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Use the given graphs of $ f $ and $ g $ to evaluate each expression, or explain why it is undefined.

(a) $ f (g(2)) $(b) $ g (f(0)) $(c) $ (f \circ g) (0) $(d) $ (g \circ f) (6) $(e) $ (g \circ g) (-2) $(f) $ (f \circ f) (4) $

(a) $g(2)=5,$ because the point (2,5) is on the graph of $g$. Thus, $f(g(2))=f(5)=4,$ because the point (5,4) is on the graph of $f$.(b) $g(f(0))=g(0)=3$(c) $(f \circ g)(0)=f(g(0))=f(3)=0$(d) $(g \circ f)(6)=g(f(6))=g(6) .$ This value is not defined, because there is no point on the graph of $g$ that has $x$ -coordinate 6.(e) $(g \circ g)(-2)=g(g(-2))=g(1)=4$(f) $(f \circ f)(4)=f(f(4))=f(2)=-2$

04:05

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

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all right. First, we're going to find part a f of G of two. So we start by finding g of to. So we find where X equals two. And then we follow that up to where it meets the G graph, and that would be at five. So g of two is five. We substitute that in, and now we're finding fo five. So we find X equals five and then we go up to where that meets the F graph, and that's at four. So our answer is for now we do something like that for part B. We start by finding f of zero. So we find X equals zero and we see where that matches the F graph, and that's at a height of zero. F of zero is zero. We substitute that in. Now we're finding g of zero. And so we go over to X equals zero and go up to where that meets the G graph. And that's at a height of three. Next, let's to part c f of g of zero. And we can write that this way if we want to f of g of zero. So the first thing we want to find is G of zero. So we go to where X equals zero and we find out on the G graph and that's three. Substitute that in, and now we have f of three. So we go over to where X equals three and we find that on the F graph, and that's at a height of zero. So the answer is zero. And now, for part D g of F six, that can be written as g of f of six like this. So the first thing we're finding his f of six. So we go over to where x equal six and we go up on the F graph and we get six for the height. So we substitute that in, and now we're finding g of six. However, six is not in the domain of G. It doesn't go that far, So this one is undefined. All right, here's part E. We can rewrite this as g of g of negative, too. And so on the inside, we're finding g of negative too. So we go over to where X equals negative to and find that on the G graph, and that's one so we substitute that in, and now we're finding g of one. So we go over to where X equals one, and we find that on the G graph. And that's right about four. So the answer is for and now for FFR four, which we can rewrite this way ff before we find the inside first effort for So we go over to where X equals four and we find that on the F graph. And that is it's supposed to be two minds a little low, but if you look at the actual graph, it's too. So we're going to substitute that in. And now we're looking for f of to. So we go to where X equals two and we find that on the F graph, and that is negative two. So there's our answer.

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