## Issy14 Group Title 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. 17 days ago 17 days ago

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1. geerky42 Group Title

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

2. Issy14 Group Title

@geerky42 one second going to post the image

3. Issy14 Group Title

|dw:1404857483770:dw|

4. Issy14 Group Title

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.

5. Issy14 Group Title

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

6. Issy14 Group Title

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

7. Issy14 Group Title

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

8. Issy14 Group Title

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

9. Issy14 Group Title

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

10. Issy14 Group Title

help @jdoe0001

11. geerky42 Group Title

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

12. geerky42 Group Title

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

13. Issy14 Group Title

with the problem given

14. Issy14 Group Title

it comes with the graph and the instructions.

15. Issy14 Group Title

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

16. Issy14 Group Title

@geerky42

17. geerky42 Group Title

yeah, that's how I figure it out

18. Issy14 Group Title

alright, is that the usual approach?

19. Issy14 Group Title

am I correct on g(x)?

20. geerky42 Group Title

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?

21. Issy14 Group Title

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)

22. Issy14 Group Title

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

23. geerky42 Group Title

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

24. geerky42 Group Title

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

25. geerky42 Group Title

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?

26. geerky42 Group Title

what does "estimate the derivative" mean?

27. Issy14 Group Title

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

28. geerky42 Group Title

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.

29. Issy14 Group Title

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

30. satellite73 Group Title

i think you are thinking too hard for this one

31. satellite73 Group Title

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

32. satellite73 Group Title

similarly $$g(2)=2$$ so $$g(g(2))=g(2)=2$$

33. satellite73 Group Title

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

34. satellite73 Group Title

ok that was a mistake!!

35. satellite73 Group Title

$\left(g(g(x)\right)'=g'(g(x))g'(x)$

36. satellite73 Group Title

then $w'(1)=g'(g(1))\times g'(1)=-1\times -1=1$

37. febylailani Group Title

|dw:1404869501887:dw|

38. satellite73 Group Title

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

39. Issy14 Group Title

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