[tex]f:\mathbb{R}\to \mathbb{R},\quad f(x) = ax+b \\\\\\ f^{-1}(x) = f(x)\\ \\\Rightarrow \begin{cases} x = ay+b \\ y = ax+b \end{cases}\Bigg|\Rightarrow y = a(ay+b)+b \Rightarrow\\ \\\\ \Rightarrow y=a^2y+ab+b \\ \\ \Rightarrow \begin{cases}a^2 = 1 \\ ab+b = 0 \end{cases}\Rightarrow\begin{cases}a = \pm 1 \\ b(a+1)= 0 \end{cases}\Bigg| \Rightarrow[/tex]
[tex]\Rightarrow (a = \pm 1\,\wedge\,b = 0)\,\vee\, (a=\pm 1\,\wedge \,a = -1) \\ \\ \Rightarrow (a = \pm 1\,\wedge\,b = 0)\,\vee\, (a=-1) \\ \\ \Rightarrow (a,b) = \Big\{(\pm 1, 0);(-1,\mathbb{R})\Big\}\\ \\\\ \Rightarrow \boxed{(a,b) = \Big\{( 1, 0);(-1,\mathbb{R})\Big\}}[/tex]
