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anonymous

  • one year ago

If f(x) = 3a|4x – 4| – ax, where a is some constant, find f ′(1). Please help me guys, I give medals @preethat @phi

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  1. anonymous
    • one year ago
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    @phi could you help me

  2. freckles
    • one year ago
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    I remember this question and there was something fishy about it.. if a=0 then f(x)=0 then f'(x)=0 and so f'(1)=0 but what if a is not 0 then we have to consider 4x-4>0 and 4x-4<0 if 4x-4>0 then x-1>0 and then so f(x)=3a(4)(x-1)-ax if x>1 find f' for this side and if 4x-4<0 then x-1<0 and then so f(x)=3a(-4)(x-1)-ax if x<1 find f' for this one see if f'(1) matches for both sides

  3. phi
    • one year ago
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    the derivative of | f(x) | is f(x) f'(x)/|f(x)|

  4. freckles
    • one year ago
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    I say it was fishy because of the choices that came with this problem

  5. freckles
    • one year ago
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    something about A) 1 B) 0 C) -1 D) can't remember

  6. freckles
    • one year ago
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    I don't know if those were the choices actually

  7. anonymous
    • one year ago
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    Yeah, but could you explain me what they are actually asking, I mean I not want just the answer i was to know how to get answer because I really want to pass this exam

  8. freckles
    • one year ago
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    they are asking for f'(1)

  9. freckles
    • one year ago
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    I broken into a piecewise function

  10. anonymous
    • one year ago
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    so that means that first I have to find the derivate function and then plug 1?

  11. freckles
    • one year ago
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    @phi gave you something where you don't need piecewise function

  12. freckles
    • one year ago
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    yes

  13. phi
    • one year ago
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    f(x) = 3a|4x – 4| – ax, or f(x)= 12a | x -1| -ax the derivative is 12 a (x-1)/| x-1| - a

  14. freckles
    • one year ago
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    hopefully the a is not zero right it should really specify that the constant is not zero

  15. phi
    • one year ago
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    notice at x=1 we have an undefined quantity and the best we can do is write a limit, which is different for 1- and 1+

  16. freckles
    • one year ago
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    well that already says f' doesn't exist at x=1 we don't have to do the limit thing

  17. anonymous
    • one year ago
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    those are the only options, but I am not sure whether to pick 1 or not enough 0 not enough information e 1

  18. freckles
    • one year ago
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    I would say not enough info because we don't know if a is 0 or not

  19. freckles
    • one year ago
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    like phi just show the f'(1) doesn't exist when a isn't 0 but when a=0 ,f(x)=0 and so f'(x)=0 and so f'(1)=0

  20. freckles
    • one year ago
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    you get two difference answers for two difference cases of a

  21. anonymous
    • one year ago
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    true, I will select it and I'll see if the answer is correct, @phi said something similar, but i don't think they gave me enough information to solve the problem

  22. anonymous
    • one year ago
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    is that a modulus or a bracket?

  23. freckles
    • one year ago
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    if you want me to explain the piecewise function more I can by the way @Joseluess

  24. freckles
    • one year ago
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    if you want

  25. anonymous
    • one year ago
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    yes please, I am a little confused

  26. freckles
    • one year ago
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    modulus @14mdaz

  27. freckles
    • one year ago
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    \[|x|=\left[\begin{matrix}x & \text{ if } x>0 \\ -x & \text{ if } x<0 \\ 0 & \text{ if } x=0\end{matrix}\right]\] sorry don't know how to type a pretty piecewise function had to use the matrix code have you ever seen this and do you understand this?

  28. anonymous
    • one year ago
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    yes I know what you're talking about

  29. freckles
    • one year ago
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    |-2|=-(-2) I used row 2 since x=-2<0 |2|=2 I used row 1 since x=2>0 |0|=0 used row 3 since x=0 anyways just in case you didn't know that now I could replace all those x's with any function and solve any resulting inequalities that occur for example I see we have |4x-4| so I'm going to replace all the x's with (4x-4)'s

  30. freckles
    • one year ago
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    \[|4x-4|=\left[\begin{matrix}4x-4 & \text{ if } 4x-4>0 \\ -(4x-4) & \text{ if } 4x-4<0 \\ 0 & \text{ if } 4x-4=0\end{matrix}\right]\]

  31. freckles
    • one year ago
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    now straightening this up a bit \[|4x-4|=\left[\begin{matrix}4x-4 & \text{ if } x>1 \\ -(4x-4) & \text{ if } x<1 \\ 0 & \text{ if } x=1\end{matrix}\right] \]

  32. freckles
    • one year ago
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    but when x>1 what is the derivative of |4x-4| aka 4x-4 since |4x-4|=4x-4 for when x>1

  33. freckles
    • one year ago
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    \[(|4x-4|)'=\left[\begin{matrix}(4x-4)' & \text{ if } x>1 \\ -(4x-4)' & \text{ if } x<1 \\ ? & \text{ if } x=1\end{matrix}\right] \\ =...\] I left that one thing as a ? because we have to see if left derivative=right derivative (as x approaches 1 from both sides)

  34. freckles
    • one year ago
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    \[(|4x-4|)'=\left[\begin{matrix}(4x-4)' & \text{ if } x>1 \\ -(4x-4)' & \text{ if } x<1 \\ ? & \text{ if } x=1\end{matrix}\right] \\ (|4x-4|)'=\left[\begin{matrix}4 & \text{ if } x>1 \\ -4 & \text{ if } x<1 \\ \text{ does not exist (since*)} & \text{ if } x=1\end{matrix}\right] \\ \\ \text{ since } f'(1^+) \ \neq f'(1^-) \\ \text{ that is } 4 \neq -4 \]

  35. freckles
    • one year ago
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    now this already tells you that your function f(x)=3a|4x-4|-ax if a isn't 0 that you have f'(1) doesn't exist because (|4x-4|)' evaluated at x=1 doesn't exist but now if a=0 your function is f(x)=3(0)|4x-4|-0x=0-0=0 your function is f(x)=0 and so f'(x)=0 since derivative of a constant is 0 now f'(x)=0 does say it doesn't matter what x we have the f' value will always be 0 so f'(1)=0 if a=0

  36. freckles
    • one year ago
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    so summary of it all f'(1) can be 0 or doesn't exist

  37. freckles
    • one year ago
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    the information that doesn't allow us to say which is all dependent on what a is

  38. anonymous
    • one year ago
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    Okay, so that's why the answer it not enough right? okay I understand it better thanks

  39. freckles
    • one year ago
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    np

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