\(f(g(x)) = f\left(-\displaystyle\frac{1}{2}x+\displaystyle\frac{3}{2}\right) = 3-2\left(-\displaystyle\frac{1}{2}x+\displaystyle\frac{3}{2}\right) =3+x-3=x\)
\(g(f(x)) = g\left(3-2x\right) = -\displaystyle\frac{1}{2}(3-2x)+\displaystyle\frac{3}{2} = -\displaystyle\frac{3}{2}+x+\displaystyle\frac{3}{2}=x\)
La funció vermella és \(f(x)=3-2x\)
La funció blava és \(g(x)=-\displaystyle\frac{1}{2}x+\displaystyle\frac{3}{2}\)
La línia discontínua verda és la bisectriu del primer i tercer quadrant (\(y=x\)). Quan busquem la funció inversa fem un bescanvi d'eixos \(x\leftrightarrow y\). És com un gir al voltant d'aquest eix.
\(f(f(x)) = f\left(\displaystyle\frac{1}{x}\right) = \displaystyle\frac{1}{\displaystyle\frac{1}{x}}=1:\displaystyle\frac{1}{x}=\displaystyle\frac{x}{1}=x\)
Hi ha una certa relació amb el nom que donem a la inversa \(f(x)=\displaystyle\frac{1}{x}=x^{-1}\)
| a) Troba l’expressió algèbrica i el domini de: \(g\circ f\), \(f\circ g\), \(f\circ f\), \(g\circ g\), \(f^{-1}\) i \(g^{-1}\) |
| b) Comprova que les funcions \(f^{-1}\) i \(g^{-1}\) inverses de \(f\) i \(g\) respectivament. |
\( (g\circ f)(x)=g(f(x)) = g\left(\displaystyle\frac{x+4}{x+2}\right) = \displaystyle\frac{3\displaystyle\frac{x+4}{x+2}}{\displaystyle\frac{x+4}{x+2}+1} = \displaystyle\frac{3\displaystyle\frac{x+4}{x+2}}{\displaystyle\frac{2x+6}{x+2}} = \displaystyle\frac{3(x+4)}{x+2} \)
\( D_{g\circ f} =\left\{x\in\mathbb{R}\mid x+2\neq0\right\} =\left\{x\in\mathbb{R}\mid x\neq-2\right\} =\mathbb{R} - \left\{-2\right\} =(-\infty,-2)\cup(-2,+\infty) \)
\( (f\circ g)(x)=f(g(x) = f\left(\displaystyle\frac{3x}{x+1}\right) = \displaystyle\frac{\displaystyle\frac{3x}{x+1}+4}{\displaystyle\frac{3x}{x+1}+2} = \displaystyle\frac{\displaystyle\frac{7x+4}{x+1}}{\displaystyle\frac{5x+2}{x+1}} = \displaystyle\frac{7x+4}{5x+2} \)
\( D_{f\circ g} =\left\{x\in\mathbb{R}\mid 5x+2\neq0\right\} =\left\{x\in\mathbb{R}\mid x\neq-\displaystyle\frac{2}{5}\right\} =\mathbb{R} - \left\{-\displaystyle\frac{2}{5}\right\} =\left(-\infty,-\displaystyle\frac{2}{5}\right)\cup\left(-\displaystyle\frac{2}{5},+\infty\right) \)
\( (f\circ f)(x)=g(f(x)) = f\left(\displaystyle\frac{x+4}{x+2}\right) = \displaystyle\frac{\displaystyle\frac{x+4}{x+2}+4}{\displaystyle\frac{x+4}{x+2}+2} = \displaystyle\frac{\displaystyle\frac{5x+12}{x+2}}{\displaystyle\frac{3x+8}{x+2}} = \displaystyle\frac{5x+12}{3x+8} \)
\( D_{f\circ f} =\left\{x\in\mathbb{R}\mid 3x+8\neq0\right\} =\left\{x\in\mathbb{R}\mid x\neq-\displaystyle\frac{8}{3}\right\} =\mathbb{R} - \left\{-\displaystyle\frac{8}{3}\right\} =\left(-\infty,-\displaystyle\frac{8}{3}\right)\cup\left(-\displaystyle\frac{8}{3},+\infty\right) \)
