Find the values of k for which 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
when x=k with (x-k) covered,
then comparing coefficients
Then i inserted these values into the equation the tried to differentiate it with the quotient rule. This is where i get stuck.
August 10th 2009, 09:35 PM
Not sure how it would work out, but I'd be tempted to differentiate using the product rule, the various terms being: , , .
Then I'd see what values of 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 .
Splitting into partial fractions is all very well, but I thought this was only usually done when integrating. In this case you're differentiating.
August 10th 2009, 09:46 PM
in the book this was the method they gave. but in any case, thanks!
August 10th 2009, 11:25 PM
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
so i'm stuck, what am i supposed to do with this?
August 11th 2009, 02:48 AM
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. ^^