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Finding points on the graph

  1. Sep 13, 2011 #1
    I have absolutely no idea how to solve this problem and I would really appreciate same help.

    For the function, find the point(s) on the graph a which the tangent line has slope 5

    y=1/3x^3-2x^2+8x+14

    Any help with this would br greatly appreciated
     
  2. jcsd
  3. Sep 13, 2011 #2

    Dick

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    Do you know what a derivative is and what it might have to do with a slope? You really have to make a better start than that.
     
  4. Sep 14, 2011 #3

    Ray Vickson

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    Use brackets. I really cannot figure out whether you mean [tex] y = \frac{1}{3} x^3 - 2 x^2 + 8x + 14[/tex] or [tex] \frac{1}{3x^3} - 2x^2 + 8x + 14[/tex] or [tex] y =\frac{1}{3x^3 - 2x^2 + 8x + 14} . [/tex] Strictly according to standard rules, what you wrote is the first of these three possibilities. If you don't want to use Latex you can just write (1/3)x^3 - 2x^2 + ... or 1/(3x^3) - 2x^2 + ... or 1/(3x^3 - 2x^2 + ...)

    RGV
     
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