# Thread: Find the value of k

1. ## 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

2. 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.

3. in the book this was the method they gave. but in any case, thanks!

4. 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?

5. 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. ^^