Results 1 to 5 of 5

Math Help - determining the maximum and minimum

  1. #1
    Junior Member
    Joined
    Mar 2008
    Posts
    30

    determining the maximum and minimum

    does the function y= (x-1)/x^2 have a maximum or a minimum because when i conduct a derivative test there signs change from positive to negative at x=2, but when i graph the function on a graphing calculator there doesn't seem to be a maximum
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Dec 2008
    Posts
    319
    The function does have a maximum, and we may calculate the derivative as follows:

    \begin{aligned}<br />
\frac{d}{dx}\frac{x-1}{x^2}&=\frac{d}{dx}(x^{-1}-x^{-2})\\<br />
&=\frac{d}{dx}(x^{-1})-\frac{d}{dx}(x^{-2})\\<br />
&=-x^{-2}+2x^{-3}.<br />
\end{aligned}<br />

    You are correct that the derivative changes sign at x=2, and as the function decreases after x=2, a maximum occurs here.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by wisezeta View Post
    does the function y= (x-1)/x^2 have a maximum or a minimum because when i conduct a derivative test there signs change from positive to negative at x=2, but when i graph the function on a graphing calculator there doesn't seem to be a maximum
    recheck your graph in closer window ... the max is subtle.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Mar 2008
    Posts
    30
    Ok, but what about the 3rd quadrant. Would there be an extrema there as well? It is similar to the extrema at x=2
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by wisezeta View Post
    Ok, but what about the 3rd quadrant. Would there be an extrema there as well? It is similar to the extrema at x=2
    no extrema in quad III ... y' = \frac{2-x}{x^3} < 0 for all x < 0
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: August 21st 2011, 12:12 PM
  2. Replies: 14
    Last Post: May 7th 2011, 05:58 PM
  3. Determining radius if material is a maximum
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 1st 2011, 02:13 PM
  4. Determining the Maximum
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: August 11th 2010, 08:48 AM
  5. Replies: 8
    Last Post: May 9th 2010, 04:05 AM

Search Tags


/mathhelpforum @mathhelpforum