if-f-f-x-x-2-x-1-find-f-0-

Question Number 171583 by mr W last updated on 18/Jun/22

i f f (f (x)) = x^{2} - x + 1, f i n d f (0) = ?

Commented by infinityaction last updated on 18/Jun/22

f (f (f (x))) = f {(x)}^{2} - f (x) + 1

f (f (0)) = 1

s o

f (1) = f {(0)}^{2} - f (0) + 1

t h a t i s

f {(0)}^{2} - f (0) - f (1) + 1 = 0 \dots . . 1

r e p e a t i n g t h e p r o c e s s s i n c e

f (f (1)) = 1

f (1) = f {(1)}^{2} - f (1) + 1

{[f (1) - 1]}^{2} = 0

f (1) = 1

p u t t h i s v a l u e i n {e q}^{n} . 1

f (0) {f (0) - 1} = 0

f (0) = 1 o r f (0) = 0

f (0) = 1

a n d

f (f (0)) = f (0)

f (f (0)) = 0 r e j e c t e d

i t i s r i g h t ? ? ? ?

Commented by mr W last updated on 18/Jun/22

v e r y n i c e s i r!

Commented by infinityaction last updated on 18/Jun/22

t h a n k y o u s i r

Answered by mr W last updated on 18/Jun/22

f (f (x)) = x^{2} - x + 1 > 0

f (f (f (x))) = f^{2} (x) - f (x) + 1

f (x^{2} - x + 1) = f^{2} (x) - f (x) + 1

w i t h x = 1 :

f (1) = f^{2} (1) - f (1) + 1 \Rightarrow f (1) = 1

w i t h x = 0 :

f (1) = f^{2} (0) - f (0) + 1 = 1 \Rightarrow f (0) = 1 o r 0 (r e j e c t e d)

Commented by infinityaction last updated on 18/Jun/22

a m a z i n g s o l u t i o n s i r

Answered by cortano1 last updated on 18/Jun/22

Facebook Tweet Pin