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Question Number 173678 by mathlove last updated on 16/Jul/22
if f(x) is 2^(nd) digre function f(x−1)+f(x)+f(x+1)=x^2 +1 then faind f(2)=?
$${if}\:{f}\left({x}\right)\:{is}\:\mathrm{2}^{{nd}} \:{digre}\:{function}\:\:\: \\ $$$${f}\left({x}−\mathrm{1}\right)+{f}\left({x}\right)+{f}\left({x}+\mathrm{1}\right)={x}^{\mathrm{2}} +\mathrm{1} \\ $$$${then}\:{faind}\:\:{f}\left(\mathrm{2}\right)=? \\ $$
Answered by Rasheed.Sindhi last updated on 16/Jul/22
f(x)=ax^2 +bx+c (say) f(x−1)=a(x−1)^2 +b(x−1)+c =ax^2 −2ax+a+bx−b+c =ax^2 +(b−2a)x+a−b+c f(x+1)=a(x+1)^2 +b(x+1)+c =ax^2 +2ax+a+bx+b+c =ax^2 +(b+2a)x+a+b+c f(x−1)+f(x)+f(x+1) =(ax^2 +(b−2a)x+a−b+c) +(ax^2 +bx+c) +(ax^2 +(b+2a)x+a+b+c) =x^2 +1 3ax^2 +3bx+2a+3c=x^2 +0x+1 3a=1,3b=0,2a+3c=1 a=1/3 ,b=0, 2(1/3)+3c=1 c=1/9 f(x)=ax^2 +bx+c=(1/3)x^2 +(1/9)=((3x^2 +1)/9) f(2)=((3(2)^2 +1)/9)=((13)/9)
$${f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c}\:\left({say}\right) \\ $$$${f}\left({x}−\mathrm{1}\right)={a}\left({x}−\mathrm{1}\right)^{\mathrm{2}} +{b}\left({x}−\mathrm{1}\right)+{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={ax}^{\mathrm{2}} −\mathrm{2}{ax}+{a}+{bx}−{b}+{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={ax}^{\mathrm{2}} +\left({b}−\mathrm{2}{a}\right){x}+{a}−{b}+{c} \\ $$$${f}\left({x}+\mathrm{1}\right)={a}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +{b}\left({x}+\mathrm{1}\right)+{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={ax}^{\mathrm{2}} +\mathrm{2}{ax}+{a}+{bx}+{b}+{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={ax}^{\mathrm{2}} +\left({b}+\mathrm{2}{a}\right){x}+{a}+{b}+{c} \\ $$$$ \\ $$$${f}\left({x}−\mathrm{1}\right)+{f}\left({x}\right)+{f}\left({x}+\mathrm{1}\right) \\ $$$$\:\:\:=\left({ax}^{\mathrm{2}} +\left({b}−\mathrm{2}{a}\right){x}+{a}−{b}+{c}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:+\left({ax}^{\mathrm{2}} +{bx}+{c}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left({ax}^{\mathrm{2}} +\left({b}+\mathrm{2}{a}\right){x}+{a}+{b}+{c}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={x}^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{3}{ax}^{\mathrm{2}} +\mathrm{3}{bx}+\mathrm{2}{a}+\mathrm{3}{c}={x}^{\mathrm{2}} +\mathrm{0}{x}+\mathrm{1} \\ $$$$\mathrm{3}{a}=\mathrm{1},\mathrm{3}{b}=\mathrm{0},\mathrm{2}{a}+\mathrm{3}{c}=\mathrm{1} \\ $$$${a}=\mathrm{1}/\mathrm{3}\:,{b}=\mathrm{0},\:\mathrm{2}\left(\mathrm{1}/\mathrm{3}\right)+\mathrm{3}{c}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{c}=\mathrm{1}/\mathrm{9} \\ $$$${f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c}=\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{9}}=\frac{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{9}} \\ $$$${f}\left(\mathrm{2}\right)=\frac{\mathrm{3}\left(\mathrm{2}\right)^{\mathrm{2}} +\mathrm{1}}{\mathrm{9}}=\frac{\mathrm{13}}{\mathrm{9}} \\ $$
Commented by mathlove last updated on 16/Jul/22
a lot of thinks heve you launge life
$${a}\:{lot}\:{of}\:{thinks}\:{heve}\:{you}\:{launge}\:{life} \\ $$
Commented by Rasheed.Sindhi last updated on 16/Jul/22
ThanX, you′ve too sir!
