calculus

what are the inflection points of 5x^2-45/x^2-25

  1. 0
  2. 0
asked by marissa
  1. what is the numerator and what is the denominator?

    posted by Damon
  2. assuming the usual sloppiness, we have

    y=(5x^2-45)/(x^2-25)
    y'= -160x/(x^2-25)^2
    y" = 160(3x^2+25)/(x^2-25)^3


    since y" is never zero, there are no inflection points, as can be confirmed here:

    http://www.wolframalpha.com/input/?i=%285x^2-45%29%2F%28x^2-25%29

    posted by Steve

Respond to this Question

First Name

Your Response

Similar Questions

  1. Calculus

    The number of people who donated to a certain organization between 1975 and 1992 can be modeled by the equation D(t)=-10.61t^(3)+208.808t^(2)-168.202t+9775.234 donors, where t is the number of years after 1975. Find the inflection
  2. 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
  3. calculus

    Use a graph of f(x) = 3 e^{-8 x^2} to estimate the x-values of any critical points and inflection points of f(x). critical points x= Inflection points x= Next, use derivatives to find the x-values of any critical points and
  4. Calculus

    Suppose f is continuous on [0, 6] and satisfies the following x 0 3 5 6 f -1 4 -1 -3 f' 5 0 -8 0 f" -1 -3 DNE 3 x 0
  5. 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".
  6. Math URGENT REALLY FAST HELP

    find the inflection points of the function: f9x)=x^2(ln(x)) Now I don't think there are any inflection points because both intervals are concave up. Am I right, or could someone help me if this is wrong?
  7. 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=
  8. 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
  9. Calculus

    If f(x) is a continuous function with f"(x)=-5x^2(2x-1)^2(3x+1)^3 , find the set of values for x for which f(x) has an inflection point. A. {0,-1/3,1/2} B. {-1/3,1/2} C. {-1/3} D. {1/2} E. No inflection points
  10. calculus

    the graph (x^5-15*x^3 +10) is sketched without axes and not to scale. I found f'(x), set it to zero and got x=0, x=3, x=-3. these are critical numbers? I plugged them into original equation to find y (?) and got (0,10),3,-152)(-3,
  11. calculus

    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'')

More Similar Questions