Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 143650 by Rankut last updated on 16/Jun/21

If  the function f and g are defined  on the set of real numbers,are such  that gof(x)=((2x−5)/(3x+7))   and   g(x)=((3x+2)/(x−5)).  find  an expression for  f(x)

$${If}\:\:{the}\:{function}\:{f}\:{and}\:{g}\:{are}\:{defined} \\ $$$${on}\:{the}\:{set}\:{of}\:{real}\:{numbers},{are}\:{such} \\ $$$${that}\:\boldsymbol{{gof}}\left(\boldsymbol{{x}}\right)=\frac{\mathrm{2}\boldsymbol{{x}}−\mathrm{5}}{\mathrm{3}\boldsymbol{{x}}+\mathrm{7}}\:\:\:{and}\: \\ $$$$\boldsymbol{{g}}\left(\boldsymbol{{x}}\right)=\frac{\mathrm{3}\boldsymbol{{x}}+\mathrm{2}}{\boldsymbol{{x}}−\mathrm{5}}. \\ $$$$\boldsymbol{\mathrm{find}}\:\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{expression}}\:\boldsymbol{\mathrm{for}}\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right) \\ $$

Answered by ajfour last updated on 16/Jun/21

((3f+2)/(f−5))=((2x−5)/(3x+7))  ⇒ 9xf+21f+6x+14    =2xf−5f−10x+25  ⇒ f(x)=((11−16x)/(7x+26))

$$\frac{\mathrm{3}{f}+\mathrm{2}}{{f}−\mathrm{5}}=\frac{\mathrm{2}{x}−\mathrm{5}}{\mathrm{3}{x}+\mathrm{7}} \\ $$$$\Rightarrow\:\mathrm{9}{xf}+\mathrm{21}{f}+\mathrm{6}{x}+\mathrm{14} \\ $$$$\:\:=\mathrm{2}{xf}−\mathrm{5}{f}−\mathrm{10}{x}+\mathrm{25} \\ $$$$\Rightarrow\:{f}\left({x}\right)=\frac{\mathrm{11}−\mathrm{16}{x}}{\mathrm{7}{x}+\mathrm{26}} \\ $$

Answered by Olaf_Thorendsen last updated on 16/Jun/21

g(x) = ((3x+2)/(x−5))  (x−5)g(x) = 3x+2  x(g(x)−3) = 5g(x)+2  x = ((5g(x)+2)/(g(x)−3))  ⇒ g^(−1) (x) = ((5x+2)/(x−3))  f(x) = g^(−1) ogof(x) = ((5(((2x−5)/(3x+7)))+2)/((((2x−5)/(3x+7)))−3))  f(x) = g^(−1) ogof(x) = ((16x−11)/(−7x−26))  f(x) = g^(−1) ogof(x) = −((16x−11)/(7x+26))

$${g}\left({x}\right)\:=\:\frac{\mathrm{3}{x}+\mathrm{2}}{{x}−\mathrm{5}} \\ $$$$\left({x}−\mathrm{5}\right){g}\left({x}\right)\:=\:\mathrm{3}{x}+\mathrm{2} \\ $$$${x}\left({g}\left({x}\right)−\mathrm{3}\right)\:=\:\mathrm{5}{g}\left({x}\right)+\mathrm{2} \\ $$$${x}\:=\:\frac{\mathrm{5}{g}\left({x}\right)+\mathrm{2}}{{g}\left({x}\right)−\mathrm{3}} \\ $$$$\Rightarrow\:{g}^{−\mathrm{1}} \left({x}\right)\:=\:\frac{\mathrm{5}{x}+\mathrm{2}}{{x}−\mathrm{3}} \\ $$$${f}\left({x}\right)\:=\:{g}^{−\mathrm{1}} \mathrm{o}{g}\mathrm{o}{f}\left({x}\right)\:=\:\frac{\mathrm{5}\left(\frac{\mathrm{2}{x}−\mathrm{5}}{\mathrm{3}{x}+\mathrm{7}}\right)+\mathrm{2}}{\left(\frac{\mathrm{2}{x}−\mathrm{5}}{\mathrm{3}{x}+\mathrm{7}}\right)−\mathrm{3}} \\ $$$${f}\left({x}\right)\:=\:{g}^{−\mathrm{1}} \mathrm{o}{g}\mathrm{o}{f}\left({x}\right)\:=\:\frac{\mathrm{16}{x}−\mathrm{11}}{−\mathrm{7}{x}−\mathrm{26}} \\ $$$${f}\left({x}\right)\:=\:{g}^{−\mathrm{1}} \mathrm{o}{g}\mathrm{o}{f}\left({x}\right)\:=\:−\frac{\mathrm{16}{x}−\mathrm{11}}{\mathrm{7}{x}+\mathrm{26}} \\ $$

Answered by TheHoneyCat last updated on 16/Jun/21

  let x∈R  and y=f(x) _((if it exists))     (x−5)g(x)=3x+2  so (y−5)g(y)=3y+2  thus (y−5)(2x−5)=(3y+2)(3x+7)  ⇔(7x+26)y=11−16x    so, f:  { ((R\{((−26)/7)}),→,R),(x,→,((11−16x)/(26+7x))) :}    I hope I did not make any calculation mistake  but the reasonning is the right one.  ;•)

$$ \\ $$$$\mathrm{let}\:{x}\in\mathbb{R} \\ $$$$\mathrm{and}\:{y}={f}\left({x}\right)\:_{\left(\mathrm{if}\:\mathrm{it}\:\mathrm{exists}\right)} \\ $$$$ \\ $$$$\left({x}−\mathrm{5}\right){g}\left({x}\right)=\mathrm{3}{x}+\mathrm{2} \\ $$$$\mathrm{so}\:\left({y}−\mathrm{5}\right){g}\left({y}\right)=\mathrm{3}{y}+\mathrm{2} \\ $$$$\mathrm{thus}\:\left({y}−\mathrm{5}\right)\left(\mathrm{2}{x}−\mathrm{5}\right)=\left(\mathrm{3}{y}+\mathrm{2}\right)\left(\mathrm{3}{x}+\mathrm{7}\right) \\ $$$$\Leftrightarrow\left(\mathrm{7}{x}+\mathrm{26}\right){y}=\mathrm{11}−\mathrm{16}{x} \\ $$$$ \\ $$$$\mathrm{so},\:{f}:\:\begin{cases}{\mathbb{R}\backslash\left\{\frac{−\mathrm{26}}{\mathrm{7}}\right\}}&{\rightarrow}&{\mathbb{R}}\\{{x}}&{\rightarrow}&{\frac{\mathrm{11}−\mathrm{16}{x}}{\mathrm{26}+\mathrm{7}{x}}}\end{cases} \\ $$$$ \\ $$$${I}\:{hope}\:{I}\:{did}\:{not}\:{make}\:{any}\:{calculation}\:{mistake} \\ $$$${but}\:{the}\:{reasonning}\:{is}\:{the}\:{right}\:{one}. \\ $$$$\left.;\bullet\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com