Question Number 198431 by hardmath last updated on 19/Oct/23
$$\mathrm{f}\::\:\mathbb{R}\:\rightarrow\:\mathbb{R} \\ $$$$\mathrm{f}\:\left(\mathrm{3x}\:−\:\mathrm{1}\right)\:=\:\mathrm{x}\:+\:\mathrm{5} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:? \\ $$
Answered by AST last updated on 20/Oct/23
![f[3((x/3)+(1/3))−1]=((x+1)/3)+5⇒f(x)=((x+16)/3)](https://www.tinkutara.com/question/Q198433.png)
$${f}\left[\mathrm{3}\left(\frac{{x}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\right)−\mathrm{1}\right]=\frac{{x}+\mathrm{1}}{\mathrm{3}}+\mathrm{5}\Rightarrow{f}\left({x}\right)=\frac{{x}+\mathrm{16}}{\mathrm{3}} \\ $$
Commented by Frix last updated on 20/Oct/23
$${f}\left(\mathrm{3}\left(\frac{{x}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\right)−\mathrm{1}\right)=\frac{{x}+\mathrm{1}}{\mathrm{3}}+\mathrm{5}=\frac{{x}+\mathrm{16}}{\mathrm{3}} \\ $$
Commented by AST last updated on 20/Oct/23
$$\checkmark \\ $$
Answered by Rasheed.Sindhi last updated on 20/Oct/23
$$\mathrm{f}\:\left(\mathrm{3x}\:−\:\mathrm{1}\right)\:=\:\mathrm{x}\:+\:\mathrm{5}\:;\:\mathrm{f}\left(\mathrm{x}\right)\:=\:? \\ $$$${Let}\:\mathrm{3}{x}−\mathrm{1}={y}\Rightarrow{x}=\frac{{y}+\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{f}\left({y}\right)=\frac{{y}+\mathrm{1}}{\mathrm{3}}+\mathrm{5}=\frac{{y}+\mathrm{16}}{\mathrm{3}} \\ $$$${Replacing}\:{y}\:{by}\:{x}: \\ $$$$\mathrm{f}\left({x}\right)=\frac{{x}+\mathrm{16}}{\mathrm{3}} \\ $$
Answered by Rasheed.Sindhi last updated on 20/Oct/23
$${f}\left(\mathrm{3}{x}−\mathrm{1}\right)={x}+\mathrm{5}=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{3}{x}+\mathrm{15}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{3}{x}−\mathrm{1}+\mathrm{16}\right) \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{3}}\left({x}+\mathrm{16}\right) \\ $$
Commented by hardmath last updated on 20/Oct/23
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{professors} \\ $$