$$\mathbb{T}\boldsymbol{\mathrm{han}}\mathbb{X},\:\:\mathrm{you}'\mathrm{ve}\:\mathrm{too}\:\mathrm{sir}! \\ $$
Commented by Tawa11 last updated on 16/Jul/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
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)
$$\begin{array}{|c|}{\mathrm{AnOther}\:\:\mathrm{Way}}\\\hline\end{array} \\ $$$${Let}\:{f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$$ \\ $$$${f}\left(−\mathrm{2}\right)=\mathrm{4}{a}−\mathrm{2}{b}+{c}……..\left({i}\right) \\ $$$${f}\left(−\mathrm{1}\right)={a}−{b}+{c}………..\left({ii}\right) \\ $$$${f}\left(\mathrm{0}\right)={c}…………………….\left({iii}\right) \\ $$$${f}\left(\mathrm{1}\right)={a}+{b}+{c}…………….\left({iv}\right) \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{4}{a}+\mathrm{2}{b}+{c}………..\left({v}\right) \\ $$$$ \\ $$$$\left({i}\right)+\left({ii}\right)+\left({iii}\right): \\ $$$${f}\left(−\mathrm{2}\right)+{f}\left(−\mathrm{1}\right)+{f}\left(\mathrm{0}\right)=\left(−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{5}{a}−\mathrm{3}{b}+\mathrm{3}{c}=\mathrm{2}………………{A} \\ $$$$\left({ii}\right)+\left({iii}\right)+\left({iv}\right): \\ $$$${f}\left(−\mathrm{1}\right)+{f}\left(\mathrm{0}\right)+{f}\left(\mathrm{1}\right)=\mathrm{0}^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{2}{a}+\mathrm{3}{c}=\mathrm{1}……………………..{B} \\ $$$$\left({iii}\right)+\left({iv}\right)+\left({v}\right): \\ $$$${f}\left(\mathrm{0}\right)+{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)=\mathrm{1}^{\mathrm{2}} +\mathrm{1}=\mathrm{2} \\ $$$$\mathrm{5}{a}+\mathrm{3}{b}+\mathrm{3}{c}=\mathrm{2}…………………{C} \\ $$$$ \\ $$$${C}−{A}:\:\mathrm{6}{b}=\mathrm{0}\Rightarrow{b}=\mathrm{0} \\ $$$$\left.\begin{matrix}{{A}={C}:\:\:\:\:\mathrm{5}{a}+\mathrm{3}{c}=\mathrm{2}}\\{{B}:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{a}+\mathrm{3}{c}=\mathrm{1}}\end{matrix}\right\}\Rightarrow\begin{cases}{\mathrm{3}{a}=\mathrm{1}\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{3}}}\\{{c}=\frac{\mathrm{1}}{\mathrm{9}}}\end{cases} \\ $$$${f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c}=\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{2}} +\mathrm{0}{x}+\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{9}} \\ $$
Commented by mathlove last updated on 16/Jul/22
pleas sir see this work f(x−1)+f(x)+f(x−1)=f(2) f(x−1+x+x+1)=f(2) f(3x)=f(2)⇒3x=2⇒x=(2/3) we have x^2 +1 ((2/3))^2 +1=(4/9)+1=((13)/9)
$${pleas}\:{sir}\:{see}\:{this}\:{work} \\ $$$${f}\left({x}−\mathrm{1}\right)+{f}\left({x}\right)+{f}\left({x}−\mathrm{1}\right)={f}\left(\mathrm{2}\right) \\ $$$${f}\left({x}−\mathrm{1}+{x}+{x}+\mathrm{1}\right)={f}\left(\mathrm{2}\right) \\ $$$${f}\left(\mathrm{3}{x}\right)={f}\left(\mathrm{2}\right)\Rightarrow\mathrm{3}{x}=\mathrm{2}\Rightarrow{x}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${we}\:{have}\:{x}^{\mathrm{2}} +\mathrm{1} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{1}=\frac{\mathrm{4}}{\mathrm{9}}+\mathrm{1}=\frac{\mathrm{13}}{\mathrm{9}} \\ $$
Commented by Rasheed.Sindhi last updated on 16/Jul/22
I didn′t understand: f(x−1)+f(x)+f(x+1) =^(?) f(x−1+x+x+1)=^(?) f(2) Could you determine f(3) etc using above approach?
