Results 1 to 5 of 5

Math Help - Find the value of k

  1. #1
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338

    Find the value of k

    Find the values of k for which \frac{x}{(x+1)^2(x-k)} has one stationary value.
    I attempted to differentiate the equation using the cover-up and quotient rule, but the further i go the longer and more messy the result becomes, it'll take too long to post all my workings here
    First:
    y= \frac{x}{(x+1)^2(x-k)}= \frac{Ax+B}{(x+1)^2}+ \frac{C}{x-k}
    when x=k with (x-k) covered,  C=\frac{k}{(k+1)^2}
     x=(Ax+B)(x-k)+(\frac{k}{(k+1)^2})(x^2+2x+1)
    then comparing coefficients
    A+\frac{k}{(k+1)^2}=0
    A=-\frac{k}{(k+1)^2}
    -kB+\frac{k}{(k+1)^2}=0
    B=\frac{1}{(k+1)^2}
    Then i inserted these values into the equation the tried to differentiate it with the quotient rule. This is where i get stuck.
    thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    736
    Thanks
    1
    Not sure how it would work out, but I'd be tempted to differentiate using the product rule, the various terms being:
    x, \frac 1 {\left({x+1}\right)^2}, \frac  1 {x-k}.

    Then I'd see what values of k make it so that what you get has a repeated root, i.e. only one value of the resulting equation equal to zero. It will probably be a quadratic in k.

    Splitting into partial fractions is all very well, but I thought this was only usually done when integrating. In this case you're differentiating.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338
    in the book this was the method they gave. but in any case, thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338
    I tried working it out using the product rule then equated that with 0. and tried to solve for x. Is that what you meant? because I got some values for x in terms of k in like
    \frac{k \pm \sqrt{k^2-8k}}{4}
    so i'm stuck, what am i supposed to do with this?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jul 2009
    Posts
    397
    Hi arze

    I think you are on the right track. I'm sure you got this equation : 2x^2 - kx + k = 0

    You are supposed to find the values of k, not the values of x so don't use quadratic formula.

    Instead, because the curve has one stationary value, set the discriminant = 0. If the discriminant = 0, the equation will have one root (only one value of x). Thus, the curve only has one stationary point. ^^
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: March 22nd 2011, 04:57 PM
  2. Replies: 2
    Last Post: July 5th 2010, 08:48 PM
  3. Replies: 1
    Last Post: February 17th 2010, 03:58 PM
  4. Replies: 0
    Last Post: June 16th 2009, 12:43 PM
  5. Replies: 2
    Last Post: April 6th 2009, 08:57 PM

/mathhelpforum @mathhelpforum