if-f-x-x-2-bx-c-f-f-1-f-f-2-0-and-f-1-f-2-find-f-0-

Question Number 198166 by mr W last updated on 12/Oct/23

i f f (x) = x^{2} + b x + c

f (f (1)) = f (f (2)) = 0 a n d f (1) \neq f (2)

f i n d f (0) = ?

Answered by mr W last updated on 12/Oct/23

f (1) = 1^{2} + b \times 1 + c = b + c + 1

f (2) = 2^{2} + b \times 2 + c = 2 b + c + 4

f (1) a n d f (2) a r e t h e r o o t s o f f (x) = 0 .

f (1) + f (2) = - b

\Rightarrow b + c + 1 + 2 b + c + 4 = - b

\Rightarrow b = - \frac{2 c + 5}{4}

f (1) f (2) = c

\Rightarrow (b + c + 1) (2 b + c + 4) = c

\Rightarrow (- \frac{2 c + 5}{4} + c + 1) (- \frac{2 c + 5}{2} + c + 4) = c

\Rightarrow c = - \frac{3}{2}

f (0) = c = - \frac{3}{2} ✓

Answered by Frix last updated on 12/Oct/23

f (x) = x^{2} - \frac{x}{2} - \frac{3}{2}

\Rightarrow

f (0) = - \frac{3}{2}

[Solving this system for a, b, c, d :

f (1) = c \land f (2) = d \land f (c) = 0 \land f (d) = 0]

Commented by AST last updated on 12/Oct/23

F o r f (x) = x^{2} - 3 x + 3 \underline{+} i;

f (1) + f (2) = 3 = - b; (s i n c e f (1) a n d f (2) a r e z e r o s)

f (1) + f (2) w i l l n o t b e r e a l (= 3)

Commented by Frix last updated on 12/Oct/23

Thank you, I corrected .

Answered by AST last updated on 12/Oct/23

f (f (1)) = f (1 + b + c)

= 1 + 2 b^{2} + c^{2} + 3 b + 3 c + 3 b c = 0

f (1) f (2) = c \Rightarrow (1 + b + c) (4 + 2 b + c) = c

\Rightarrow 4 + 6 b + 2 b^{2} + 4 c + 3 b c + c^{2} = 0 \Rightarrow

(1 + 2 b^{2} + c^{2} + 3 b + 3 c + 3 b c) + 3 + 3 b + c = 0 \Rightarrow 3 b + c = - 3

\Rightarrow f (f (2)) = f (4 + 2 b + c) = f (1 - b)

f (1) + f (2) = - b \Rightarrow 1 + b + c + 1 - b = 2 + c = - b

\Rightarrow c + b = - 2 \Rightarrow b = - \frac{1}{2} \Rightarrow c = - \frac{3}{2}

f (x) = x^{2} - \frac{1}{2} x - \frac{3}{2} \Rightarrow f (0) = c = - \frac{3}{2}

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