calculus

posted by .

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.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. calculus

    Find all relative extrema and points of inflection of the function: f(x) = sin (x/2) 0 =< x =< 4pi =< is supposed to be less than or equal to. I can find the extrema, but the points of inflection has me stumped. The inflection …
  2. calculus

    Find any relative extrema and any points of inflection if they exist of f(x)=x^2+ ln x^2 showing calculus. Please show work in detail so I can follow. Thanks. The answer is no relative max or min and the points of inflection are (1,1) …
  3. cal

    Sketch the graph and show all local extrema and inflection points. f(x)= 1/x^2-2x-8 I believe the ans is "Relative max: [1,-1/9]" and No inflection points".
  4. Calculus A

    Find all relative extrema and points of inflection for the following function... h(X)= X^2+5X+4/ X-1 min= max= inflection points=
  5. APCalculus

    For the function y=4-2x^2+1/6x^4 find the following: Domain x and y intercepts Vertical asymptotes Horizontal asymptotes Symmetry F'(x) Critical numbers Increasing f(x) Decreasing f(x) Extrema F"(x) Possible points of inflection Concave …
  6. APCalculus

    For the function y=2sin(x)-cos^2(x) on [0, 2pi] find the following: Domain x and y intercepts Vertical asymptotes Horizontal asymptotes Symmetry F'(x) Critical numbers Increasing f(x) Decreasing f(x) Extrema F"(x) Possible points of …
  7. Relative Extrema: AP Calculus

    Consider the function f(x)= (3/4)x^4-x^3-3x^2+6x Find the relative extrema for f(x); be sure to label each as a maximum or minimum. You do not need to find function values; just find the x-values. Determine the interval(s) where f(x) …
  8. calculus

    Identify any intercepts, relative extrema, points of inflection, and asymptotes. (If an answer does not exist, enter DNE.) y = x^3 + 3x^2 + 3x + 2 x-intercept (x, y) = y-intercept (x, y) = relative maximum (x, y) = relative minimum …
  9. Calculus AB

    Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x) = 3x(1/3) + 6x(4/3). You must justify your answer using an analysis of f '(x) and f "(x). My start to a solution: x-values: (-1/8) and (-1/4) …
  10. Calculus

    Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x)=9x^1/3+9/2x^4/3. I've gotten -0.5 as the relative minimum and x=1 for the inflection points. Is this correct?

More Similar Questions