$${I}\:{didn}'{t}\:{understand}: \\ $$$${f}\left({x}−\mathrm{1}\right)+{f}\left({x}\right)+{f}\left({x}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\overset{?} {=}{f}\left({x}−\mathrm{1}+{x}+{x}+\mathrm{1}\right)\overset{?} {=}{f}\left(\mathrm{2}\right) \\ $$$${Could}\:{you}\:{determine}\:{f}\left(\mathrm{3}\right)\:{etc} \\ $$$${using}\:{above}\:{approach}? \\ $$
Commented by mathlove last updated on 17/Jul/22
thats ok wrung way
$${thats}\:{ok}\:{wrung}\:{way} \\ $$
Commented by Tawa11 last updated on 17/Jul/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
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 ⇒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: ab+ab+a+b+1+ab−a−b+1=1/k ⇒3ab=(1−2k)/k ⇒ab=((1−2k)/(3k)) 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 ⇒3k=1 ⇒k=(1/3) ∴ b^2 =((−1)/3)⇒b=(i/( (√3)))⇒a=−(i/( (√3))) f(x)=(1/3)(x−(i/( (√3))))(x+(i/( (√3))))=(1/3)(x^2 +(1/3)) f(2)=(1/3)(4+(1/3))=((13)/9)
$${f}\left({x}\right)={k}\left({x}−{a}\right)\left({x}−{b}\right) \\ $$$${f}\left({x}−\mathrm{1}\right)={k}\left({x}−\left({a}+\mathrm{1}\right)\right)\left({x}−\left({b}+\mathrm{1}\right)\right) \\ $$$${f}\left({x}+\mathrm{1}\right)={k}\left({x}−\left({a}−\mathrm{1}\right)\right)\left({x}−\left({b}−\mathrm{1}\right)\right) \\ $$$${f}\left({x}−\mathrm{1}\right)+{f}\left({x}\right)+{f}\left({x}+\mathrm{1}\right)={x}^{\mathrm{2}} +\mathrm{1} \\ $$$$\Rightarrow{k}\left({x}−{a}\right)\left({x}−{b}\right)+{k}\left({x}−\left({a}+\mathrm{1}\right)\right)\left({x}−\left({b}+\mathrm{1}\right)\right)+{k}\left({x}−\left({a}−\mathrm{1}\right)\right)\left({x}−\left({b}−\mathrm{1}\right)\right)={x}^{\mathrm{2}} +\mathrm{1} \\ $$$$ \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{constant}\:\mathrm{terms}: \\ $$$${ab}+{ab}+{a}+{b}+\mathrm{1}+{ab}−{a}−{b}+\mathrm{1}=\mathrm{1}/{k} \\ $$$$\Rightarrow\mathrm{3}{ab}=\left(\mathrm{1}−\mathrm{2}{k}\right)/{k} \\ $$$$\Rightarrow{ab}=\frac{\mathrm{1}−\mathrm{2}{k}}{\mathrm{3}{k}} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{coefficient}\:\mathrm{of}\:{x}: \\ $$$$\left({a}+{b}\right)+\left({a}+{b}+\mathrm{2}\right)+\left({a}+{b}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\mathrm{3}\left({a}+{b}\right)=\mathrm{0} \\ $$$${a}=−{b} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{coefficient}\:\mathrm{of}\:{x}^{\mathrm{2}} : \\ $$$${k}+{k}+{k}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{3}{k}=\mathrm{1} \\ $$$$\Rightarrow{k}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$$$\therefore\:{b}^{\mathrm{2}} =\frac{−\mathrm{1}}{\mathrm{3}}\Rightarrow{b}=\frac{{i}}{\:\sqrt{\mathrm{3}}}\Rightarrow{a}=−\frac{{i}}{\:\sqrt{\mathrm{3}}} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{3}}\left({x}−\frac{{i}}{\:\sqrt{\mathrm{3}}}\right)\left({x}+\frac{{i}}{\:\sqrt{\mathrm{3}}}\right)=\frac{\mathrm{1}}{\mathrm{3}}\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$${f}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{4}+\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{13}}{\mathrm{9}} \\ $$
Commented by mathlove last updated on 17/Jul/22
thinks
$${thinks} \\ $$

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