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Given-f-R-R-is-a-quadratic-polynomial-f-1-1-f-2-1-2-and-f-3-1-3-Find-f-4-

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Question Number 200167 by hardmath last updated on 15/Nov/23
Given f:R→R is a quadratic polynomial f(1) = 1 , f(2) = (1/2) and f(3) = (1/3) Find: f(4) = ?
$$\mathrm{Given}\:\:\:\mathrm{f}:\mathbb{R}\rightarrow\mathbb{R}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{quadratic}\:\mathrm{polynomial} \\ $$$$\mathrm{f}\left(\mathrm{1}\right)\:=\:\mathrm{1}\:,\:\mathrm{f}\left(\mathrm{2}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\mathrm{and}\:\:\mathrm{f}\left(\mathrm{3}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{Find}:\:\:\mathrm{f}\left(\mathrm{4}\right)\:=\:? \\ $$
Answered by witcher3 last updated on 15/Nov/23
⇔f(1)−1=2f(2)−1=3f(3)−1=0 let xf(x)−1=g(x)⇒deg(g)=3 g(1)=g(2)=g(3)=0 ⇒g(x)=a(x−1)(x−2)(x−3) xf(x)=1+g(x) ⇒1+g(0)=0⇒−6a+1=0 a=(1/6) f(x)=(1/x)((1/6)(x−3)(x−2)(x−1)+1) =g′(x)=(x^2 /6)−x+((11)/6)=f(x)
$$\Leftrightarrow\mathrm{f}\left(\mathrm{1}\right)−\mathrm{1}=\mathrm{2f}\left(\mathrm{2}\right)−\mathrm{1}=\mathrm{3f}\left(\mathrm{3}\right)−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{let}\:\mathrm{xf}\left(\mathrm{x}\right)−\mathrm{1}=\mathrm{g}\left(\mathrm{x}\right)\Rightarrow\mathrm{deg}\left(\mathrm{g}\right)=\mathrm{3} \\ $$$$\mathrm{g}\left(\mathrm{1}\right)=\mathrm{g}\left(\mathrm{2}\right)=\mathrm{g}\left(\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{g}\left(\mathrm{x}\right)=\mathrm{a}\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}−\mathrm{3}\right) \\ $$$$\mathrm{xf}\left(\mathrm{x}\right)=\mathrm{1}+\mathrm{g}\left(\mathrm{x}\right) \\ $$$$\Rightarrow\mathrm{1}+\mathrm{g}\left(\mathrm{0}\right)=\mathrm{0}\Rightarrow−\mathrm{6a}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{a}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{x}}\left(\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{x}−\mathrm{3}\right)\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}−\mathrm{1}\right)+\mathrm{1}\right) \\ $$$$=\mathrm{g}'\left(\mathrm{x}\right)=\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{6}}−\mathrm{x}+\frac{\mathrm{11}}{\mathrm{6}}=\mathrm{f}\left(\mathrm{x}\right) \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 15/Nov/23
let f(x)=ax^2 +bx+c f(1)=a+b+c=1 f(2)=4a+2b+c=(1/2) f(3)=9a+3b+c=(1/3) f(2)−f(1)=3a+b=−(1/2) 3(f(2)−f(1))=9a+3b=−(3/2) f(3)−3(f(2)−f(1))=c=(1/3)−(−(3/2))=((11)/6) f(1)⇒a+b=1−((11)/6)=−(5/6) 2a+2b=−(5/3)..................(i) f(2)⇒4a+2b=(1/2)−((11)/6)=−(4/3)......(ii) (ii)−(ii): 2a=−(4/3)−(−(5/3))=(1/3) a=(1/6) f(1): (1/6)+b+((11)/6)=1⇒b=1−2=−1 f(4)=16a+4b+c=16((1/6))+4(−1)+((11)/6) =((16−24+11)/6)=(3/6)=(1/2) ✓
