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swin2013 Group Title

Find the equation for the line tangent to the curve at the given point?

  • one year ago
  • one year ago

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  1. swin2013 Group Title
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    y=x^3+1/2x @ x =1

    • one year ago
  2. swin2013 Group Title
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    My work: y = x^3/2x + 1/2x ====> i split the fractions up dy/dx = x = 2/x^2 ====> this is where I messed up I think....

    • one year ago
  3. swin2013 Group Title
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    * dy/dx = x-2/x^2

    • one year ago
  4. swin2013 Group Title
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    yea

    • one year ago
  5. swin2013 Group Title
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    it's not really difficult lol

    • one year ago
  6. Chlorophyll Group Title
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    No, it's not! It essentially need to be precisely though !

    • one year ago
  7. Chlorophyll Group Title
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    f'(x) = ( 2x³ -1 ) / 2x²

    • one year ago
  8. swin2013 Group Title
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    well the numerator is not in parenthesis, does that make a difference?

    • one year ago
  9. swin2013 Group Title
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    and even if you take a derivative for f(x) it wouldn't be 2x^3-1 right? it would be just 2x^3?

    • one year ago
  10. Chlorophyll Group Title
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    You need to use the equation editor if you don't know how to write properly!

    • one year ago
  11. swin2013 Group Title
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    oh wow. oooooooooook

    • one year ago
  12. swin2013 Group Title
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    well either way, the derivative you have is wrong

    • one year ago
  13. swin2013 Group Title
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    @zepdrix can you please help me? :)

    • one year ago
  14. Chlorophyll Group Title
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    @swin2013 I'm not wrong at this trivial level :P

    • one year ago
  15. swin2013 Group Title
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    the derivative of x^3-1 is not 2x^3-1...

    • one year ago
  16. Chlorophyll Group Title
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    Of course not!

    • one year ago
  17. swin2013 Group Title
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    Ok then

    • one year ago
  18. Chlorophyll Group Title
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    It proves that you have no clue about the formula !!!

    • one year ago
  19. swin2013 Group Title
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    ooook. well i just quoted from what you said f'(x) was so yea

    • one year ago
  20. swin2013 Group Title
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    I came here to search for help @Chlorophyll, so i can just wait on the next person to help me

    • one year ago
  21. Chlorophyll Group Title
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    It's a FRACTION!

    • one year ago
  22. Chlorophyll Group Title
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    You're poor soul !!!

    • one year ago
  23. swin2013 Group Title
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    ok lmao

    • one year ago
  24. zepdrix Group Title
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    Hey were you able to figure this one out ok? c:

    • one year ago
  25. swin2013 Group Title
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    A little! sorry about the little mess earlier lolllll dy/dx = x - 1/2x^2

    • one year ago
  26. swin2013 Group Title
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    @zepdrix :) I'm not sure if you can still respond to this post since it's closed hahaha

    • one year ago
  27. zepdrix Group Title
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    Yes you can, I still get pop-ups. It just doesn't show up on the main list.

    • one year ago
  28. zepdrix Group Title
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    \[\large y=\frac{x^3+1}{2x}\qquad \rightarrow \qquad y=\frac{1}{2}x^2+\frac{1}{2}x^{-2}\] \[\large y'=x-x^{-3}\]I think this is what you should be getting.

    • one year ago
  29. zepdrix Group Title
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    Remember, when you subtract one from a NEGATIVE number, it actually gets LARGER in the negative direction.

    • one year ago
  30. zepdrix Group Title
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    \[\large \left(x^{-2}\right)\quad =\quad -2x^{-2-1}\quad = \quad -2x^{-3}\]Confused on this part?

    • one year ago
  31. swin2013 Group Title
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    I get how you got 1/2 x^2 but not 1/2 x^-2 i split up the fractions to kinda help me see it :/ so I got \[\frac{ x ^{3} }{ 2x } + \frac{ 1 }{ 2x }\] which is \[\frac{ 1 }{ 2 } x ^{2} + \frac{ 1 }{ 2 }x? \] :( sorry lolll

    • one year ago
  32. swin2013 Group Title
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    *1/2x^-1

    • one year ago
  33. zepdrix Group Title
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    Oh i wrote it wrong? :C woops...\[\large y=\frac{ 1 }{ 2 } x ^{2} + \frac{ 1 }{ 2 }x\] \[\large y'=x-x^{-2}\]

    • one year ago
  34. swin2013 Group Title
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    Yeeeee Yeah I got it :)) Thankss!

    • one year ago
  35. swin2013 Group Title
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    well idk does that look right? y' = x - 1/x^2?

    • one year ago
  36. zepdrix Group Title
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    Yes. Understand how to proceed from here or no?

    • one year ago
  37. swin2013 Group Title
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    Yes I do, thank youss!!!

    • one year ago
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