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Critical points

  1. Mar 13, 2005 #1
    hi guys i'm stuck on this, i keep thinking i have the right answeres but i can't get them.
    the function is f(x): -2x^3 + 39x^2 - 180x + 1
    i need to list all of the critical points, and hten indicate where it is increasing and where it is decreasing.

    I set the derivative -6x^2 + 78x - 180 = 0 and then solved for x. this is correct right. by doing so i ended up with critical points at -2 and 15. Then made a number line and figured that the slope before -2 was neg, b/w -2 and 15 it was pos, and after 15 it was neg. so my interval of increasing was (-2,15) and the interval of decreasing was (-inf, -2) U (15,inf) where am i going wrong - PLEASE HELP ME
     
  2. jcsd
  3. Mar 13, 2005 #2

    dextercioby

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    Did u compute the second derivative,too...?To find the nature of those 2 points,u must evaluate the second derivative in those points.

    Daniel.
     
  4. Mar 13, 2005 #3

    mathman

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    You made a mistake in solving the quadratic. The roots are 3 and 15.
     
  5. Mar 13, 2005 #4
    Yes i determined the second derivative as -12x + 78. Then using the second derivative rule plugged in 15, and -2 telling me that -2 was a local min and 15 was a local max. DO you know if 15, -2 are the only critical points in my problem????
     
  6. Mar 13, 2005 #5
    i don't see how u got 3 and 15 ??
     
  7. Mar 13, 2005 #6

    dextercioby

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    The roots are 3 & 10.

    Daniel.
     
  8. Mar 13, 2005 #7

    cronxeh

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    -6x^2 + 78x - 180 = 0

    x^2 -13x + 30 = 0
    x=(13+/-sqrt(169-4*1*30))/2
    x=(13+/-7)/2

    x=20/2 = 10
    x=6/2= 3
     
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