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Question Number 142052 by Rankut last updated on 26/May/21

Given that fog(x)=((2x−1)/x) and g(x)=5x+2, Find f(x).

$$\boldsymbol{\mathrm{Given}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{fog}}\left(\boldsymbol{\mathrm{x}}\right)=\frac{\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{1}}{\boldsymbol{\mathrm{x}}}\:\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{g}}\left(\boldsymbol{\mathrm{x}}\right)=\mathrm{5}\boldsymbol{\mathrm{x}}+\mathrm{2}, \\ $$$$\boldsymbol{\mathrm{Find}}\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right). \\ $$$$ \\ $$

Answered by iloveisrael last updated on 26/May/21

⇒f(5x+2)=((2x−1)/x) ⇒f(5x+2)=2−(1/x) ⇒f(x)=2−(1/((((x−2)/5))))=2−(5/(x−2)) ⇒f(x)=((2x−9)/(x−2))

$$\Rightarrow{f}\left(\mathrm{5}{x}+\mathrm{2}\right)=\frac{\mathrm{2}{x}−\mathrm{1}}{{x}} \\ $$$$\Rightarrow{f}\left(\mathrm{5}{x}+\mathrm{2}\right)=\mathrm{2}−\frac{\mathrm{1}}{{x}} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{2}−\frac{\mathrm{1}}{\left(\frac{{x}−\mathrm{2}}{\mathrm{5}}\right)}=\mathrm{2}−\frac{\mathrm{5}}{{x}−\mathrm{2}} \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\mathrm{2}{x}−\mathrm{9}}{{x}−\mathrm{2}} \\ $$

Commented by Rankut last updated on 26/May/21

how did you get (((x−2)/5))

$${how}\:{did}\:{you}\:{get}\:\left(\frac{{x}−\mathrm{2}}{\mathrm{5}}\right) \\ $$

Answered by EDWIN88 last updated on 26/May/21

f(g(x))=f(5x+2)=((2x−1)/x) let 5x+2=u⇒x=((u−2)/5) we get f(u)=((2(((u−2)/5))−1)/((((u−2)/5))))= ((2u−4−5)/(u−2)) f(u)=((2u−9)/(u−2)) ⇒ so f(x)=((2x−9)/(x−2))

$$\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)=\mathrm{f}\left(\mathrm{5x}+\mathrm{2}\right)=\frac{\mathrm{2x}−\mathrm{1}}{\mathrm{x}} \\ $$$$\:\mathrm{let}\:\mathrm{5x}+\mathrm{2}=\mathrm{u}\Rightarrow\mathrm{x}=\frac{\mathrm{u}−\mathrm{2}}{\mathrm{5}} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{f}\left(\mathrm{u}\right)=\frac{\mathrm{2}\left(\frac{\mathrm{u}−\mathrm{2}}{\mathrm{5}}\right)−\mathrm{1}}{\left(\frac{\mathrm{u}−\mathrm{2}}{\mathrm{5}}\right)}=\:\frac{\mathrm{2u}−\mathrm{4}−\mathrm{5}}{\mathrm{u}−\mathrm{2}} \\ $$$$\:\mathrm{f}\left(\mathrm{u}\right)=\frac{\mathrm{2u}−\mathrm{9}}{\mathrm{u}−\mathrm{2}}\:\Rightarrow\:\mathrm{so}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{2x}−\mathrm{9}}{\mathrm{x}−\mathrm{2}} \\ $$

Answered by mathmax by abdo last updated on 26/May/21

fog(x)=((2x−1)/5) ⇒f(g(x))=((2x−1)/5) ⇒f(5x+2)=((2x−1)/x) let 5x+2=t ⇒x=((t−2)/5) ⇒f(t)=2−(1/((t−2)/5)) =2−(5/(t−2))=((2t−4−5)/(t−2)) =((2t−9)/(t−2)) ⇒f(x)=((2x−9)/(x−2))

$$\mathrm{fog}\left(\mathrm{x}\right)=\frac{\mathrm{2x}−\mathrm{1}}{\mathrm{5}}\:\Rightarrow\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)=\frac{\mathrm{2x}−\mathrm{1}}{\mathrm{5}}\:\Rightarrow\mathrm{f}\left(\mathrm{5x}+\mathrm{2}\right)=\frac{\mathrm{2x}−\mathrm{1}}{\mathrm{x}} \\ $$$$\mathrm{let}\:\mathrm{5x}+\mathrm{2}=\mathrm{t}\:\Rightarrow\mathrm{x}=\frac{\mathrm{t}−\mathrm{2}}{\mathrm{5}}\:\Rightarrow\mathrm{f}\left(\mathrm{t}\right)=\mathrm{2}−\frac{\mathrm{1}}{\frac{\mathrm{t}−\mathrm{2}}{\mathrm{5}}}\:=\mathrm{2}−\frac{\mathrm{5}}{\mathrm{t}−\mathrm{2}}=\frac{\mathrm{2t}−\mathrm{4}−\mathrm{5}}{\mathrm{t}−\mathrm{2}} \\ $$$$=\frac{\mathrm{2t}−\mathrm{9}}{\mathrm{t}−\mathrm{2}}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{2x}−\mathrm{9}}{\mathrm{x}−\mathrm{2}} \\ $$

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