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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 −2ax+a+bx−b+c +ax^2 +bx+c +ax^2 +2ax+a+bx+b+c =x^2 +1 3ax^2 +3bx+2a+3c=x^2 +1 3a=1⇒a=1/3 3b=0⇒b=0 2a+3c=1⇒2(1/3)+3c=1 ⇒c=1/9 f(x)=(1/3)x^2 +(1/9) f(2)=(1/3)(2)^2 +(1/9)=(4/3)+(1/9)=((13)/9)
$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c} \\ $$$$\mathrm{f}\left(\mathrm{x}−\mathrm{1}\right)+\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}+\mathrm{1}\right)=\mathrm{x}^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{f}\left(\mathrm{2}\right)=? \\ $$$$\: \\ $$$$\mathrm{a}\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{b}\left(\mathrm{x}−\mathrm{1}\right)+\mathrm{c} \\ $$$$\:\:\:\:\:\:+\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{a}\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{b}\left(\mathrm{x}+\mathrm{1}\right)+\mathrm{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{x}^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{ax}^{\mathrm{2}} −\cancel{\mathrm{2ax}}+\mathrm{a}+\mathrm{bx}−\cancel{\mathrm{b}}+\mathrm{c} \\ $$$$\:\:\:\:\:+\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{ax}^{\mathrm{2}} +\cancel{\mathrm{2ax}}+\mathrm{a}+\mathrm{bx}+\cancel{\mathrm{b}}+\mathrm{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{x}^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{3ax}^{\mathrm{2}} +\mathrm{3bx}+\mathrm{2a}+\mathrm{3c}=\mathrm{x}^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{3a}=\mathrm{1}\Rightarrow\mathrm{a}=\mathrm{1}/\mathrm{3} \\ $$$$\mathrm{3b}=\mathrm{0}\Rightarrow\mathrm{b}=\mathrm{0} \\ $$$$\mathrm{2a}+\mathrm{3c}=\mathrm{1}\Rightarrow\mathrm{2}\left(\mathrm{1}/\mathrm{3}\right)+\mathrm{3c}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{c}=\mathrm{1}/\mathrm{9} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$\mathrm{f}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{2}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{9}}=\frac{\mathrm{4}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{9}}=\frac{\mathrm{13}}{\mathrm{9}} \\ $$
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)
$${f}\left({x}−\mathrm{1}\right)+{f}\left({x}+\mathrm{1}\right)+{f}\left({x}\right)= \\ $$$${a}\left[\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({x}+\mathrm{1}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} \right]+{b}\left[\mathrm{3}{x}\right]+\mathrm{3}{c} \\ $$$$={a}\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}\right)+\mathrm{3}{bx}+\mathrm{3}{c} \\ $$$$\begin{cases}{\mathrm{3}{ax}^{\mathrm{2}} +\mathrm{3}{bx}+\left(\mathrm{2}{a}+\mathrm{3}{c}\right)={x}^{\mathrm{2}} +\mathrm{1}}\\{{a}=\frac{\mathrm{1}}{\mathrm{3}}\:;\:\:{b}=\mathrm{0}\:;\:\:\:\:{c}=\frac{\mathrm{1}}{\mathrm{9}}\:}\end{cases} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{9}} \\ $$$${donc}\:\:\:\:\:\:\:{f}\left(\mathrm{2}\right)=\frac{\mathrm{13}}{\mathrm{9}} \\ $$
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
