if f(x) is 2^(nd) digre function f(x−1)+f(x)+f(x+1)=x^2 +1 then faind f(2)=?


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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
	$determinant (((AnOther Way))) Let f(x)=ax^2 +bx+c f(−2)=4a−2b+c........(i) f(−1)=a−b+c...........(ii) f(0)=c.........................(iii) f(1)=a+b+c................(iv) f(2)=4a+2b+c...........(v) (i)+(ii)+(iii): f(−2)+f(−1)+f(0)=(−1)^2 +1 5a−3b+3c=2..................A (ii)+(iii)+(iv): f(−1)+f(0)+f(1)=0^2 +1 2a+3c=1..........................B (iii)+(iv)+(v): f(0)+f(1)+f(2)=1^2 +1=2 5a+3b+3c=2.....................C C−A: 6b=0⇒b=0 {: ((A=C: 5a+3c=2)),((B: 2a+3c=1)) }⇒ { ((3a=1⇒a=(1/3))),((c=(1/9))) :} f(x)=ax^2 +bx+c=(1/3)x^2 +0x+(1/9) =((3x^2 +1)/9)$
	$\begin{matrix} AnOther Way \end{matrix}$ $L e t f (x) = {a x}^{2} + b x + c$ $f (- 2) = 4 a - 2 b + c . . . . . . . . (i)$ $f (- 1) = a - b + c . . . . . . . . . . . (i i)$ $f (0) = c . . . . . . . . . . . . . . . . . . . . . . . . . (i i i)$ $f (1) = a + b + c . . . . . . . . . . . . . . . . (i v)$ $f (2) = 4 a + 2 b + c . . . . . . . . . . . (v)$ $(i) + (i i) + (i i i) :$ $f (- 2) + f (- 1) + f (0) = {(- 1)}^{2} + 1$ $5 a - 3 b + 3 c = 2 . . . . . . . . . . . . . . . . . . A$ $(i i) + (i i i) + (i v) :$ $f (- 1) + f (0) + f (1) = 0^{2} + 1$ $2 a + 3 c = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . B$ $(i i i) + (i v) + (v) :$ $f (0) + f (1) + f (2) = 1^{2} + 1 = 2$ $5 a + 3 b + 3 c = 2 . . . . . . . . . . . . . . . . . . . . . C$ $C - A : 6 b = 0 \Rightarrow b = 0$ $\begin{matrix} A = C : 5 a + 3 c = 2 \\ B : 2 a + 3 c = 1 \end{matrix}} \Rightarrow {\begin{cases} 3 a = 1 \Rightarrow a = \frac{1}{3} \\ c = \frac{1}{9} \end{cases}$ $f (x) = {a x}^{2} + b x + c = \frac{1}{3} x^{2} + 0 x + \frac{1}{9}$ $= \frac{3 x^{2} + 1}{9}$
	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

	$t h i n k s$
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