to cancel us. Well, which leaves me with just X. Now, while we're here, let's figure out the domain of this composite function. When I check the domain of a composite function, I have to check two places. First G of X was my input into F. So I have to have a valid input into that outermost function. So I have to look at the domain of that. In their most function, G of X. There are no restrictions. There could be anything I want. So we're good so far. Now we have to look at the composite function as a whole. In this case, that's X again, No restrictions. So my domain is all real numbers. Negative. Infinity to positive infinity. Okay, let's check out the other one g of f of X. Now. My outermost function is G. So we're gonna go to RG function, which is X minus 7/3. What's going in place of that X f of X? So come over here. Look at F of X and that is three X plus seven. Well, let's simplify a bit. In my numerator, I have a plus seven and a minus seven so they cancel. And then a three divided by a three cancels is well, so all I'm left with once again is X. And let's figure out our domain for my ex function. Just like before. I have to check that innermost function because that is being input into G. So I have to have valid inputs into that outer G function. So f of X. Well, that's three X plus seven. There are no restrictions there. X could be anything we want it to be. We also have to check the composite function as a whole. That's X again, No restrictions. I could let x be any real number. I wish so. My domain are all real numbers. Negative. Infinity to positive. Infinity

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