If f and g are the functions whose graphs are shown below, let u(x)=f(g(x)) and v(x)=g(f(x)) . find u'(3) and v'(3)

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If f and g are the functions whose graphs are shown below, let u(x)=f(g(x)) and v(x)=g(f(x)) . find u'(3) and v'(3)

Mathematics
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1 Attachment
start with finding expressions for u'(x) and v'(x)
u'(x)=f'(g(x))g'(x) v'(x)=g'(f(x))f'(x)

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yeah i dont understand this. am i just looking at 3 on the graph?
say you want u'(3), you first need g(3) because g(x) is part of the formula you need what is g(3) according to the graph?
looks like 4
looks like g(3)=10 to me each square is 2 vertically, and 1 horizontally as far as I can tell
g(3)=8 I mean, sorry
I'm not really sure about the increments on the graph though, it's a bit vague...
8, how so?
It looks to me like each square vertically is 2 (by the marking), but I guess it must be 4 since 8 is not on the graph horizontally so ok, g(3)=4 what is the value of f'(4) ? [remember we need f'(g(3))g'(3)]
also remember that f'(4) represents the slope of f(x) at x=4
f(4) is 2
we want f'(4) that is the slope of f(x) at x=4 what is the slope of the line in the graph f(x) at x=4 ?
when x is 4, f is 2 right?
yes, but that is the value of f(4) not f'(4) f'(4) is the SLOPE, not the value of the function do you remember how to find the slope of a straight line from algebra?
ohh
y2-y1/x2-x1?
actually sorry, they are the same in this case, but that is pure luck! lol notice that the slope (rise over run) is also 2 !
that is what f' means, the slope of f so similarly what is the last piece of the formula we need ? u'(3)=f'(g(3))g'(3) and we still need g'(3) what is it?
rise over run right? 4/3? no?
at x=3 it looks to me like g has a slope of -1
how did you figure that out?
look at the tail last portion of g(x), the far right portion of the top graph. it dips downward it seems to go down one unit for every unit it goes to the right rise/run=-1/1=-1
here's a sketch of g|dw:1329536578080:dw|
ok i see now i see
here's the part from x=2 to x=4
|dw:1329536693848:dw|in your graph you can see it goes one over and one down on that region
(I'm not gonna draw the little squares in...)
i see that thanks
so you have all the pieces now u'(3)=f'(g(3))g'(3) you know f'(g(3)) and g'(3), so multiply them to find u'(3)
what is g(3) ?
4 slope is -1
right, g(3)=4 and what is f'(g(3)) ?
f is also 4 right? so that makes 4 x 4?
no, we want f'(g(3))=f'(4) because g(3)=3, right ? what is f'(4) ?
the slope of at 4 is 3
you are thinking about the idea correctly, but it looks like 2 to me
ok the gradients on this graph is crap
yeah, I agree with that, but the problem only seems to make sense if we count each square as 1...|dw:1329537340258:dw|so we want the slope of the upward portion of f
4/2
2
right, looks that way so what is u'(3)=f'(g(3))g'(3) ?
(2)(4) (4)
there should be no 4 in the answer what is f'(g(3)) ?
(we just figured it out above)
2
yes, and what is g'(3) ?
-1
so f'(g(3))g'(3) is what?
(2) (-1)
u'(3)=f'(g(3))g'(3)=-2 yep :D
thanks for the help
so now you get to try for v'(3) ! good luck :)
I got zero
yep!

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