calculus
posted by Tina .
Find constants a and b in the function f(x)=ax^b/(ln(x)) such that f(19)=1 and the function has a local minimum at x=19. What is a and b?

Just plug and chug:
f = ax^b/lnx
1 = a*19^b / ln 19
a * 19^b = 2.944
f' = (abx^(b1) * lnx)  ax^(b1))/ln^2(x)
= [ax^(b1) * (b*ln x1)]/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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