let f be function given by f(x)= Ln(x)/x for all x> 0. the dervative of f is given by f'(x)= (1 - Ln(x))/x squared.
a) write equation for the line tangent to the graph of f at x=e squared
b) Find the x-coordinate of the critical point of f. Determine wheter this point is a relative minimum, a relative maximum or neither for the function f.
c) the graph of the function f has exactly one point of inflation. find the x coordinates of this point.
(d) find lim f(x) x to the 0+
calc - Reiny, Monday, May 19, 2008 at 8:54am
a) when x=e^2, y = lne^2/e^2 = 2/e^2 and
y' = (1-lne^2)/e^4 = -1/e^4
so the equation is x + (e^4)y = c
but (e^2,2/e^2) lies on it, so
e^2 + e^4(2/e^2) = c
c = 3e^2
the equation is x + (e^4)y = 3e^2
( I am using a quick way to find the equation of a line based on the fact that if the slope is -A/B, then the equation is Ax + By = C )
calc - Reiny, Monday, May 19, 2008 at 8:57am
b) set f'(x) = 0
(1- lnx)/x^2 = 0
1 - ln x = 0
ln x = 1
x = e then y = ln e/e = 1/e
the critical point is (1,1/e)
To determine if it is a max/min I will do c) first
calc - Reiny, Monday, May 19, 2008 at 9:06am
c) for f''(x) I got
(-3 + 2lnx)/x^3
setting that equal to zero ...
2lnx = 3
lnx = 3/2
x = 4.4817 and y = .3347
so there is a point of inflection at (4.48,0.33)
Back to b)
when you plug x=e^2 into f'' you get
(-3+ 2lne^2)/e^6 which is positive.
Therefore the critical point from b) is a minimum point of the function.