Posted by **sarah** on Monday, May 19, 2008 at 3:16am.

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.

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