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November 26, 2014

November 26, 2014

Posted by **Tina** on Wednesday, October 5, 2011 at 1:15pm.

- calculus -
**Steve**, Wednesday, October 5, 2011 at 2:34pmJust plug and chug:

f = ax^b/lnx

1 = a*19^b / ln 19

a * 19^b = 2.944

f' = (abx^(b-1) * lnx) - ax^(b-1))/ln^2(x)

= [ax^(b-1) * (b*ln x-1)]/ln^2(x)

to get f'=0, we need b*lnx = 1

b*ln19 = 1

b = 1/ln19

a* 19^b = 2.944

a*19^(1/ln19) = a * 19^.3396 = a*2.718 = 2.944

a = 1.083

So, if my math is right, f(x) = 1.083*x^.3396/lnx

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