if-f-x-is-2-nd-digre-function-f-x-1-f-x-f-x-1-x-2-1-then-faind-f-2-

Question Number 173678 by mathlove last updated on 16/Jul/22

i f f (x) i s 2^{n d} d i g r e f u n c t i o n

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

t h e n f a i n d f (2) = ?

Answered by Rasheed.Sindhi last updated on 16/Jul/22

f (x) = {a x}^{2} + b x + c (s a y)

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

= {a x}^{2} - 2 a x + a + b x - b + c

= {a x}^{2} + (b - 2 a) x + a - b + c

f (x + 1) = a {(x + 1)}^{2} + b (x + 1) + c

= {a x}^{2} + 2 a x + a + b x + b + c

= {a x}^{2} + (b + 2 a) x + a + b + c

f (x - 1) + f (x) + f (x + 1)

= ({a x}^{2} + (b - 2 a) x + a - b + c)

+ ({a x}^{2} + b x + c)

+ ({a x}^{2} + (b + 2 a) x + a + b + c)

= x^{2} + 1

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

3 a = 1, 3 b = 0, 2 a + 3 c = 1

a = 1 / 3, b = 0, 2 (1 / 3) + 3 c = 1

c = 1 / 9

f (x) = {a x}^{2} + b x + c = \frac{1}{3} x^{2} + \frac{1}{9} = \frac{3 x^{2} + 1}{9}

f (2) = \frac{3 {(2)}^{2} + 1}{9} = \frac{13}{9}

Commented by mathlove last updated on 16/Jul/22

a l o t o f t h i n k s h e v e y o u l a u n g e l i f e

Commented by Rasheed.Sindhi last updated on 16/Jul/22

T han X, {you}^{'} ve too sir!

Commented by Tawa11 last updated on 16/Jul/22

Great sir

Answered by Rasheed.Sindhi last updated on 16/Jul/22

\begin{matrix} AnOther Way \end{matrix}

L e t f (x) = {a x}^{2} + b x + c

f (- 2) = 4 a - 2 b + c \dots \dots . . (i)

f (- 1) = a - b + c \dots \dots \dots . . (i i)

f (0) = c \dots \dots \dots \dots \dots \dots \dots \dots . (i i i)

f (1) = a + b + c \dots \dots \dots \dots \dots . (i v)

f (2) = 4 a + 2 b + c \dots \dots \dots . . (v)

(i) + (i i) + (i i i) :

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

5 a - 3 b + 3 c = 2 \dots \dots \dots \dots \dots \dots A

(i i) + (i i i) + (i v) :

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

2 a + 3 c = 1 \dots \dots \dots \dots \dots \dots \dots \dots . . B

(i i i) + (i v) + (v) :

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

5 a + 3 b + 3 c = 2 \dots \dots \dots \dots \dots \dots \dots C

C - A : 6 b = 0 \Rightarrow b = 0

\begin{matrix} A = C : 5 a + 3 c = 2 \\ B : 2 a + 3 c = 1 \end{matrix}} \Rightarrow {\begin{cases} 3 a = 1 \Rightarrow a = \frac{1}{3} \\ c = \frac{1}{9} \end{cases}

f (x) = {a x}^{2} + b x + c = \frac{1}{3} x^{2} + 0 x + \frac{1}{9}

= \frac{3 x^{2} + 1}{9}

Commented by mathlove last updated on 16/Jul/22

p l e a s s i r s e e t h i s w o r k

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

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

f (3 x) = f (2) \Rightarrow 3 x = 2 \Rightarrow x = \frac{2}{3}

w e h a v e x^{2} + 1

{(\frac{2}{3})}^{2} + 1 = \frac{4}{9} + 1 = \frac{13}{9}

Commented by Rasheed.Sindhi last updated on 16/Jul/22

I {d i d n}^{'} t u n d e r s t a n d :

f (x - 1) + f (x) + f (x + 1)

\overset{?}{=} f (x - 1 + x + x + 1) \overset{?}{=} f (2)

C o u l d y o u d e t e r m i n e f (3) e t c

u s i n g a b o v e a p p r o a c h ?

Commented by mathlove last updated on 17/Jul/22

t h a t s o k w r u n g w a y

Commented by Tawa11 last updated on 17/Jul/22

Great sir

Answered by pablo1234523 last updated on 16/Jul/22

f (x) = k (x - a) (x - b)

f (x - 1) = k (x - (a + 1)) (x - (b + 1))

f (x + 1) = k (x - (a - 1)) (x - (b - 1))

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

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

sum of constant terms :

a b + a b + a + b + 1 + a b - a - b + 1 = 1 / k

\Rightarrow 3 a b = (1 - 2 k) / k

\Rightarrow a b = \frac{1 - 2 k}{3 k}

sum of coefficient of x :

(a + b) + (a + b + 2) + (a + b - 2) = 0

3 (a + b) = 0

a = - b

sum of coefficient of x^{2} :

k + k + k = 1

\Rightarrow 3 k = 1

\Rightarrow k = \frac{1}{3}

∴ b^{2} = \frac{- 1}{3} \Rightarrow b = \frac{i}{\sqrt{3}} \Rightarrow a = - \frac{i}{\sqrt{3}}

f (x) = \frac{1}{3} (x - \frac{i}{\sqrt{3}}) (x + \frac{i}{\sqrt{3}}) = \frac{1}{3} (x^{2} + \frac{1}{3})

f (2) = \frac{1}{3} (4 + \frac{1}{3}) = \frac{13}{9}

Commented by mathlove last updated on 17/Jul/22

t h i n k s