Question Number 82701 by jagoll last updated on 23/Feb/20
$$\mathrm{2}{f}\left({x}−\mathrm{1}\right)\:+\mathrm{3}{f}\left({x}+\mathrm{1}\right)\overset{\:} {\:}=\:\mathrm{3}{x}^{\mathrm{2}} −\mathrm{5}{x} \\ $$$${find}\:{f}\left({x}\right) \\ $$
Answered by TANMAY PANACEA last updated on 23/Feb/20
![f(x)=ax^2 +bx+c 2f(x−1)+3f(x+1) 2[a(x−1)^2 +b(x−1)+c]+3[a(x+1)^2 +b(x+1)+c] =[2ax^2 −4ax+2a+2bx−2b+2c+3ax^2 +6ax+3a+3bx+3b+3c] =x^2 (5a)+x(2a+5b)+(2a−2b+2c+3a+3b+3c) 5a=3→a=(3/5) 2a+5b=−5 2×(3/5)+5b=−5 5b=((−31)/5)→b=((−31)/(25)) 5a+b+5c=0 5×(3/5)+((−31)/(25))+5c=0 75−31+125c=0 →c=((−44)/(125)) f(x)=(3/5)x^2 −((31)/(25))x−((44)/(125)) pls check](https://www.tinkutara.com/question/Q82705.png)
$${f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$$\mathrm{2}{f}\left({x}−\mathrm{1}\right)+\mathrm{3}\boldsymbol{{f}}\left({x}+\mathrm{1}\right) \\ $$$$\mathrm{2}\left[{a}\left({x}−\mathrm{1}\right)^{\mathrm{2}} +{b}\left({x}−\mathrm{1}\right)+{c}\right]+\mathrm{3}\left[{a}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +{b}\left({x}+\mathrm{1}\right)+{c}\right] \\ $$$$=\left[\mathrm{2}{ax}^{\mathrm{2}} −\mathrm{4}{ax}+\mathrm{2}{a}+\mathrm{2}{bx}−\mathrm{2}{b}+\mathrm{2}{c}+\mathrm{3}{ax}^{\mathrm{2}} +\mathrm{6}{ax}+\mathrm{3}{a}+\mathrm{3}{bx}+\mathrm{3}{b}+\mathrm{3}{c}\right] \\ $$$$={x}^{\mathrm{2}} \left(\mathrm{5}{a}\right)+{x}\left(\mathrm{2}{a}+\mathrm{5}{b}\right)+\left(\mathrm{2}{a}−\mathrm{2}{b}+\mathrm{2}{c}+\mathrm{3}{a}+\mathrm{3}{b}+\mathrm{3}{c}\right) \\ $$$$\mathrm{5}{a}=\mathrm{3}\rightarrow{a}=\frac{\mathrm{3}}{\mathrm{5}}\:\:\:\mathrm{2}{a}+\mathrm{5}{b}=−\mathrm{5} \\ $$$$\mathrm{2}×\frac{\mathrm{3}}{\mathrm{5}}+\mathrm{5}{b}=−\mathrm{5} \\ $$$$\mathrm{5}{b}=\frac{−\mathrm{31}}{\mathrm{5}}\rightarrow\boldsymbol{{b}}=\frac{−\mathrm{31}}{\mathrm{25}} \\ $$$$\mathrm{5}\boldsymbol{{a}}+\boldsymbol{{b}}+\mathrm{5}\boldsymbol{{c}}=\mathrm{0} \\ $$$$\mathrm{5}×\frac{\mathrm{3}}{\mathrm{5}}+\frac{−\mathrm{31}}{\mathrm{25}}+\mathrm{5}\boldsymbol{{c}}=\mathrm{0} \\ $$$$\mathrm{75}−\mathrm{31}+\mathrm{125}\boldsymbol{{c}}=\mathrm{0}\:\:\:\:\rightarrow\boldsymbol{{c}}=\frac{−\mathrm{44}}{\mathrm{125}} \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\frac{\mathrm{3}}{\mathrm{5}}\boldsymbol{{x}}^{\mathrm{2}} −\frac{\mathrm{31}}{\mathrm{25}}\boldsymbol{{x}}−\frac{\mathrm{44}}{\mathrm{125}}\:\boldsymbol{{pls}}\:\boldsymbol{{check}} \\ $$
Commented by jagoll last updated on 23/Feb/20
$${o}\:{o}\:{ok}\:{sir}.\:{thanks}\:{much} \\ $$
Commented by TANMAY PANACEA last updated on 23/Feb/20
$${i}\:{have}\:{assumed}\:{because}\:{right}\:{hand}\:{side}\:{show} \\ $$$${x}^{\mathrm{2}} \\ $$
Commented by jagoll last updated on 23/Feb/20
$${sir},\:{where}\:{do}\:{we}\:{suppose}\:{that}\: \\ $$$${the}\:{function}\:{f}\left({x}\right)\:{is}\:{quadratic} \\ $$$${function}? \\ $$
Commented by mr W last updated on 23/Feb/20
$${to}\:{jagoll}\:{sir}: \\ $$$${you}\:{may}\:{assume}\:{f}\left({x}\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{k}} {x}^{{k}} \\ $$$${but}\:{as}\:{soon}\:{as}\:{Af}\left({x}−\mathrm{1}\right)+{Bf}\left({x}+\mathrm{1}\right)\:{is} \\ $$$${polynonmial}\:{second}\:{order},\:{you}\:{will} \\ $$$${automatically}\:{get}\:{a}_{{k}} =\mathrm{0}\:{for}\:{k}\geqslant\mathrm{3}. \\ $$
Commented by mr W last updated on 23/Feb/20
$${it}\:{is}\:{the}\:{same}\:{method}\:{to}\:{use}\:{for}\:{solving} \\ $$$${questions}\:{like}: \\ $$$${if}\:{a}_{\mathrm{1}} =\mathrm{1}\:{and} \\ $$$$\mathrm{3}{a}_{{n}+\mathrm{1}} +\mathrm{2}{a}_{{n}−\mathrm{1}} =\mathrm{3}{n}^{\mathrm{2}} −\mathrm{5}{n} \\ $$$${find}\:{a}_{{n}} =? \\ $$
Commented by jagoll last updated on 23/Feb/20
$${oo}\:{yes}.\:{i}\:{remember}\:{the}\:{method}\: \\ $$$${that}\:{you}\:{explained}\:{a}\:{few}\:{days}\:{ago} \\ $$