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March 3, 2015

March 3, 2015

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

- Calculus -
**Reiny**, Monday, October 26, 2009 at 12:58pmf'(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 1:11amI 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 2:10amOh, okay. I found out I had to use the quadratic formula. Thanks for the help!

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