Find a and b such that y=(x+a)/(x+b)^2 has a local minimum/maximum at x=M and a point of inflection at x=N. (I need hints please)

Question

anonymous · Answer

1/ take y' and let y'=0 , replace x =M into the expression of y'=0 (you may have 1 equation)
2/ take y" and let y"=0, replace x =N, you may have the 2nd equation
2equations, 2 unknowns a and b.
--> it's solvable

anonymous · Answer

$$y' = \frac{ (x+b)^2 - (2)(x+a)(x+b) }{ (x+b)^4 }$$

anonymous · Answer

my mind tells me I can factor out in there but my sleepy sleepy mind is in doubt and researching

anonymous · Answer

$$y'=\dfrac{-2a+b-x}{(b+x)^3}$$
no condition since we have x = M not -b, --> y'=0 iff -2a+b -M =0 (this is the first equation)
$$y"=\dfrac{2(3a-2b+x)}{(b+x)^4}$$ so that y"=0 iff 3a-2b+N=0 (this is the second one)
solve for a, b from those equations

anonymous · Answer

show me your work, please. The answers should be respect to M, N

anonymous · Answer

I got a= N - 2M and b= 2N - 3M

anonymous · Answer

I just solved a twice verifying and got a = N - 2M

anonymous · Answer

Oh, I am sorry, you are correct