Results 1 to 6 of 6

Math Help - Math help wanted :) (Finding stationary points etc.)

  1. #1
    Member
    Joined
    Jul 2008
    Posts
    212

    Math help wanted :) (Finding stationary points etc.)

    Let
    f(x) = 3x^3 - 18 squarerootx
    for 0 equal/less x equal/less 4.


    (a) Find any local maxima, minima or points of inflexion.
    (b) Sketch the function, clearly marking and labeling the domain and range
    and the global maximum and minimum.

    -18 is not little, it's just a regular -18.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
     y= 3x^3-18\sqrt{x}

     \dfrac{dy}{dx}= 9x^2-\dfrac{9}{\sqrt{x}}

    For min/max values solve for

     \dfrac{dy}{dx}= 0

     9x^2-\dfrac{9}{\sqrt{x}}=0
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2008
    Posts
    212
    So x is 1?
    and then what do I do?

    Sorry I wrote 9 before that was a typo
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    x \neq 9

    Once you find the true value for x you then sub that back into the equation y.

    Spoiler:
    x = 1
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jul 2008
    Posts
    212
    Yup I know it equals 1 sorry it was a typo before. So subbing 1 back into the original equation I got =-15 so that means there is a local maximum there? My textbook says to find the 2nd derivative and then sub the x = 1 into that derivative. if y'' < 0 it is a maximum, etc etc
    So what is the 2nd derivative? 18x... what is the rest? Square roots confuse me!
    Last edited by brumby_3; August 31st 2009 at 08:36 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,004
    Thanks
    1657
    Quote Originally Posted by brumby_3 View Post
    Yup I know it equals 1 sorry it was a typo before. So subbing 1 back into the original equation I got =-15 so that means there is a local maximum there?
    What makes you think it is a maximum?
    My textbook says to find the 2nd derivative and then sub the x = 1 into that derivative. if y'' < 0 it is a maximum, etc etc
    So what is the 2nd derivative? 18x... what is the rest? Square roots confuse me!
    Then think \sqrt{x}= x^{\frac{1}{2}}.
    That, I suspect, was how pickslides got the first derivative: y= 3x^3- 18x^\frac{1}{2} so y'= 3(3x^{3-1})- 18\left(\frac{1}{2}x^{\frac{1}{2}- 1}\right)= 9x^2- 9x^{-\frac{1}{2}}.
    (And, of course, x^{-\frac{1}{2}}= \frac{1}{\sqrt{x}}.)

    Now, y''= 9(2x^{2-1})- 9(-\frac{1}{2})x^{-\frac{1}{2}-1}
    What is that?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. finding the stationary points?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 6th 2012, 06:43 PM
  2. Finding stationary points
    Posted in the Calculus Forum
    Replies: 6
    Last Post: September 2nd 2011, 06:25 AM
  3. Finding stationary points.
    Posted in the Calculus Forum
    Replies: 11
    Last Post: October 14th 2010, 02:07 PM
  4. Finding stationary points
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 21st 2010, 07:14 AM
  5. Finding coordinates of stationary points???
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 5th 2007, 11:53 AM

Search Tags


/mathhelpforum @mathhelpforum