[tex]f:\mathbb{R}\rightarrow\mathbb{R} \: \: \: ,f(x)=ax+b[/tex]
[tex]a)(f \circ f)(x) = x + 4 \: \: \: ,\forall\:x\:\in\:\mathbb{R}[/tex]
[tex](f \circ f)(x) = f(f(x))[/tex]
[tex]f(f(x)) = f(ax + b) = a(ax + b) + b = {a}^{2} x + ab + b[/tex]
[tex] {a}^{2} x + ab + b = x + 4[/tex]
[tex] = > {a}^{2} = 1 = > a = \pm1[/tex]
[tex]=>ab + b = 4[/tex]
[tex]b(a + 1) = 4[/tex]
[tex]b = \frac{4}{a + 1} [/tex]
[tex]pt. \: a = - 1[/tex]
[tex]b = \frac{4}{ - 1 + 1} = \frac{4}{0} \: nu \: are \: sens[/tex]
[tex]pt. \: a = 1[/tex]
[tex]b = \frac{4}{1 + 1} = \frac{4}{2} = 2[/tex]
[tex] = > f(x) = x + 2[/tex]
[tex]b)(f \circ f)(x) = 4x + 3 \: \: \: ,\forall\:x\:\in\:\mathbb{R}[/tex]
[tex]f(f(x)) = {a}^{2} x + ab + b[/tex]
[tex] {a}^{2} x + ab + b = 4x + 3[/tex]
[tex] = > {a}^{2} = 4 = > a = \pm \sqrt{4} = > a = \pm2[/tex]
[tex]=>ab + b = 3[/tex]
[tex]b(a + 1) = 3[/tex]
[tex]b = \frac{3}{a + 1} [/tex]
[tex]pt. \: a = - 2[/tex]
[tex]b = \frac{3}{ - 2 + 1} = \frac{3}{ - 1} = - 3[/tex]
[tex]pt. \: a = 2[/tex]
[tex]b = \frac{3}{2 + 1} = \frac{3}{3} = 1 [/tex]
[tex] = > f_{1}(x)= - 2x - 3[/tex]
[tex] = > f_{2}(x)=2x+ 1[/tex]
