For the function F(x) = inx/x^2 , find the approximate location of the critical point in the interval (0, 5).

Answer 1

F ' (x) = (x^2(1/x) - lnx(2x) )/x^4
= (x - 2x lnx)/x^4
= 0 for critical values

x - 2x lnx = 0
x(1 - 2lnx) = 0
x = 0 , but for lnx to be defined, x > 0
or
1-2lnx = 0
lnx = 1/2 , x = √e
x = appr 1.65
y = (1/2) / (1.65)^2 = appr .18

the critical point is appr (1.65 , 0.15)