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Question Number 153583 by pete last updated on 08/Sep/21

If fog(x)=((2x−1)/x) and g(x)=5x+2, find f(x).

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

Commented by otchereabdullai@gmail.com last updated on 08/Sep/21

nice question

$$\mathrm{nice}\:\mathrm{question} \\ $$

Commented by benhamimed last updated on 08/Sep/21

g(x)=y ; x=((y−2)/5) f(y)=((2((y−2)/5)−1)/((y−2)/5))=((2y−4−5)/(y−2))=((2y−9)/(y−2)) f(x)=((2x−9)/(x−2))

$${g}\left({x}\right)={y}\:\:\:\:;\:{x}=\frac{{y}−\mathrm{2}}{\mathrm{5}} \\ $$$${f}\left({y}\right)=\frac{\mathrm{2}\frac{{y}−\mathrm{2}}{\mathrm{5}}−\mathrm{1}}{\frac{{y}−\mathrm{2}}{\mathrm{5}}}=\frac{\mathrm{2}{y}−\mathrm{4}−\mathrm{5}}{{y}−\mathrm{2}}=\frac{\mathrm{2}{y}−\mathrm{9}}{{y}−\mathrm{2}} \\ $$$${f}\left({x}\right)=\frac{\mathrm{2}{x}−\mathrm{9}}{{x}−\mathrm{2}} \\ $$$$ \\ $$

Commented by pete last updated on 08/Sep/21

Thanks very much sir

$$\mathrm{Thanks}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir} \\ $$

Answered by liberty last updated on 08/Sep/21

f(g(x))=2−(1/x) f(5x+2)=2−(1/x) f(((5x+2−2)/5))=2−(1/((x−2)/5)) f(x)=2−(5/(x−2))=((2x−4−5)/(x−2))=((2x−9)/(x−2))

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

Commented by pete last updated on 08/Sep/21

Thank you sir

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by Rasheed.Sindhi last updated on 08/Sep/21

fog(x)=((2x−1)/x);g(x)=5x+2;f(x)=? f(5x+2)=((2x−1)/x) Let 5x+2=y⇒x=((y−2)/5) f(y)=((2(((y−2)/5))−1)/((y−2)/5))=((2y−4−5)/(y−2))=((2y−9)/(y−2)) y←x: f(x)=((2x−9)/(x−2))

$$\:{fog}\left({x}\right)=\frac{\mathrm{2}{x}−\mathrm{1}}{{x}};\mathrm{g}\left({x}\right)=\mathrm{5}{x}+\mathrm{2};{f}\left({x}\right)=? \\ $$$${f}\left(\mathrm{5}{x}+\mathrm{2}\right)=\frac{\mathrm{2}{x}−\mathrm{1}}{{x}} \\ $$$${Let}\:\mathrm{5}{x}+\mathrm{2}={y}\Rightarrow{x}=\frac{{y}−\mathrm{2}}{\mathrm{5}} \\ $$$${f}\left({y}\right)=\frac{\mathrm{2}\left(\frac{{y}−\mathrm{2}}{\mathrm{5}}\right)−\mathrm{1}}{\frac{{y}−\mathrm{2}}{\mathrm{5}}}=\frac{\mathrm{2}{y}−\mathrm{4}−\mathrm{5}}{{y}−\mathrm{2}}=\frac{\mathrm{2}{y}−\mathrm{9}}{{y}−\mathrm{2}} \\ $$$${y}\leftarrow{x}: \\ $$$${f}\left({x}\right)=\frac{\mathrm{2}{x}−\mathrm{9}}{{x}−\mathrm{2}} \\ $$

Commented by pete last updated on 08/Sep/21

Thank you sir

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by amin96 last updated on 08/Sep/21

fog(x)=2−(1/x) f(x)=((ax+b)/(cx+d)) fog(x)=((5ax+2a+b)/(5cx+2c+d))=((2x−1)/x) ⇒ { ((a=(2/5))),((2a+b=−1)),((c=(1/5))),((2c+d=0)) :} b=(9/5) d=((−2)/5) a=(2/5) c=(1/5) f(x)=(((2/5)x−(9/5))/((x/5)−(2/5)))=(((2x−9)/5)/((x/5)−(2/5)))=((2x−9)/(x−2))

$${fog}\left({x}\right)=\mathrm{2}−\frac{\mathrm{1}}{{x}}\:\:\:\:{f}\left({x}\right)=\frac{{ax}+{b}}{{cx}+{d}} \\ $$$${fog}\left({x}\right)=\frac{\mathrm{5}{ax}+\mathrm{2}{a}+{b}}{\mathrm{5}{cx}+\mathrm{2}{c}+{d}}=\frac{\mathrm{2}{x}−\mathrm{1}}{{x}}\:\:\Rightarrow\:\:\begin{cases}{{a}=\frac{\mathrm{2}}{\mathrm{5}}}\\{\mathrm{2}{a}+{b}=−\mathrm{1}}\\{{c}=\frac{\mathrm{1}}{\mathrm{5}}}\\{\mathrm{2}{c}+{d}=\mathrm{0}}\end{cases} \\ $$$${b}=\frac{\mathrm{9}}{\mathrm{5}}\:\:\:\:\:{d}=\frac{−\mathrm{2}}{\mathrm{5}}\:\:{a}=\frac{\mathrm{2}}{\mathrm{5}}\:\:{c}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$${f}\left({x}\right)=\frac{\frac{\mathrm{2}}{\mathrm{5}}{x}−\frac{\mathrm{9}}{\mathrm{5}}}{\frac{{x}}{\mathrm{5}}−\frac{\mathrm{2}}{\mathrm{5}}}=\frac{\frac{\mathrm{2}{x}−\mathrm{9}}{\mathrm{5}}}{\frac{{x}}{\mathrm{5}}−\frac{\mathrm{2}}{\mathrm{5}}}=\frac{\mathrm{2}{x}−\mathrm{9}}{{x}−\mathrm{2}} \\ $$

Answered by amin96 last updated on 08/Sep/21

g(x)=5x+2 ⇒ g^(−1) (x)=((x−2)/5) fof(x)=((2x−1)/x) ⇒ fogog^(−1) (x)=f(x) f(x)=((2(((x−2))/5)−1)/((x−2)/5))=((2x−4−5)/(x−2))=((2x−9)/(x−2))

$${g}\left({x}\right)=\mathrm{5}{x}+\mathrm{2}\:\:\:\Rightarrow\:\:{g}^{−\mathrm{1}} \left({x}\right)=\frac{{x}−\mathrm{2}}{\mathrm{5}} \\ $$$${fof}\left({x}\right)=\frac{\mathrm{2}{x}−\mathrm{1}}{{x}}\:\:\Rightarrow\:\:{fogog}^{−\mathrm{1}} \left({x}\right)={f}\left({x}\right) \\ $$$${f}\left({x}\right)=\frac{\mathrm{2}\frac{\left({x}−\mathrm{2}\right)}{\mathrm{5}}−\mathrm{1}}{\frac{{x}−\mathrm{2}}{\mathrm{5}}}=\frac{\mathrm{2}{x}−\mathrm{4}−\mathrm{5}}{{x}−\mathrm{2}}=\frac{\mathrm{2}{x}−\mathrm{9}}{{x}−\mathrm{2}} \\ $$

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