Results 1 to 2 of 2

Math Help - more extreme values

  1. #1
    Member >_<SHY_GUY>_<'s Avatar
    Joined
    Jan 2008
    From
    CA
    Posts
    181

    more extreme values

    i just don't get this now...i got the x-coordinate, but maybe im doing something wrong to find the y coordinate

    y=2x^x-8x+9

    and how would i find the extremas of:
    y=x^3 + x^2 - 8x + 5
    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
    29
    Quote Originally Posted by >_<SHY_GUY>_< View Post
    y=2x^x-8x+9

    and how would i find the extremas of:
    y=x^3 + x^2 - 8x + 5
    y=x^3 + x^2 - 8x + 5

    find extremas of this function make the derivative equal to zero.

    \frac{dy}{dx}=3x^2 + 2x - 8

    \frac{dy}{dx}=0

    3x^2 + 2x - 8 = 0

    Now solve for x using the quadratic formula.

    x = \frac{-2\pm\sqrt{2^2-4\times3\times-8}}{2\times3}

    sub these values back into y to get your answer...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Extreme Values
    Posted in the Calculus Forum
    Replies: 7
    Last Post: November 8th 2009, 04:27 AM
  2. extreme values
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 6th 2009, 10:30 PM
  3. Help finding the extreme values
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 19th 2009, 10:02 AM
  4. Extreme Values
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 1st 2008, 02:05 PM
  5. extreme values
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 18th 2007, 12:45 AM

Search Tags


/mathhelpforum @mathhelpforum