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

Question Number 1147 by prakash jain last updated on 04/Jul/15

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

f (0) = ?

Commented by prakash jain last updated on 04/Jul/15

f (f (0)) = 1

Commented by 123456 last updated on 04/Jul/15

f ∙ f = x^{2} - x + 1

= x^{2} - 2 \times \frac{1}{2} \times x + \frac{1}{4} + 1 - \frac{1}{4}

= {(x - \frac{1}{2})}^{2} + \frac{3}{4}

= \frac{{(2 x - 1)}^{2} + 3}{4}

Δ = {(- 1)}^{2} - 4 (1) (1) = 1 - 4 = - 3

x = \frac{1 \pm ı \sqrt{3}}{2}

Answered by 123456 last updated on 04/Jul/15

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

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

f (f (0)) = 1

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

f (f (1)) = 1

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

f (1) = 1

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

f (0) = 0 \lor f (0) = 1

Commented by prakash jain last updated on 05/Jul/15

If f (0) = 0

f (f (0)) = 1 \Rightarrow f (0) = 1

Is this a contradictio n ?

Commented by 123456 last updated on 05/Jul/15

yes :)

then f (0) = 1

Commented by prakash jain last updated on 16/Jul/15

In the question the function is defined as

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

So

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

Commented by Rasheed Soomro last updated on 12/Aug/15

D e a r P a r k a s h

T H A N K S f o r g u i d a n c e . I w a s w r o n g . I d i d n o t u n d e r s t a n d

t h e l o g i c o f M r 123456 a t f i r s t! B u t n o w I d o (w i t h y o u r h e l p) .

A p p r e c i a t i o n f o r 123456