Results 1 to 4 of 4

Thread: impossible stationary points

  1. #1
    New Member
    Join Date
    Feb 2018
    HSC
    2018
    Gender
    Male
    Posts
    6
    Rep Power
    1

    impossible stationary points

    Hi I would like some assistance on this question Thanks

    find any turning point on the curve y=(x-3)√(4-x) and determine their nature.

  2. #2
    Junior Member OkDen's Avatar
    Join Date
    Mar 2016
    HSC
    2018
    Gender
    Male
    Location
    49°51′S 128°34′W
    Posts
    116
    Rep Power
    2

    Re: impossible stationary points

    Quote Originally Posted by cos16mh View Post
    Hi I would like some assistance on this question Thanks

    find any turning point on the curve y=(x-3)√(4-x) and determine their nature.
    For Turning Points, dy/dx = 0

    To find derivative of y=(x-3)√(4-x) use Product Rule.

    So, dy/dx = (-3x+11)/2√(-x+4)
    Since, dy/dx = 0, 0 =(-3x+11)/2√(-x+4)

    Which gives: x = 11/3
    Sub into original function: y=(x-3)√(4-x), giving y = (2√3)/9

    To find nature of Turning Point, sub in turning points (11/3 , (2√3)/9) to the second derivative to determine nature.

    positive = minimum
    negative = maximium

    Or you can graph it, or test coordinates.

    HSC 2018
    MX1 - BIO - CHE - PHY - ADV

  3. #3
    New Member
    Join Date
    Feb 2018
    HSC
    2018
    Gender
    Male
    Posts
    6
    Rep Power
    1

    Re: impossible stationary points

    Quote Originally Posted by OkDen View Post


    For Turning Points, dy/dx = 0

    To find derivative of y=(x-3)√(4-x) use Product Rule.

    So, dy/dx = (-3x+11)/2√(-x+4)
    Since, dy/dx = 0, 0 =(-3x+11)/2√(-x+4)

    Which gives: x = 11/3
    Sub into original function: y=(x-3)√(4-x), giving y = (2√3)/9

    To find nature of Turning Point, sub in turning points (11/3 , (2√3)/9) to the second derivative to determine nature.

    positive = minimum
    negative = maximium

    Or you can graph it, or test coordinates.

    Its the second derivative that I'm struggling with

  4. #4
    Junior Member fan96's Avatar
    Join Date
    May 2017
    HSC
    2018
    Gender
    Male
    Location
    NSW
    Posts
    117
    Rep Power
    1

    Re: impossible stationary points

    Quote Originally Posted by cos16mh View Post
    Its the second derivative that I'm struggling with
    Then test the first derivative around the turning points.
    HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]

Thread Information

Users Browsing this Thread

There are currently 1 users browsing this thread. (0 members and 1 guests)

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •