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October 1, 2014

October 1, 2014

Posted by **bill nye** on Tuesday, March 10, 2009 at 5:11pm.

- calculus -
**drwls**, Tuesday, March 10, 2009 at 7:44pmYou wrote << f(0)=4 and f(0)=3,>> but

both cannot be true. One of the f(x) functions must be the derivative, f'(x). Which is it?

Anyway, all you have to do is integrate your f''(x) once and use the value of f'(x) at x=0 to get the first cosntant, and then integrate it again and use the value of f(0)at x=0 to get the second constant

- calculus -
**bill nye**, Tuesday, March 10, 2009 at 8:03pmoh im sorry it is f(0)=4 and f'(0)=3

- calculus -
**bill nye**, Tuesday, March 10, 2009 at 8:29pmand it's f''(x)=9x+6sin(x)

sorry

- calculus -
**bill nye**, Thursday, March 12, 2009 at 12:29ami got the answer 1.5x^3-6sinx+9x+4

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