f-x-1-f-x-2-4-f-x-1-f-x-f-0-0-

Question Number 57020 by 121194 last updated on 28/Mar/19

{[f (x + 1) - f (x)]}^{2} = 4 [f (x) - 1]

f (x) = ?

- - - - - - - - - - - - -

f (0) = 0

Commented by kaivan.ahmadi last updated on 28/Mar/19

t h e i n f o r m a t i o n i s n o t e n o u g h

Commented by kaivan.ahmadi last updated on 28/Mar/19

i f f (x) = k = c o n s t a n t \Rightarrow

4 (f (x) - 1) = 0 \Rightarrow f (x) = 1

i f f (x) = a x + b \Rightarrow f (x + 1) = a x + a + b \Rightarrow

a^{2} = 4 (a x + b - 1) \Rightarrow 4 a x + 4 b - a^{2} - 4 = 0 \Rightarrow

w e c a n t d e t e r m i n e a, b

Commented by prakash jain last updated on 29/Mar/19

x = 0

f {(1)}^{2} = - 4 \Rightarrow no such function exists