Question

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.

Answer 1

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

Answer 2

@geerky42
one second going to post the image

Answer 3

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.

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

Help in need of a mentor please. @Taylor<3sRin @myko @ganeshie8

Answer 9

help @jdoe0001

Answer 10

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

Answer 11

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

Answer 12

with the problem given

Answer 13

it comes with the graph and the instructions.

Answer 14

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

Answer 15

@geerky42

Answer 16

yeah, that's how I figure it out

Answer 17

alright, is that the usual approach?

Answer 18

am I correct on g(x)?

Answer 19

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?

Answer 20

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)

Answer 21

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

Answer 22

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 \(f'(x) = h'(~j(x)~)~j'(x)\)

Answer 23

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

Answer 24

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?

Answer 25

what does "estimate the derivative" mean?

Answer 26

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

Answer 27

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.

Answer 28

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

Answer 29

i think you are thinking too hard for this one

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

ok that was a mistake!!

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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