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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.
well, we need to know what g(x) is. were you given function of g?
@geerky42 one second going to post the image
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.
If you have time , I could really use assistance on this question. @TuringTest
@tkhunny If you have time, I could really use assistance on this question
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
I know the chain rule by heart, I don't know how to apply here. f'(x)= f'(g(x)) x g'(x)
Help in need of a mentor please. @Taylor<3sRin @myko @ganeshie8
hmm, wait. exactly what do you need help with?
it comes with the graph and the instructions.
How did you find f(x). ? did you just transform it from the original function |x|
yeah, that's how I figure it out
alright, is that the usual approach?
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?
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)
that's the problem and then they give you the graph
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 \(f'(x) = h'(~j(x)~)~j'(x)\)
PS: \[\large\dfrac{d}{dx}|x| = \dfrac{x}{|x|}\]
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?
what does "estimate the derivative" mean?
basically, when they ask us to do that they want an approximate value not the absolute value
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.
it's ok, I'm confused as well, thank you for your help.
i think you are thinking too hard for this one
from the picture we see \(g(1)=3\) and \(g(3)=1\) so \(g(g(1))=g(3)=1\)
similarly \(g(2)=2\) so \(g(g(2))=g(2)=2\)
as for the derivative, \[\left(g(g(x)\right)'=g(g(x))g'(x)\] by the chain rule
ok that was a mistake!!
\[\left(g(g(x)\right)'=g'(g(x))g'(x)\]
then \[w'(1)=g'(g(1))\times g'(1)=-1\times -1=1\]
|dw:1404869501887:dw|
which should not surprise you since \(g(g(x))=x\)
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