\( (g\circ g)(x)=f(g(x) = f\left(\displaystyle\frac{3x}{x+1}\right) = \displaystyle\frac{3\displaystyle\frac{3x}{x+1}}{\displaystyle\frac{3x}{x+1}+1} = \displaystyle\frac{\displaystyle\frac{9x}{x+1}}{\displaystyle\frac{4x+1}{x+1}} = \displaystyle\frac{9x}{4x+1} \)
\( D_{g\circ g} =\left\{x\in\mathbb{R}\mid 4x+1\neq0\right\} =\left\{x\in\mathbb{R}\mid x\neq-\displaystyle\frac{1}{4}\right\} =\mathbb{R} - \left\{-\displaystyle\frac{1}{4}\right\} =\left(-\infty,-\displaystyle\frac{1}{4}\right)\cup\left(-\displaystyle\frac{1}{4},+\infty\right) \)
\(f^{-1}(y)=x\) i \(y=\displaystyle\frac{x+4}{x+2}\)
\(y(x+2)=x+4\) \(yx+2y=x+4\) \(yx-x=4-2y\) \(x=f^{-1}(y)=\displaystyle\frac{4-2y}{y-1}\) \[f^{-1}(x)=\frac{4-2x}{x-1}\]
\( D_{f^{-1}} =\left\{x\in\mathbb{R}\mid x-1\neq0\right\} =\left\{x\in\mathbb{R}\mid x\neq+1\right\} =\mathbb{R} - \left\{+1\right\} =\left(-\infty,1\right)\cup\left(1,+\infty\right) \)
\(g^{-1}(y)=x\) i \(y=\displaystyle\frac{3x}{x+1}\)
\(y(x+1)=3x\) \(yx+y=3x\) \(yx-3x=-y\) \(x=g^{-1}(y)=\displaystyle\frac{-y}{y-3}\) \[g^{-1}(x)=-\frac{x}{x-3}\]
\( D_{g^{-1}} =\left\{x\in\mathbb{R}\mid x-3\neq0\right\} =\left\{x\in\mathbb{R}\mid x\neq+3\right\} =\mathbb{R} - \left\{+3\right\} =\left(-\infty,3\right)\cup\left(3,+\infty\right) \)
\( f^{-1}(f(x)) = f^{-1}\left(\displaystyle\frac{x+4}{x+2}\right) = \displaystyle\frac{4-2\displaystyle\frac{x+4}{x+2}}{\displaystyle\frac{x+4}{x+2}-1} = \displaystyle\frac{\displaystyle\frac{4x+8-2(x+4)}{x+2}}{\displaystyle\frac{2}{x+2}} = \displaystyle\frac{\displaystyle\frac{2x}{x+2}}{\displaystyle\frac{2}{x+2}} = x \)
\( f(f^{-1}(x)) = f\left(\displaystyle\frac{4-2x}{x-1}\right) = \displaystyle\frac{\displaystyle\frac{4-2x}{x-1}+4}{\displaystyle\frac{4-2x}{x-1}+2} = \displaystyle\frac{\displaystyle\frac{4-2x+4x-4}{x-1}}{\displaystyle\frac{4-2x+2x-2}{x-1}} = \displaystyle\frac{\displaystyle\frac{2x}{x-1}}{\displaystyle\frac{2}{x-1}} = x \)
\( g^{-1}(g(x)) = g^{-1}\left(\displaystyle\frac{3x}{x+1}\right) = -\displaystyle\frac{\displaystyle\frac{3x}{x+1}}{\displaystyle\frac{3x}{x+1}-3} = -\displaystyle\frac{\displaystyle\frac{3x}{x+1}}{\displaystyle\frac{3x-3x-3}{x+1}} = -\displaystyle\frac{\displaystyle\frac{3x}{x+1}}{\displaystyle\frac{-3}{x+1}} = x \)
\( g(g^{-1}(x)) = g\left(-\displaystyle\frac{x}{x-3}\right) = \displaystyle\frac{-3\displaystyle\frac{x}{x-3}}{-\displaystyle\frac{x}{x-3}+1} = \displaystyle\frac{-\displaystyle\frac{3x}{x-3}}{\displaystyle\frac{-x+x-3}{x-3}} = \displaystyle\frac{-\displaystyle\frac{3x}{x-3}}{\displaystyle\frac{-3}{x-3}} = x \)