$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{ax}}^{\mathrm{2}} +\boldsymbol{{bx}}+\boldsymbol{{c}} \\ $$$$\boldsymbol{{f}}\left(\mathrm{0}\right)=\boldsymbol{{c}} \\ $$$$\boldsymbol{{f}}\left(\mathrm{1}\right)=\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}} \\ $$$$\boldsymbol{{f}}\left(\mathrm{2}\right)=\mathrm{4}\boldsymbol{{a}}+\mathrm{2}\boldsymbol{{b}}+\boldsymbol{{c}} \\ $$$$\boldsymbol{{f}}\left(\mathrm{3}\right)=\mathrm{9}\boldsymbol{{a}}+\mathrm{3}\boldsymbol{{b}}+\boldsymbol{{c}} \\ $$$$\boldsymbol{{f}}\left(−\mathrm{1}\right)=\boldsymbol{{a}}−\boldsymbol{{b}}+\boldsymbol{{c}}\:\:\boldsymbol{{f}}\left(−\mathrm{1}\right)+\boldsymbol{{f}}\left(\mathrm{0}\right)+\boldsymbol{{f}}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$=>\mathrm{2}\boldsymbol{{a}}+\mathrm{2}\boldsymbol{{b}}+\boldsymbol{{c}}=\mathrm{1}\:\left(\mathrm{1}\right) \\ $$$$\boldsymbol{{f}}\left(\mathrm{0}\right)+\boldsymbol{{f}}\left(\mathrm{1}\right)+\boldsymbol{{f}}\left(\mathrm{2}\right)=\boldsymbol{{c}}+\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}+\mathrm{4}\boldsymbol{{a}}+\mathrm{2}\boldsymbol{{b}}+\boldsymbol{{c}}=\mathrm{2} \\ $$$$=>\mathrm{5}\boldsymbol{{a}}+\mathrm{3}\boldsymbol{{b}}+\mathrm{3}\boldsymbol{{c}}=\mathrm{2}\:\left(\mathrm{2}\right) \\ $$$$\boldsymbol{{f}}\left(\mathrm{1}\right)+\boldsymbol{{f}}\left(\mathrm{2}\right)+\boldsymbol{{f}}\left(\mathrm{3}\right)=\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}+\mathrm{4}\boldsymbol{{a}}+\mathrm{2}\boldsymbol{{b}}+\boldsymbol{{c}}+\mathrm{9}\boldsymbol{{a}}+\mathrm{3}\boldsymbol{{b}}+\boldsymbol{{c}}=\mathrm{5} \\ $$$$\mathrm{14}\boldsymbol{{a}}+\mathrm{6}\boldsymbol{{b}}+\mathrm{3}\boldsymbol{{c}}=\mathrm{5}\:\left(\mathrm{3}\right) \\ $$$$\begin{cases}{\mathrm{2}\boldsymbol{{a}}+\mathrm{2}\boldsymbol{{b}}+\boldsymbol{{c}}=\mathrm{1}}\\{\mathrm{14}\boldsymbol{{a}}+\mathrm{6}\boldsymbol{{b}}+\mathrm{3}\boldsymbol{{c}}=\mathrm{5}}\end{cases}\mathrm{5}\boldsymbol{{a}}+\mathrm{3}\boldsymbol{{b}}+\mathrm{3}\boldsymbol{{c}}=\mathrm{2} \\ $$$$\boldsymbol{{c}}=\mathrm{1}−\mathrm{2}\boldsymbol{{a}}−\mathrm{2}\boldsymbol{{b}}\: \\ $$$$\mathrm{5}\boldsymbol{{a}}+\mathrm{3}\boldsymbol{{b}}+\mathrm{3}−\mathrm{6}\boldsymbol{{a}}−\mathrm{6}\boldsymbol{{b}}=\mathrm{2} \\ $$$$=>−\boldsymbol{{a}}−\mathrm{3}\boldsymbol{{b}}=−\mathrm{1}\:\left(\mathrm{4}\right) \\ $$$$\mathrm{14}\boldsymbol{{a}}+\mathrm{6}\boldsymbol{{b}}+\mathrm{3}−\mathrm{6}\boldsymbol{{a}}−\mathrm{6}\boldsymbol{{b}}=\mathrm{5} \\ $$$$=>\mathrm{8}\boldsymbol{{a}}=\mathrm{2}=>\boldsymbol{{a}}=\frac{\mathrm{1}}{\mathrm{4}}\blacksquare{Moivre} \\ $$$$\boldsymbol{{b}}=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{2}}{\mathrm{9}} \\ $$$$\boldsymbol{{c}}=\mathrm{1}−\frac{\mathrm{4}}{\mathrm{9}}−\frac{\mathrm{2}}{\mathrm{4}}=\mathrm{1}−\frac{\mathrm{16}+\mathrm{18}}{\mathrm{36}}=\mathrm{1}−\frac{\mathrm{34}}{\mathrm{36}}=\mathrm{1}−\frac{\mathrm{17}}{\mathrm{18}}=\frac{\mathrm{1}}{\mathrm{18}} \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{{x}}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{9}}\boldsymbol{{x}}+\frac{\mathrm{1}}{\mathrm{18}} \\ $$$$\boldsymbol{{f}}\left(\mathrm{2}\right)=\mathrm{1}+\frac{\mathrm{4}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{18}}=\mathrm{1}+\frac{\mathrm{9}}{\mathrm{18}}=\frac{\mathrm{27}}{\mathrm{18}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\boldsymbol{{f}}\left(\mathrm{2}\right)=\frac{\mathrm{3}}{\mathrm{2}}\bigstar\:{celebre}\:{cedric}\:{junior} \\ $$$$ \\ $$
Commented by Rasheed.Sindhi last updated on 10/Feb/23
Good approach! (But check for errors.)
$${Good}\:{approach}! \\ $$$$\left({But}\:{check}\:{for}\:{errors}.\right) \\ $$
Commented by CElcedricjunior last updated on 12/Feb/23
ok thank!
$$\boldsymbol{{ok}}\:\boldsymbol{{thank}}! \\ $$

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