Homework Help Forum: Calculus

Posted by Z32 on Monday, October 26, 2009 at 12:51pm.

Find the inflection points for f(x) = 12x^5 + 45x^4 - 360x^3 + 7

• Calculus - Reiny, Monday, October 26, 2009 at 12:58pm

  f'(x) = 60x^4 + 180x^3 - 1080x^2 f''(x) = 240x^3 + 540x^2 - 2160x = 0 for points of inflection 240x^3 + 540x^2 - 2160x = 0 divide by 60 4x^3 + 9x^2 - 36x = 0 x(4x^2 + 9x - 36) = 0 Can you take it from there ? There are 3 solutions for x

  
  

• Calculus - Z32, Tuesday, October 27, 2009 at 1:11am

  I got 0 for one of my inflections since a 0 in the place of the "x"(4x^2 + 9x - 36) would make it zero. But I'm not sure what would need to go into the other x's to make it 0. I tried 4/9 9/36 and 36/9 but I couldn't get it.

  
  

• Calculus - Z32, Tuesday, October 27, 2009 at 2:10am

  Oh, okay. I found out I had to use the quadratic formula. Thanks for the help!

  
  

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