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Posted by 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 - , 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 - , 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 - , 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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