calculus question (urgent)

f(x) = k (sqrt(x)) - lnx
f'(x) = (1/2)kx^(-1/2) - (1/x)
f"(x) = (-1/4)kx^(-3/2) + x^(-2)
for the curve f'(x) = 0, it may be a max. point or min. point.
f'(1) = (1/2)k(1)^(-1/2) - 1/(1) = 0
∴ k = 2
when k = 2
f"(1) = (-1/4)(2)(1)^(-3/2) + (1)^(-2)
= 1/2 ＞ 0
f has a relative min.
for the graph of f has a point of inflection on the x-axis
f"(x) = (-1/4)kx^(-3/2) + x^(-2) = 0 ......(1)
f(x) = k (sqrt(x)) - lnx = 0......(2)
from (1) : x^(-2)[(-1/4)ksqrt(x) + 1) = 0
k = 4/sqrt(x),
put k = 4/sqrt(x) into (2), lnx = 4, x = e^4
and k = 4/sqrt(x) = 4/e^2

2009-12-03 13:07:29 補充：
a.

slope of the line tangent to the graph of f at x=e^2 given by f(x)=lnx/x
given f'(x) = [1 -lnx]/x^2
f'(e^2) = [1 -lne^2]/(e^2)^2 = -1/e^4
f(e^2)=lne^2/e^2 = 2/e^2
the required equation : y - 2/e^2 = (-1/e^4)(x - e^2)
x + (e^2)y - 2e^2 = 0

2009-12-03 13:07:54 補充：
b.

given f'(x) = [1 -lnx]/x^2

f'(x) = 0 , ∴ lnx =1

x = e, f has a critical point.

f'(x) = [1 -lnx]/x^2

f"(x) = [x^2(-1/x) - 2x(1 - lnx)]/x^4

= [-3x +2xlnx]/x^4

f"(e) = [-3e +2elne]/e^4

= -e/e^4 ＜ 0
when x = e, f has a relative max point.

2009-12-03 13:08:24 補充：
c.
when f"(x) =0, f has Point of inflection,
[-3x +2xlnx]/x^4 = 0
∴ x[3 -2lnx] = 0
for x ＞ 0, ln x = 3/2
x = e^(1.5), f has Point of inflection