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