Răspuns:
A(-3 ; 3) ; B(1 ; -1)
Explicație pas cu pas:
f : R -> R , f(x) = x²+x-3
x = -f(x) =>
x = -x²-x+3 =>
x²+2x-3 = 0 <=> x²+3x-x-3 = 0 =>
x(x+3)-(x+3) = 0 => (x+3)(x-1) = 0 =>
x₁ = -3 ; x₂ = 1
A(-3 ; 3) ; B(1 ; -1)
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Verificare:
x₁ = -3 => f(-3) = (-3)²-3-3 = 9-6 = 3 = -x₁ = corect
x₂ = 1 => f(1) = 1²+1-3 = -1 = -x₂ = corect