Composite function

[man111]

If [tex]f(x) = x^2-2x[/tex]. Then no. of distinct value of [tex]c[/tex] in [tex]f(f(f(f(c)))) = 3[/tex]

[portokal]

If [tex]f(x) = x^2-2x[/tex]. Then no. of distinct value of [tex]c[/tex] in [tex]f(f(f(f(c)))) = 3[/tex]

if [tex]f(f(f(f(c)))) = 3[/tex]
means that [tex]f(f(f(c)))=-1[/tex] and therefor [tex]f(f(c))=1[/tex] ... [tex]f(c)=1\pm \sqrt{2}[/tex] ... [tex]c^2-2c=1\pm \sqrt{2}[/tex] ... but in one of the cases D<0 so there is only one solution and i think the eq is not that difficult ..

[mkmarinov]

Could it not be that [tex]f(f(f(c)))=3[/tex] ?
Then [tex]f(f(c))=3 \cup f(f(c))=-1[/tex]
If [tex]f(f(c))=3 => f(c)=3 \cup f(c)=-1[/tex], which yields 3 solutions.
If [tex]f(f(c))=-1 => f(c)=1[/tex], which yields 2 more solutions.

I still think that there should be a more elegant way of solving this problem. Partly reminds me of the second problem in this year's Bulgarian National Olympiad.

[portokal]

Could it not be that [tex]f(f(f(c)))=3[/tex] ?
Then [tex]f(f(c))=3 \cup f(f(c))=-1[/tex]
If [tex]f(f(c))=3 => f(c)=3 \cup f(c)=-1[/tex], which yields 3 solutions.
If [tex]f(f(c))=-1 => f(c)=1[/tex], which yields 2 more solutions.

I still think that there should be a more elegant way of solving this problem. Partly reminds me of the second problem in this year's Bulgarian National Olympiad.

you are right... it was the first idea that occured to my mind and i thought that if i can solve it in 30 sec then the solution should be either wrong or incomplete ... i was right
i am going to think about the problem for a lil while