$${let}\:{f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$$\:\:\:\:\:\:{f}\left(\mathrm{1}\right)={a}+{b}+{c}=\mathrm{1} \\ $$$$\:\:\:\:\:{f}\left(\mathrm{2}\right)=\mathrm{4}{a}+\mathrm{2}{b}+{c}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:{f}\left(\mathrm{3}\right)=\mathrm{9}{a}+\mathrm{3}{b}+{c}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${f}\left(\mathrm{2}\right)−{f}\left(\mathrm{1}\right)=\mathrm{3}{a}+{b}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{3}\left({f}\left(\mathrm{2}\right)−{f}\left(\mathrm{1}\right)\right)=\mathrm{9}{a}+\mathrm{3}{b}=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${f}\left(\mathrm{3}\right)−\mathrm{3}\left({f}\left(\mathrm{2}\right)−{f}\left(\mathrm{1}\right)\right)={c}=\frac{\mathrm{1}}{\mathrm{3}}−\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\mathrm{11}}{\mathrm{6}} \\ $$$${f}\left(\mathrm{1}\right)\Rightarrow{a}+{b}=\mathrm{1}−\frac{\mathrm{11}}{\mathrm{6}}=−\frac{\mathrm{5}}{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{a}+\mathrm{2}{b}=−\frac{\mathrm{5}}{\mathrm{3}}………………\left({i}\right) \\ $$$${f}\left(\mathrm{2}\right)\Rightarrow\mathrm{4}{a}+\mathrm{2}{b}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{11}}{\mathrm{6}}=−\frac{\mathrm{4}}{\mathrm{3}}……\left({ii}\right) \\ $$$$\left({ii}\right)−\left({ii}\right):\:\:\:\mathrm{2}{a}=−\frac{\mathrm{4}}{\mathrm{3}}−\left(−\frac{\mathrm{5}}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${f}\left(\mathrm{1}\right):\:\:\:\frac{\mathrm{1}}{\mathrm{6}}+{b}+\frac{\mathrm{11}}{\mathrm{6}}=\mathrm{1}\Rightarrow{b}=\mathrm{1}−\mathrm{2}=−\mathrm{1} \\ $$$${f}\left(\mathrm{4}\right)=\mathrm{16}{a}+\mathrm{4}{b}+{c}=\mathrm{16}\left(\frac{\mathrm{1}}{\mathrm{6}}\right)+\mathrm{4}\left(−\mathrm{1}\right)+\frac{\mathrm{11}}{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{16}−\mathrm{24}+\mathrm{11}}{\mathrm{6}}=\frac{\mathrm{3}}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{2}}\:\checkmark \\ $$
Answered by Rasheed.Sindhi last updated on 15/Nov/23
letf(x)= ax^2 +bx+c determinant (((1)),a,b,c),( , ,a,(a+b)),( ,a,(a+b),(f(1)=a+b+c=1))) a=1−b−c determinant (((2)),(1−b−c),b,c),( , ,(2−2b−2c),(4−2b−4c)),( ,(1−b−c),(2−b−2c),(f(2)=4−2b−3c=(1/2)))) −4+2b+3c=−(1/2)⇒b=((7/2)−3c)/2=(7/4)−(3/2)c a=1−b−c=1−((7/4)−(3/2)c)−c=−(3/4)+(c/2) determinant (((3)),(−(3/4)+(c/2)),((7/4)−(3/2)c),c),( , ,(−(9/4)+((3c)/2)),(−(3/2))),( ,(−(3/4)+(c/2)),(−(1/2)),(f(3)=−(3/2)+c=(1/3)))) c=(1/3)+(3/2)=((11)/6) a=−(3/4)+(c/2)=−(3/4)+(1/2)∙((11)/6)=((−9+11)/(12))=(2/(12))=(1/6) b=(7/4)−(3/2)c=(7/4)−(3/2)(((11)/6))=((7−11)/4)=−1 f(x)=(1/6)x^2 −x+((11)/6) f(4)=? determinant (((4)),(1/6),(−1),((11)/6)),( , ,(2/3),(−(4/3))),( ,(1/6),(−(1/3)),(f(4)=(1/2))))
