Results 1 to 3 of 3

Math Help - calculus inflection, increasing, decreasing

  1. #1
    Member
    Joined
    Sep 2005
    Posts
    136

    calculus inflection, increasing, decreasing

    1.On what interval the function f(x)=x/(x^2+1) increasing?

    (a)[-1,1](b)[-inf,1] (c) [-1,2] (d) [-inf, 2]

    2. find the interval on which the function f(x)=x+cosx 0<=x<=2pi is increasing
    (a)(pi/3,5pi/3)
    (b)5pi/3,2pi)
    (c) (pi/2, 3pi/2)
    (d)(0,2pi,)
    (e)none
    3.
    How many pointsof inflection does f(x) =x^3e^(-x) have?
    (a)0
    (b)1
    (c)2
    (d)3
    (e) none
    4. consider the function f(x)=x^2 in interval [0,1/2].According to mean value theorem thre must be a number c in (0,1/2) such that f '(c)is equal to particular value d. what is d?
    (a) 3/2
    (b)1
    (c) 1/2
    (d)2
    (e) none
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by bobby77 View Post
    4. consider the function f(x)=x^2 in interval [0,1/2].According to mean value theorem thre must be a number c in (0,1/2) such that f '(c)is equal to particular value d. what is d?
    (a) 3/2
    (b)1
    (c) 1/2
    (d)2
    (e) none
    f'(c)=\frac{f(1/2)-f(0)}{1/2-0}
    Thus,
    f'(c)=\frac{1/4}{1/2}=\frac{1}{2}
    Thus that number is d=1/2
    How many pointsof inflection does f(x) =x^3e^(-x) have?
    Then,
    y'=3x^2e^{-x}-x^3e^{-x}
    y''=6xe^{-x}-3x^2e^{-x}-3x^2e^{-x}+x^3e^{-x}
    Thus,
    6xe^{-x}-6x^2e^{-x}+x^3e^{-x}=0
    Thus,
    6x-6x^2+x^3=0
    Thus,
    x(x^2-6x+6)=0
    The quadradic has discriminant,
    6^2-4\cdot 6>0 so it has two solution (non-trivial) thus, in total 3 solutions.
    Corresponding to 3 inflections points.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,911
    Thanks
    775
    Hello, Bobby!

    3. How many points of inflection does f(x) = x^3e^{-x} have?

    . . . (a)\;0\qquad(b)\;1\qquad(c)\;2\qquad(d)\;3\qquad(e  )\text{ none}

    Points of inflection occur where f''(x) = 0

    We have: . f'(x)\;=\;-x^3e^{-x} + 3x^2e^{-x}
    Then: . f''(x)\;=\;x^3e^{-x} - 3x^2e^{-x} - 3x^2e^{-x} + 6xe^{-x} \;=\;x^3e^{-x} - 6x^2e^{-x} + 6xe^{-x}

    We have: . xe^{-x}(x^2 - 6x + 6) \:=\:0

    Hence, the roots are: . x\,=\,0
    . . and: . x^2 - 6x + 6 \:=\:0\quad\Rightarrow\quad x \,=\,3\pm\sqrt{3}

    Therefore, there are three inflection points . . . answer (d).

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: October 29th 2009, 09:02 PM
  2. increasing or decreasing
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 14th 2009, 08:02 PM
  3. Increasing/Decreasing
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 16th 2009, 07:06 PM
  4. Increasing/Decreasing and Abs Max and Min
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 28th 2009, 06:57 PM
  5. increasing decreasing
    Posted in the Calculus Forum
    Replies: 3
    Last Post: August 6th 2007, 02:29 PM

Search Tags


/mathhelpforum @mathhelpforum