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Question Number 158039 by HongKing last updated on 30/Oct/21

f (((x + 1)/2)) = x + 2 ⇒ f(x) = ?

$$\mathrm{f}\:\left(\frac{\mathrm{x}\:+\:\mathrm{1}}{\mathrm{2}}\right)\:=\:\mathrm{x}\:+\:\mathrm{2}\:\:\Rightarrow\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:? \\ $$

Answered by puissant last updated on 30/Oct/21

y=((x+1)/2) → x=2y−1 ⇒ f(x) = (2x−1)+2 = 2x+1

$${y}=\frac{{x}+\mathrm{1}}{\mathrm{2}}\:\rightarrow\:{x}=\mathrm{2}{y}−\mathrm{1} \\ $$$$\Rightarrow\:{f}\left({x}\right)\:=\:\left(\mathrm{2}{x}−\mathrm{1}\right)+\mathrm{2}\:=\:\mathrm{2}{x}+\mathrm{1} \\ $$

Answered by yeti123 last updated on 30/Oct/21

f(((x + 1)/2)) = x + 1 + 1 y = x + 1 ⇒ f((y/2)) = y + 1 f(((2y)/2)) = 2y + 1 f(y) = 2y + 1 f(x) = 2x + 1

$${f}\left(\frac{{x}\:+\:\mathrm{1}}{\mathrm{2}}\right)\:=\:{x}\:+\:\mathrm{1}\:+\:\mathrm{1} \\ $$$${y}\:=\:{x}\:+\:\mathrm{1}\:\Rightarrow \\ $$$${f}\left(\frac{{y}}{\mathrm{2}}\right)\:=\:{y}\:+\:\mathrm{1} \\ $$$${f}\left(\frac{\mathrm{2}{y}}{\mathrm{2}}\right)\:=\:\mathrm{2}{y}\:+\:\mathrm{1} \\ $$$${f}\left({y}\right)\:=\:\mathrm{2}{y}\:+\:\mathrm{1} \\ $$$${f}\left({x}\right)\:=\:\mathrm{2}{x}\:+\:\mathrm{1} \\ $$

Commented by Rasheed.Sindhi last updated on 30/Oct/21

f(((x + 1)/2)) = x + 1 + 1 f(((x + 1)/2)) =2( ((x + 1 + 1)/2))=2(((x+1)/2))+1 Replacing ((x+1)/2) by x f(x)=2x+1

$${f}\left(\frac{{x}\:+\:\mathrm{1}}{\mathrm{2}}\right)\:=\:{x}\:+\:\mathrm{1}\:+\:\mathrm{1} \\ $$$${f}\left(\frac{{x}\:+\:\mathrm{1}}{\mathrm{2}}\right)\:=\mathrm{2}\left(\:\frac{{x}\:+\:\mathrm{1}\:+\:\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}\left(\frac{{x}+\mathrm{1}}{\mathrm{2}}\right)+\mathrm{1} \\ $$$${Replacing}\:\frac{{x}+\mathrm{1}}{\mathrm{2}}\:{by}\:{x} \\ $$$${f}\left({x}\right)=\mathrm{2}{x}+\mathrm{1} \\ $$$$ \\ $$

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