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Question Number 118740 by Jamshidbek2311 last updated on 19/Oct/20

f(x+2)+f(x−1)=2x^2 +14 f(x)=?

$${f}\left({x}+\mathrm{2}\right)+{f}\left({x}−\mathrm{1}\right)=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{14} \\ $$$${f}\left({x}\right)=? \\ $$

Commented by PRITHWISH SEN 2 last updated on 19/Oct/20

x=x+1 f(x+3)+f(x)=2x^2 +4x+16 .....(i) x=x−2 f(x)+f(x−3)= 2x^2 −8x+22...(ii) adding (i) & (ii) f(x+3)+f(x−3)+2f(x)= 4f(x)+2×3^2 = 4x^2 −4x+38 ⇒f(x)=((4x^2 −4x+38−2×3^2 )/4) = x^2 −x+5

$$\mathrm{x}=\mathrm{x}+\mathrm{1} \\ $$$$\mathrm{f}\left(\mathrm{x}+\mathrm{3}\right)+\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{16}\:.....\left(\mathrm{i}\right) \\ $$$$\mathrm{x}=\mathrm{x}−\mathrm{2} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}−\mathrm{3}\right)=\:\mathrm{2x}^{\mathrm{2}} −\mathrm{8x}+\mathrm{22}...\left(\mathrm{ii}\right) \\ $$$$\mathrm{adding}\:\left(\mathrm{i}\right)\:\&\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{f}\left(\mathrm{x}+\mathrm{3}\right)+\mathrm{f}\left(\mathrm{x}−\mathrm{3}\right)+\mathrm{2f}\left(\mathrm{x}\right)=\:\mathrm{4f}\left(\mathrm{x}\right)+\mathrm{2}×\mathrm{3}^{\mathrm{2}} =\:\mathrm{4x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{38} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{4x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{38}−\mathrm{2}×\mathrm{3}^{\mathrm{2}} }{\mathrm{4}}\:=\:\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{5} \\ $$

Answered by bemath last updated on 19/Oct/20

let f(x) = px^2 +qx+r f(x+2)=px^2 +4px+4p+qx+2q+r = px^2 +(4p+q)x+4p+2q+r f(x−1)=px^2 −2px+p+qx−q+r =px^2 +(−2p+q)x+p−q+r ⇒f(x+2)+f(x−1)= 2px^2 +(2p+2q)x+5p+q+2r = 2x^2 +14 { ((2p=2⇒p=1)),((2p+2q=0; q=−1 )),((5p+q+2r=14; 4+2r=14 ; r=5)) :} ∴ f(x)=x^2 −x+5

$${let}\:{f}\left({x}\right)\:=\:{px}^{\mathrm{2}} +{qx}+{r} \\ $$$${f}\left({x}+\mathrm{2}\right)={px}^{\mathrm{2}} +\mathrm{4}{px}+\mathrm{4}{p}+{qx}+\mathrm{2}{q}+{r} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:{px}^{\mathrm{2}} +\left(\mathrm{4}{p}+{q}\right){x}+\mathrm{4}{p}+\mathrm{2}{q}+{r} \\ $$$${f}\left({x}−\mathrm{1}\right)={px}^{\mathrm{2}} −\mathrm{2}{px}+{p}+{qx}−{q}+{r} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={px}^{\mathrm{2}} +\left(−\mathrm{2}{p}+{q}\right){x}+{p}−{q}+{r} \\ $$$$\Rightarrow{f}\left({x}+\mathrm{2}\right)+{f}\left({x}−\mathrm{1}\right)= \\ $$$$\mathrm{2}{px}^{\mathrm{2}} +\left(\mathrm{2}{p}+\mathrm{2}{q}\right){x}+\mathrm{5}{p}+{q}+\mathrm{2}{r}\:=\:\mathrm{2}{x}^{\mathrm{2}} +\mathrm{14} \\ $$$$\begin{cases}{\mathrm{2}{p}=\mathrm{2}\Rightarrow{p}=\mathrm{1}}\\{\mathrm{2}{p}+\mathrm{2}{q}=\mathrm{0};\:{q}=−\mathrm{1}\:}\\{\mathrm{5}{p}+{q}+\mathrm{2}{r}=\mathrm{14};\:\mathrm{4}+\mathrm{2}{r}=\mathrm{14}\:;\:{r}=\mathrm{5}}\end{cases} \\ $$$$\therefore\:{f}\left({x}\right)={x}^{\mathrm{2}} −{x}+\mathrm{5}\: \\ $$$$ \\ $$

Commented by Jamshidbek2311 last updated on 19/Oct/20

wrong.

$${wrong}. \\ $$

Commented by benjo_mathlover last updated on 19/Oct/20

checking f(x)=x^2 −x+5 → { ((f(x+2)=x^2 +4x+4−x−2+5=x^2 +3x+7)),((f(x−1)=x^2 −2x+1−x+1+5=x^2 −3x+7)) :} then f(x+2)+f(x−1)=2x^2 +14 Correct.

$${checking}\: \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} −{x}+\mathrm{5}\:\rightarrow\begin{cases}{{f}\left({x}+\mathrm{2}\right)={x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}−{x}−\mathrm{2}+\mathrm{5}={x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{7}}\\{{f}\left({x}−\mathrm{1}\right)={x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}−{x}+\mathrm{1}+\mathrm{5}={x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{7}}\end{cases} \\ $$$${then}\:{f}\left({x}+\mathrm{2}\right)+{f}\left({x}−\mathrm{1}\right)=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{14} \\ $$$$\boldsymbol{{C}}{o}\boldsymbol{{rrect}}. \\ $$

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