Post a New Question

calculus

posted by on .

find the relative extrema and points of inflection, if possible, of y = xln((x^4)/8)

Any help would be awesome...thanks!
(i know that the relative extrema come from critical values from y' and points of inflection come from y'')

  • calculus - ,

    dy/dx = y' = ln((x^4)/8) + x(4/x)
    = ln ( (x^4)/8 ) + 4

    y '' = 4/x

    for relative extrema, y' = 0

    ln ((x^4)/8) = 0

    x^4 / 8 = e^0 = 1
    x^4 = 8
    x = ± 8^(1/4)
    plug in those values into y = ...

    for pts of inflection, 4/x = 0 ----> no solution
    thus, no points of inflection

  • calculus - ,

    And yet, y'' changes from - to + as x increases through 0, so while y is discontinuous at x=0, there appears to be an inflection point there.

    To get the limit of y at x=0,
    y(0) = 0*(-oo)

    setting t=1/x and using L'Hopital's rule a bit, we can see that the limit is 0.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question