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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.


Răspuns :

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

[tex]a)(f \circ f \circ f)(1) + (f \circ f \circ f)(3) = ?[/tex]

[tex](f \circ f \circ f)(1) = f(f(f(1)))[/tex]

[tex]f(f(f(1))) = f(f( {1}^{2} - 4 \times 1 + 3)) [/tex]

[tex] = f(f(1 - 4 + 3)) = f(f(0)) [/tex]

[tex] = f( {0}^{2} - 4 \times 0 + 3) = f(0 - 0 + 3) [/tex]

[tex] = f(3) = {3}^{2} - 4 \times 3 + 3 = 9 - 12 + 3 = 0[/tex]

[tex](f \circ f \circ f)(3) = f(f(f(3)))[/tex]

[tex]f(f(f(3))) = f(f( {3}^{2} - 4 \times 3 + 3))[/tex]

[tex] = f(f(9 - 12 + 3)) = f(f(0))[/tex]

[tex] = f( {0}^{2} - 4 \times 0 + 3) = f(0 - 0 + 3) [/tex]

[tex] = f(3) = {3}^{2} - 4 \times 3 + 3 = 9 - 12 + 3 = 0[/tex]

[tex](f \circ f \circ f)(1) + (f \circ f \circ f)(3) = 0 + 0 = 0[/tex]

[tex]b)(g \circ f)(x) = - 3[/tex]

[tex]g(f(x)) = - 3[/tex]

[tex]g( {x}^{2} - 4x + 3) = - 3[/tex]

[tex]2( {x}^{2} - 4x + 3) - 3 = - 3[/tex]

[tex]2( {x}^{2} - 4x + 3) = - 3 + 3[/tex]

[tex]2( {x}^{2} - 4x + 3) = 0 \: | \div 2[/tex]

[tex] {x}^{2} - 4x + 3 = 0[/tex]

[tex] {x}^{2} - 3x - x + 3 = 0[/tex]

[tex]x(x - 3) - (x - 3) = 0[/tex]

[tex](x - 3)(x - 1) = 0[/tex]

[tex]x - 3 = 0 = > x_{1} = 3[/tex]

[tex]x - 1 = 0 = > x_{2} = 1[/tex]

[tex]c)(f \circ g)(x) = 0[/tex]

[tex]f(g(x)) = 0[/tex]

[tex]f(2x - 3) = 0[/tex]

[tex] {(2x - 3)}^{2} - 4(2x - 3) + 3 = 0[/tex]

[tex]4 {x}^{2} - 12x + 9 - 8x + 12 + 3 = 0[/tex]

[tex]4 {x}^{2} - 20x + 24 = 0 \: | \div 4[/tex]

[tex] {x}^{2} - 5x + 6 = 0[/tex]

[tex] {x}^{2} - 3x - 2x + 6 = 0[/tex]

[tex]x(x - 3) - 2(x - 3) = 0[/tex]

[tex](x - 3)(x - 2) = 0[/tex]

[tex]x - 3 = 0 = > x_{1}= 3[/tex]

[tex]x - 2 = 0 = > x_{2} = 2[/tex]