Question Number 86030 by oustmuchiya@gmail.com last updated on 26/Mar/20
$${Tbe}\:{function}\:\boldsymbol{\mathrm{f}}\:{and}\:\boldsymbol{\mathrm{g}}\:{are}\:{defined}\:{by}\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{3}\:{and}\:\boldsymbol{\mathrm{g}}\left(\boldsymbol{\mathrm{x}}\right)=\mathrm{3}\boldsymbol{\mathrm{x}}. \\ $$$$\boldsymbol{\mathrm{F}}{ind}\:\left(\boldsymbol{\mathrm{a}}\right)\:\boldsymbol{\mathrm{f}}^{−\mathrm{1}} \left(\boldsymbol{\mathrm{x}}\right)\:\:\:\:\left(\boldsymbol{\mathrm{b}}\right)\:\boldsymbol{\mathrm{gf}}\left(\boldsymbol{\mathrm{x}}\right)\:\:\:\left(\boldsymbol{\mathrm{c}}\right)\:\boldsymbol{\mathrm{gf}}\left(\mathrm{2}\right) \\ $$
Commented by Rio Michael last updated on 26/Mar/20
![f: x → 2x−3 and g : x → 3x (a) let y = f(x) ⇒ y = 2x−3 or x = 2y − 3 ⇒ y = f^(−1) (x) = ((x + 3)/2) , x ∈ R (b) gf(x) = g[f(x)] = 3(2x−3) = 6x−9 (c) gf(2) = 6(2)−9 = 3](https://www.tinkutara.com/question/Q86051.png)
$$\:{f}:\:{x}\:\rightarrow\:\mathrm{2}{x}−\mathrm{3}\:\:\:\mathrm{and}\:\mathrm{g}\::\:{x}\:\rightarrow\:\mathrm{3}{x} \\ $$$$\:\left(\mathrm{a}\right)\:\mathrm{let}\:{y}\:=\:{f}\left({x}\right)\:\:\Rightarrow\:\:{y}\:=\:\mathrm{2}{x}−\mathrm{3}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{or}\:\:{x}\:=\:\mathrm{2}{y}\:−\:\mathrm{3}\:\Rightarrow\:\:{y}\:=\:{f}^{−\mathrm{1}} \left({x}\right)\:=\:\frac{{x}\:+\:\mathrm{3}}{\mathrm{2}}\:,\:{x}\:\in\:\mathbb{R} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{g}{f}\left({x}\right)\:=\:\mathrm{g}\left[{f}\left({x}\right)\right]\:=\:\mathrm{3}\left(\mathrm{2}{x}−\mathrm{3}\right)\:=\:\mathrm{6}{x}−\mathrm{9} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{gf}\left(\mathrm{2}\right)\:=\:\mathrm{6}\left(\mathrm{2}\right)−\mathrm{9}\:=\:\mathrm{3} \\ $$