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Question Number 186751 by 073 last updated on 09/Feb/23

	Answered by Rasheed.Sindhi last updated on 09/Feb/23

	$f (x) = {ax}^{2} + bx + c$ $f (x - 1) + f (x) + f (x + 1) = x^{2} + 1$ $f (2) = ?$ $a {(x - 1)}^{2} + b (x - 1) + c$ $+ {ax}^{2} + bx + c$ $+ a {(x + 1)}^{2} + b (x + 1) + c$ $= x^{2} + 1$ ${ax}^{2} - 2 ax + a + bx - b + c$ $+ {ax}^{2} + bx + c$ $+ {ax}^{2} + 2 ax + a + bx + b + c$ $= x^{2} + 1$ ${3 ax}^{2} + 3 bx + 2 a + 3 c = x^{2} + 1$ $3 a = 1 \Rightarrow a = 1 / 3$ $3 b = 0 \Rightarrow b = 0$ $2 a + 3 c = 1 \Rightarrow 2 (1 / 3) + 3 c = 1$ $\Rightarrow c = 1 / 9$ $f (x) = \frac{1}{3} x^{2} + \frac{1}{9}$ $f (2) = \frac{1}{3} {(2)}^{2} + \frac{1}{9} = \frac{4}{3} + \frac{1}{9} = \frac{13}{9}$
	Answered by a.lgnaoui last updated on 09/Feb/23
	$f(x−1)+f(x+1)+f(x)= a[(x−1)^2 +(x+1)^2 +x^2 ]+b[3x]+3c =a(3x^2 +2)+3bx+3c { ((3ax^2 +3bx+(2a+3c)=x^2 +1)),((a=(1/3) ; b=0 ; c=(1/9) )) :} f(x)=(1/3)x^2 +(1/9) donc f(2)=((13)/9)$
	$f (x - 1) + f (x + 1) + f (x) =$ $a [{(x - 1)}^{2} + {(x + 1)}^{2} + x^{2}] + b [3 x] + 3 c$ $= a (3 x^{2} + 2) + 3 b x + 3 c$ ${\begin{cases} 3 {a x}^{2} + 3 b x + (2 a + 3 c) = x^{2} + 1 \\ a = \frac{1}{3}; b = 0; c = \frac{1}{9} \end{cases}$ $f (x) = \frac{1}{3} x^{2} + \frac{1}{9}$ $d o n c f (2) = \frac{13}{9}$
	Answered by CElcedricjunior last updated on 10/Feb/23
	$f(x)=ax^2 +bx+c f(0)=c f(1)=a+b+c f(2)=4a+2b+c f(3)=9a+3b+c f(−1)=a−b+c f(−1)+f(0)+f(1)=1 =>2a+2b+c=1 (1) f(0)+f(1)+f(2)=c+a+b+c+4a+2b+c=2 =>5a+3b+3c=2 (2) f(1)+f(2)+f(3)=a+b+c+4a+2b+c+9a+3b+c=5 14a+6b+3c=5 (3) { ((2a+2b+c=1)),((14a+6b+3c=5)) :}5a+3b+3c=2 c=1−2a−2b 5a+3b+3−6a−6b=2 =>−a−3b=−1 (4) 14a+6b+3−6a−6b=5 =>8a=2=>a=(1/4)■Moivre b=(1/3)(1−(1/3))=(2/9) c=1−(4/9)−(2/4)=1−((16+18)/(36))=1−((34)/(36))=1−((17)/(18))=(1/(18)) f(x)=(1/4)x^2 +(2/9)x+(1/(18)) f(2)=1+(4/9)+(1/(18))=1+(9/(18))=((27)/(18))=(3/2) f(2)=(3/2)★ celebre cedric junior$
	$f (x) = {a x}^{2} + b x + c$ $f (0) = c$ $f (1) = a + b + c$ $f (2) = 4 a + 2 b + c$ $f (3) = 9 a + 3 b + c$ $f (- 1) = a - b + c f (- 1) + f (0) + f (1) = 1$ $=> 2 a + 2 b + c = 1 (1)$ $f (0) + f (1) + f (2) = c + a + b + c + 4 a + 2 b + c = 2$ $=> 5 a + 3 b + 3 c = 2 (2)$ $f (1) + f (2) + f (3) = a + b + c + 4 a + 2 b + c + 9 a + 3 b + c = 5$ $14 a + 6 b + 3 c = 5 (3)$ ${\begin{cases} 2 a + 2 b + c = 1 \\ 14 a + 6 b + 3 c = 5 \end{cases} 5 a + 3 b + 3 c = 2$ $c = 1 - 2 a - 2 b$ $5 a + 3 b + 3 - 6 a - 6 b = 2$ $=> - a - 3 b = - 1 (4)$ $14 a + 6 b + 3 - 6 a - 6 b = 5$ $=> 8 a = 2 => a = \frac{1}{4} ◼ M o i v r e$ $b = \frac{1}{3} (1 - \frac{1}{3}) = \frac{2}{9}$ $c = 1 - \frac{4}{9} - \frac{2}{4} = 1 - \frac{16 + 18}{36} = 1 - \frac{34}{36} = 1 - \frac{17}{18} = \frac{1}{18}$ $f (x) = \frac{1}{4} x^{2} + \frac{2}{9} x + \frac{1}{18}$ $f (2) = 1 + \frac{4}{9} + \frac{1}{18} = 1 + \frac{9}{18} = \frac{27}{18} = \frac{3}{2}$ $f (2) = \frac{3}{2} ★ c e l e b r e c e d r i c j u n i o r$
	Commented by Rasheed.Sindhi last updated on 10/Feb/23

	$G o o d a p p r o a c h!$ $(B u t c h e c k f o r e r r o r s .)$
	Commented by CElcedricjunior last updated on 12/Feb/23

	$o k t h a n k!$
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