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Question Number 105822 by bemath last updated on 01/Aug/20

Given f(x+(1/(2x))) = x^2 +(1/(4x^2 ))+3x+(3/(2x)) find f(2)

$$\mathcal{G}{iven}\:{f}\left({x}+\frac{\mathrm{1}}{\mathrm{2}{x}}\right)\:=\:{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} }+\mathrm{3}{x}+\frac{\mathrm{3}}{\mathrm{2}{x}} \\ $$$${find}\:{f}\left(\mathrm{2}\right)\: \\ $$

Answered by bobhans last updated on 01/Aug/20

⇒x^2 +(1/(4x^2 )) = (x+(1/(2x)))^2 −1 ⇔ f(x+(1/(2x))) = (x+(1/(2x)))^2 +3(x+(1/(2x)))−1 ⇔ f(X) = X^2 +3X−1 f(2) = 4+6−1 = 9 ▲

$$\Rightarrow{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} }\:=\:\left({x}+\frac{\mathrm{1}}{\mathrm{2}{x}}\right)^{\mathrm{2}} −\mathrm{1} \\ $$$$\Leftrightarrow\:{f}\left({x}+\frac{\mathrm{1}}{\mathrm{2}{x}}\right)\:=\:\left({x}+\frac{\mathrm{1}}{\mathrm{2}{x}}\right)^{\mathrm{2}} +\mathrm{3}\left({x}+\frac{\mathrm{1}}{\mathrm{2}{x}}\right)−\mathrm{1} \\ $$$$\Leftrightarrow\:{f}\left({X}\right)\:=\:{X}^{\mathrm{2}} +\mathrm{3}{X}−\mathrm{1}\: \\ $$$${f}\left(\mathrm{2}\right)\:=\:\mathrm{4}+\mathrm{6}−\mathrm{1}\:=\:\mathrm{9}\:\blacktriangle \\ $$

Answered by mathmax by abdo last updated on 01/Aug/20

let x+(1/(2x))=t ⇒2x^2 +1 =2tx ⇒2x^2 −2tx +1 =0 ⇒ Δ^′ =t^2 −2 ⇒x_1 =((t+(√(t^2 −2)))/2) and x_2 =((t−(√(t^2 −2)))/2) x=x_1 ⇒f(t) =(((t+(√(t^2 −2)))/2))^2 +(1/(4(((t+(√(t^2 −2)))/2))^2 )) +3×((t+(√(t^2 −2)))/2) +(3/(2(((t+(√(t^2 −2)))/2)))) ⇒ f(2) =(((2+(√2))/2))^2 +(1/((2+(√2))^2 )) +3.((2+(√2))/2) +(3/(2+(√2))) x=x_2 ⇒f(t) =(((t−(√(t^2 −2)))/2))^2 +(1/(4(((t−(√(t^2 −2)))/2))^2 )) +3×((t−(√(t^2 −2)))/2) +(3/(2(((t−(√(t^2 −2)))/2)))) ⇒ f(2) =(((2−(√2))/2))^2 +(1/(4(((2−(√2))/2))^2 )) +(3/2)(2−(√2))+(3/(2−(√2)))

$$\mathrm{let}\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2x}}=\mathrm{t}\:\Rightarrow\mathrm{2x}^{\mathrm{2}} \:+\mathrm{1}\:=\mathrm{2tx}\:\Rightarrow\mathrm{2x}^{\mathrm{2}} −\mathrm{2tx}\:+\mathrm{1}\:=\mathrm{0}\:\Rightarrow \\ $$$$\Delta^{'} \:=\mathrm{t}^{\mathrm{2}} −\mathrm{2}\:\Rightarrow\mathrm{x}_{\mathrm{1}} =\frac{\mathrm{t}+\sqrt{\mathrm{t}^{\mathrm{2}} −\mathrm{2}}}{\mathrm{2}}\:\:\mathrm{and}\:\mathrm{x}_{\mathrm{2}} =\frac{\mathrm{t}−\sqrt{\mathrm{t}^{\mathrm{2}} −\mathrm{2}}}{\mathrm{2}} \\ $$$$\mathrm{x}=\mathrm{x}_{\mathrm{1}} \:\Rightarrow\mathrm{f}\left(\mathrm{t}\right)\:=\left(\frac{\mathrm{t}+\sqrt{\mathrm{t}^{\mathrm{2}} −\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}\left(\frac{\mathrm{t}+\sqrt{\mathrm{t}^{\mathrm{2}} −\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} }\:+\mathrm{3}×\frac{\mathrm{t}+\sqrt{\mathrm{t}^{\mathrm{2}} −\mathrm{2}}}{\mathrm{2}} \\ $$$$+\frac{\mathrm{3}}{\mathrm{2}\left(\frac{\mathrm{t}+\sqrt{\mathrm{t}^{\mathrm{2}} −\mathrm{2}}}{\mathrm{2}}\right)}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{2}\right)\:=\left(\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\:+\mathrm{3}.\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}}\:+\frac{\mathrm{3}}{\mathrm{2}+\sqrt{\mathrm{2}}} \\ $$$$\mathrm{x}=\mathrm{x}_{\mathrm{2}} \:\Rightarrow\mathrm{f}\left(\mathrm{t}\right)\:=\left(\frac{\mathrm{t}−\sqrt{\mathrm{t}^{\mathrm{2}} −\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}\left(\frac{\mathrm{t}−\sqrt{\mathrm{t}^{\mathrm{2}} −\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} }\:+\mathrm{3}×\frac{\mathrm{t}−\sqrt{\mathrm{t}^{\mathrm{2}} −\mathrm{2}}}{\mathrm{2}} \\ $$$$+\frac{\mathrm{3}}{\mathrm{2}\left(\frac{\mathrm{t}−\sqrt{\mathrm{t}^{\mathrm{2}} −\mathrm{2}}}{\mathrm{2}}\right)}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{2}\right)\:=\left(\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}\left(\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} }\:+\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)+\frac{\mathrm{3}}{\mathrm{2}−\sqrt{\mathrm{2}}} \\ $$

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