# Find the value of k

• Aug 10th 2009, 08:54 PM
arze
Find the value of k
Find the values of k for which $\displaystyle \frac{x}{(x+1)^2(x-k)}$ 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:
$\displaystyle y= \frac{x}{(x+1)^2(x-k)}= \frac{Ax+B}{(x+1)^2}+ \frac{C}{x-k}$
when x=k with (x-k) covered, $\displaystyle C=\frac{k}{(k+1)^2}$
$\displaystyle x=(Ax+B)(x-k)+(\frac{k}{(k+1)^2})(x^2+2x+1)$
then comparing coefficients
$\displaystyle A+\frac{k}{(k+1)^2}=0$
$\displaystyle A=-\frac{k}{(k+1)^2}$
$\displaystyle -kB+\frac{k}{(k+1)^2}=0$
$\displaystyle B=\frac{1}{(k+1)^2}$
Then i inserted these values into the equation the tried to differentiate it with the quotient rule. This is where i get stuck.
thanks(Bow)
• Aug 10th 2009, 09:35 PM
Matt Westwood
Not sure how it would work out, but I'd be tempted to differentiate using the product rule, the various terms being:
$\displaystyle x$, $\displaystyle \frac 1 {\left({x+1}\right)^2}$, $\displaystyle \frac 1 {x-k}$.

Then I'd see what values of $\displaystyle k$ 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 $\displaystyle k$.

Splitting into partial fractions is all very well, but I thought this was only usually done when integrating. In this case you're differentiating.
• Aug 10th 2009, 09:46 PM
arze
in the book this was the method they gave. but in any case, thanks!
• Aug 10th 2009, 11:25 PM
arze
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
$\displaystyle \frac{k \pm \sqrt{k^2-8k}}{4}$
so i'm stuck, what am i supposed to do with this?
• Aug 11th 2009, 02:48 AM
songoku
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. ^^