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Differentiation II

How to solve Exercise 8H Question 18? See attached workings. I could not arrive to a solution.
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Original post by JudaicImposter
How to solve Exercise 8H Question 18? See attached workings. I could not arrive to a solution.


Towards the end. Not sure where you got this from (red bit below)

It should be (1−x)(1+x)(x−k)(x+1)4−x(x+1)2=0\dfrac{(1-x)(1+x)(x-k)}{(x+1)^4} - \dfrac{x}{(x+1)^2} = 0

Hence (1−x)(x−k)−x(x+1)=0(1-x)(x-k) - x(x+1) = 0

Now you want this to have a single solution.

(edited 5 years ago)
Original post by RDKGames
Towards the end. Not sure where you got this from (red bit below)

It should be (1−x)(1+x)(x−k)(x+1)4−x(x+1)2=0\dfrac{(1-x)(1+x)(x-k)}{(x+1)^4} - \dfrac{x}{(x+1)^2} = 0

Hence (1−x)(x−k)−x(x+1)=0(1-x)(x-k) - x(x+1) = 0

Now you want this to have a single solution.



How to find what value of k gives the equation a single solution when you don't know what x is? If there is one solution the quadratic terms of x would be eliminated.
Original post by JudaicImposter
How to find what value of k gives the equation a single solution when you don't know what x is? If there is one solution the quadratic terms of x would be eliminated.


It's a quadratic in xx... if we impose a condition that it has a single solution, then have a think what that says about the discriminant.

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