Question

Se considera funcțiile f,g:R-->R, f(x)=x²-4x+3, g(x)=2x-3.
a) Calculati: (f°f°f)(1)+(f°f°f)(3).
b) Rezolvati ecuatia: (g°f)(x)=-3.
c) Rezolvati ecuatia: (f°g)(x)=0.

Answer 1

$f,g:\mathbb{R}\rightarrow\mathbb{R} \: \: \: ,f(x)= {x}^{2} - 4x + 3 \: \: \: ,g(x) = 2x - 3$

$a)(f \circ f \circ f)(1) + (f \circ f \circ f)(3) = ?$

$(f \circ f \circ f)(1) = f(f(f(1)))$

$f(f(f(1))) = f(f( {1}^{2} - 4 \times 1 + 3))$

$= f(f(1 - 4 + 3)) = f(f(0))$

$= f( {0}^{2} - 4 \times 0 + 3) = f(0 - 0 + 3)$

$= f(3) = {3}^{2} - 4 \times 3 + 3 = 9 - 12 + 3 = 0$

$(f \circ f \circ f)(3) = f(f(f(3)))$

$f(f(f(3))) = f(f( {3}^{2} - 4 \times 3 + 3))$

$= f(f(9 - 12 + 3)) = f(f(0))$

$= f( {0}^{2} - 4 \times 0 + 3) = f(0 - 0 + 3)$

$= f(3) = {3}^{2} - 4 \times 3 + 3 = 9 - 12 + 3 = 0$

$(f \circ f \circ f)(1) + (f \circ f \circ f)(3) = 0 + 0 = 0$

$b)(g \circ f)(x) = - 3$

$g(f(x)) = - 3$

$g( {x}^{2} - 4x + 3) = - 3$

$2( {x}^{2} - 4x + 3) - 3 = - 3$

$2( {x}^{2} - 4x + 3) = - 3 + 3$

$2( {x}^{2} - 4x + 3) = 0 \: | \div 2$

${x}^{2} - 4x + 3 = 0$

${x}^{2} - 3x - x + 3 = 0$

$x(x - 3) - (x - 3) = 0$

$(x - 3)(x - 1) = 0$

$x - 3 = 0 = > x_{1} = 3$

$x - 1 = 0 = > x_{2} = 1$

$c)(f \circ g)(x) = 0$

$f(g(x)) = 0$

$f(2x - 3) = 0$

${(2x - 3)}^{2} - 4(2x - 3) + 3 = 0$

$4 {x}^{2} - 12x + 9 - 8x + 12 + 3 = 0$

$4 {x}^{2} - 20x + 24 = 0 \: | \div 4$

${x}^{2} - 5x + 6 = 0$

${x}^{2} - 3x - 2x + 6 = 0$

$x(x - 3) - 2(x - 3) = 0$

$(x - 3)(x - 2) = 0$

$x - 3 = 0 = > x_{1}= 3$

$x - 2 = 0 = > x_{2} = 2$