Question

Se f(x) = ax + b e d= - a. mostre que f(f(x)) = x
⁻⁻⁻⁻⁻⁻
Cx + d

Answer 1

$f(x)=\dfrac{ax+b}{cx+d}\\\\\\f(x)=\dfrac{ax+b}{cx-a}~~~\therefore~~~f(f(x))=\dfrac{af(x)+b}{cf(x)-a}$

Substituindo f(x) no numerador:

$af(x)+b=a\left(\dfrac{ax+b}{cx-a}\right)+b\\\\\\af(x)+b=\dfrac{a^{2}x+ab}{cx-a}+\dfrac{b(cx-a)}{cx-a}\\\\\\af(x)+b=\dfrac{a^{2}x+ab+bcx-ab}{cx-a}\\\\\\\boxed{\boxed{af(x)+b=\dfrac{(a^{2}+bc)x}{cx-a}}}$

Substituindo f(x) no denominador:

$cf(x)-a=c\left(\dfrac{ax+b}{cx-a}\right)-a\\\\\\cf(x)-a=\dfrac{acx+bc}{cx-a}+\dfrac{-a(cx-a)}{cx-a}\\\\\\cf(x)-a=\dfrac{acx+bc-acx+a^{2}}{cx-a}\\\\\\\boxed{\boxed{cf(x)-a=\dfrac{a^{2}+bc}{cx-a}}}$
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Então:

$f(f(x))=\dfrac{af(x)+b}{cf(x)-a}\\\\\\f(f(x))=\dfrac{[\frac{(a^{2}+bc)x}{cx-a}]}{[\frac{a^{2}+bc}{cx-a}]}\\\\\\f(f(x))=\dfrac{(a^{2}+bc)\cdot x}{cx-a}\cdot\dfrac{cx-a}{a^{2}+bc}\\\\\\f(f(x))=\dfrac{(a^{2}+bc)\cdot x}{a^{2}+bc}\\\\\\\boxed{\boxed{f(f(x))=x}}$

Claro que algumas restrições devem ser feitas para que possamos cancelar cx - a, mas isso vale para todo x diferente de a / c