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swin2013

  • 3 years ago

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

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  1. swin2013
    • 3 years ago
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    y=x^3+1/2x @ x =1

  2. swin2013
    • 3 years ago
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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....

  3. swin2013
    • 3 years ago
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    * dy/dx = x-2/x^2

  4. swin2013
    • 3 years ago
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    yea

  5. swin2013
    • 3 years ago
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    it's not really difficult lol

  6. Chlorophyll
    • 3 years ago
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    No, it's not! It essentially need to be precisely though !

  7. Chlorophyll
    • 3 years ago
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    f'(x) = ( 2x³ -1 ) / 2x²

  8. swin2013
    • 3 years ago
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    well the numerator is not in parenthesis, does that make a difference?

  9. swin2013
    • 3 years ago
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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?

  10. Chlorophyll
    • 3 years ago
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    You need to use the equation editor if you don't know how to write properly!

  11. swin2013
    • 3 years ago
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    oh wow. oooooooooook

  12. swin2013
    • 3 years ago
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    well either way, the derivative you have is wrong

  13. swin2013
    • 3 years ago
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    @zepdrix can you please help me? :)

  14. Chlorophyll
    • 3 years ago
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    @swin2013 I'm not wrong at this trivial level :P

  15. swin2013
    • 3 years ago
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    the derivative of x^3-1 is not 2x^3-1...

  16. Chlorophyll
    • 3 years ago
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    Of course not!

  17. swin2013
    • 3 years ago
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    Ok then

  18. Chlorophyll
    • 3 years ago
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    It proves that you have no clue about the formula !!!

  19. swin2013
    • 3 years ago
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    ooook. well i just quoted from what you said f'(x) was so yea

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

  21. Chlorophyll
    • 3 years ago
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    It's a FRACTION!

  22. Chlorophyll
    • 3 years ago
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    You're poor soul !!!

  23. swin2013
    • 3 years ago
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    ok lmao

  24. zepdrix
    • 3 years ago
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    Hey were you able to figure this one out ok? c:

  25. swin2013
    • 3 years ago
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    A little! sorry about the little mess earlier lolllll dy/dx = x - 1/2x^2

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

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

  28. zepdrix
    • 3 years ago
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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.

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

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

  31. swin2013
    • 3 years ago
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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

  32. swin2013
    • 3 years ago
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    *1/2x^-1

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

  34. swin2013
    • 3 years ago
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    Yeeeee Yeah I got it :)) Thankss!

  35. swin2013
    • 3 years ago
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    well idk does that look right? y' = x - 1/x^2?

  36. zepdrix
    • 3 years ago
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    Yes. Understand how to proceed from here or no?

  37. swin2013
    • 3 years ago
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    Yes I do, thank youss!!!

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