Question

Urgent!
Va multumesc!

Answer 1

$f:\mathbb{R}\rightarrow\mathbb{R} \: \: \: ,f(x)=ax+b$

$a)(f \circ f)(x) = x + 4 \: \: \: ,\forall\:x\:\in\:\mathbb{R}$

$(f \circ f)(x) = f(f(x))$

$f(f(x)) = f(ax + b) = a(ax + b) + b = {a}^{2} x + ab + b$

${a}^{2} x + ab + b = x + 4$

$= > {a}^{2} = 1 = > a = \pm1$

$=>ab + b = 4$

$b(a + 1) = 4$

$b = \frac{4}{a + 1}$

$pt. \: a = - 1$

$b = \frac{4}{ - 1 + 1} = \frac{4}{0} \: nu \: are \: sens$

$pt. \: a = 1$

$b = \frac{4}{1 + 1} = \frac{4}{2} = 2$

$= > f(x) = x + 2$

$b)(f \circ f)(x) = 4x + 3 \: \: \: ,\forall\:x\:\in\:\mathbb{R}$

$f(f(x)) = {a}^{2} x + ab + b$

${a}^{2} x + ab + b = 4x + 3$

$= > {a}^{2} = 4 = > a = \pm \sqrt{4} = > a = \pm2$

$=>ab + b = 3$

$b(a + 1) = 3$

$b = \frac{3}{a + 1}$

$pt. \: a = - 2$

$b = \frac{3}{ - 2 + 1} = \frac{3}{ - 1} = - 3$

$pt. \: a = 2$

$b = \frac{3}{2 + 1} = \frac{3}{3} = 1$

$= > f_{1}(x)= - 2x - 3$

$= > f_{2}(x)=2x+ 1$