- Issy14

Let w(x) = g(g(x)). Find:
(a) w(1) (b) w(2) (c) w(3)
somebody please help, I have no idea what to do.

- jamiebookeater

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- geerky42

well, we need to know what g(x) is. were you given function of g?

- Issy14

@geerky42
one second going to post the image

- Issy14

|dw:1404857483770:dw|

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## More answers

- Issy14

These are the directions:
In Problems 57–60, use Figure 3.17 and the chain rule to estimate the derivative, or state why the chain rule does not apply.
The graph of f(x) has a sharp corner at x = 2.

- Issy14

If you have time , I could really use assistance on this question. @TuringTest

- Issy14

@tkhunny If you have time, I could really use assistance on this question

- Issy14

I think i figured out g(x) = -1x+4
f(x) is not continuous it has a breat at (0,2) therefore if i try to get the slope they are going to have the same magnitude but different signs f(x)=4x or f(x)= -4x

- Issy14

I know the chain rule by heart, I don't know how to apply here. f'(x)= f'(g(x)) x g'(x)

- Issy14

- Issy14

help @jdoe0001

- geerky42

\(f(x) = -2|x-2|+4\)

- geerky42

hmm, wait. exactly what do you need help with?

- Issy14

with the problem given

- Issy14

it comes with the graph and the instructions.

- Issy14

How did you find f(x). ?
did you just transform it from the original function |x|

- Issy14

- geerky42

yeah, that's how I figure it out

- Issy14

alright, is that the usual approach?

- Issy14

am I correct on g(x)?

- geerky42

well, i don't know what's your given problems are.
are you asked to take derivative of f(x) using chain rule, or state why chain rule does not apply?

- Issy14

both
In Problems 57–60, use Figure 3.17 and the chain rule to estimate the derivative, or state why the chain rule does not apply.
The graph of f(x) has a sharp corner at x = 2.
60. Let w(x) = g(g(x)). Find:
(a) w(1) (b) w(2) (c) w(3)

- Issy14

that's the problem and then they give you the graph

- geerky42

ok i see. well, we managed to figure f(x) out by using transformation from parent function, so we have f(x) = -2|x-2|+4
we can take derivative of f(x) because it is continuous. just not differentiable at break point.
so we can let h(x) = -2x+4 and j(x) = |x-2| so we have \(f'(x) = h'(~j(x)~)~j'(x)\)

- geerky42

PS: \[\large\dfrac{d}{dx}|x| = \dfrac{x}{|x|}\]

- geerky42

on second thought, you was asked to estimate the derivative, but what we are doing is finding the actual derivative. hmm, not sure what to do here. do you have any idea?

- geerky42

what does "estimate the derivative" mean?

- Issy14

basically, when they ask us to do that they want an approximate value not the absolute value

- geerky42

huh, i am sorry, this problem confused me, because i don't see how chain rule can be apply to estimate value from given graph? so i am not sure... sorry.

- Issy14

it's ok, I'm confused as well, thank you for your help.

- anonymous

i think you are thinking too hard for this one

- anonymous

from the picture we see \(g(1)=3\) and \(g(3)=1\) so \(g(g(1))=g(3)=1\)

- anonymous

similarly \(g(2)=2\) so \(g(g(2))=g(2)=2\)

- anonymous

as for the derivative,
\[\left(g(g(x)\right)'=g(g(x))g'(x)\] by the chain rule

- anonymous

ok that was a mistake!!

- anonymous

\[\left(g(g(x)\right)'=g'(g(x))g'(x)\]

- anonymous

then
\[w'(1)=g'(g(1))\times g'(1)=-1\times -1=1\]

- anonymous

|dw:1404869501887:dw|

- anonymous

which should not surprise you since \(g(g(x))=x\)

- Issy14

then there was no need to find out the function of each one? I did that before but I thought that it was too simple. I will continue reading your answer, thank you. @satellite73 and @febylailani

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