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f(ax₁ + bx₂) = af(x₁) + bf(x₂)
1) f: x∈R → x + 2 ∈R
f(ax₁ + bx₂) = ax₁ + bx₂ + 2
af(x₁) + bf(x₂) = ax₁ + 2 + bx₂ + 2
ax₁ + bx₂ + 2 ≠ ax₁ + 2 + bx₂ + 2
np.: a = b = x₁ = x₂ = 0
ax₁ + bx₂ + 2 = 0 + 0 + 2 = 2 ≠ 4 = 0 + 2 + 0 + 2 = ax₁ + 2 + bx₂ + 2
NIE
2) f:x∈R→ cos(x)∈R
f(ax₁ + bx₂) = cos(ax₁ + bx₂)
af(x₁) + bf(x₂) = acos(x₁) + bcos(x₂)
cos(ax₁ + bx₂) ≠ acos(x₁) + bcos(x₂)
np.: a = b = 1, x₁ = x₂ = 0
cos(ax₁ + bx₂) = cos0 = 1 ≠ 2 = cos0 + cos0 = acos(x₁) + bcos(x₂)
NIE
3)f:x∈R→ √x∈R
f(ax₁ + bx₂) = √(ax₁ + bx₂)
af(x₁) + bf(x₂) = a√x₁ + b√x₂
√(ax₁ + bx₂) ≠ a√x₁ + b√x₂
np.: a = 2, b = x₁ = x₂ = 1
√(ax₁ + bx₂) = √(2 + 1) = √3 ≠ 3 = 2 + 1 = a√x₁ + b√x₂
NIE
4)f:x∈R²→x1 ∈R
x = (x₁, x₂)
f(a(x₁, x₂) + b(x₃, x₄)) = f((ax₁ + bx₃, ax₂ + bx₄)) = ax₁ + bx₃ = af((x₁, x₂)) + bf((x₃, x₄))
TAK
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