Results 1 to 9 of 9

Math Help - Critical value

  1. #1
    Newbie
    Joined
    Oct 2010
    Posts
    11

    Critical value

    Hi.
    There is no critical value for 1/(x+1)^2.Right?
    The open interval is (-∞,∞).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by wowo View Post
    Hi.
    There is no critical value for 1/(x+1)^2.Right?
    The open interval is (-∞,∞).
    Are you talking about doing a derivative test?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by wowo View Post
    Hi.
    There is no critical value for 1/(x+1)^2.Right?
    The open interval is (-∞,∞).
    Also, the interval is (-\infty,-1)\cup (-1,\infty)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2010
    Posts
    11
    Yes
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by wowo View Post
    Yes
    \displaystyle\frac{d}{dx}\left[\frac{1}{(x+1)^2}\right]=\frac{-2}{(x+1)^3}

    \displaystyle \frac{-2}{(x+1)^3}=0

    Critical value is -1
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Oct 2010
    Posts
    11
    f(x)=x/(x+1)^2
    f'(x)=1/(x+1)^2

    My bad.
    Sorry.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by wowo View Post
    f(x)=x/(x+1)^2
    f'(x)=1/(x+1)^2

    My bad.
    Sorry.
    \displaystyle f'(x)=\frac{-(x-1)}{(x+1)^2}=0\Rightarrow x-1=0 \ \mbox{and} \ x\neq -1
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Oct 2010
    Posts
    11
    Quote Originally Posted by dwsmith View Post
    \displaystyle f'(x)=\frac{-(x-1)}{(x+1)^2}=0\Rightarrow x-1=0 \ \mbox{and} \ x\neq -1
    The interval is (-∞,∞)?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by wowo View Post
    The interval is (-∞,∞)?
    No, (-\infty, -1)\cup (-1,\infty)

    If x = -1, you will be dividing by 0.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. critical value
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: July 30th 2011, 10:19 AM
  2. critical value
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: December 12th 2010, 04:06 AM
  3. t critical value
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: October 31st 2009, 10:16 PM
  4. Critical x help
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 30th 2009, 06:01 AM
  5. Critical value
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: March 28th 2008, 05:38 AM

Search Tags


/mathhelpforum @mathhelpforum