$${letf}\left({x}\right)=\:{ax}^{\mathrm{2}} +{bx}+{c} \\ $$$$\begin{array}{|c|c|c|}{\left.\mathrm{1}\right)}&\hline{{a}}&\hline{{b}}&\hline{{c}}\\{\:}&\hline{\:}&\hline{{a}}&\hline{{a}+{b}}\\{\:}&\hline{{a}}&\hline{{a}+{b}}&\hline{{f}\left(\mathrm{1}\right)={a}+{b}+{c}=\mathrm{1}}\\\hline\end{array}\: \\ $$$${a}=\mathrm{1}−{b}−{c} \\ $$$$\begin{array}{|c|c|c|}{\left.\mathrm{2}\right)}&\hline{\mathrm{1}−{b}−{c}}&\hline{{b}}&\hline{{c}}\\{\:}&\hline{\:}&\hline{\mathrm{2}−\mathrm{2}{b}−\mathrm{2}{c}}&\hline{\mathrm{4}−\mathrm{2}{b}−\mathrm{4}{c}}\\{\:}&\hline{\mathrm{1}−{b}−{c}}&\hline{\mathrm{2}−{b}−\mathrm{2}{c}}&\hline{{f}\left(\mathrm{2}\right)=\mathrm{4}−\mathrm{2}{b}−\mathrm{3}{c}=\frac{\mathrm{1}}{\mathrm{2}}}\\\hline\end{array}\: \\ $$$$−\mathrm{4}+\mathrm{2}{b}+\mathrm{3}{c}=−\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{b}=\left(\frac{\mathrm{7}}{\mathrm{2}}−\mathrm{3}{c}\right)/\mathrm{2}=\frac{\mathrm{7}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{2}}{c} \\ $$$${a}=\mathrm{1}−{b}−{c}=\mathrm{1}−\left(\frac{\mathrm{7}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{2}}{c}\right)−{c}=−\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{c}}{\mathrm{2}} \\ $$$$\begin{array}{|c|c|c|}{\left.\mathrm{3}\right)}&\hline{−\frac{\mathrm{3}}{\mathrm{4}}+\frac{{c}}{\mathrm{2}}}&\hline{\frac{\mathrm{7}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{2}}{c}}&\hline{{c}}\\{\:}&\hline{\:}&\hline{−\frac{\mathrm{9}}{\mathrm{4}}+\frac{\mathrm{3}{c}}{\mathrm{2}}}&\hline{−\frac{\mathrm{3}}{\mathrm{2}}}\\{\:}&\hline{−\frac{\mathrm{3}}{\mathrm{4}}+\frac{{c}}{\mathrm{2}}}&\hline{−\frac{\mathrm{1}}{\mathrm{2}}}&\hline{{f}\left(\mathrm{3}\right)=−\frac{\mathrm{3}}{\mathrm{2}}+{c}=\frac{\mathrm{1}}{\mathrm{3}}}\\\hline\end{array} \\ $$$$\: \\ $$$$\:{c}=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{11}}{\mathrm{6}} \\ $$$${a}=−\frac{\mathrm{3}}{\mathrm{4}}+\frac{{c}}{\mathrm{2}}=−\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\mathrm{11}}{\mathrm{6}}=\frac{−\mathrm{9}+\mathrm{11}}{\mathrm{12}}=\frac{\mathrm{2}}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${b}=\frac{\mathrm{7}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{2}}{c}=\frac{\mathrm{7}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{\mathrm{11}}{\mathrm{6}}\right)=\frac{\mathrm{7}−\mathrm{11}}{\mathrm{4}}=−\mathrm{1} \\ $$$$ \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{2}} −{x}+\frac{\mathrm{11}}{\mathrm{6}} \\ $$$${f}\left(\mathrm{4}\right)=? \\ $$$$\begin{array}{|c|c|c|}{\left.\mathrm{4}\right)}&\hline{\frac{\mathrm{1}}{\mathrm{6}}}&\hline{−\mathrm{1}}&\hline{\frac{\mathrm{11}}{\mathrm{6}}}\\{\:}&\hline{\:}&\hline{\frac{\mathrm{2}}{\mathrm{3}}}&\hline{−\frac{\mathrm{4}}{\mathrm{3}}}\\{\:}&\hline{\frac{\mathrm{1}}{\mathrm{6}}}&\hline{−\frac{\mathrm{1}}{\mathrm{3}}}&\hline{{f}\left(\mathrm{4}\right)=\frac{\mathrm{1}}{\mathrm{2}}}\\\hline\end{array} \\ $$$$ \\ $$

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