Question

Ma poate ajuta cienva va rog frumos? Ex 8 cu rezolvarea completa daca se poate ca sa mi pot da seama cum s a ajuns la rezultat...

Answer 1

Răspuns:

Explicație pas cu pas:

a) f : R --> R ; f(x) = 5x/3 - ³⁾2 = (5x-6)/3

∩ Ox <=> f(x) = 0 => (5x-6)/3 = 0 => x = 6/5 => A (6/5 ; 0)

∩ Oy <=> x = 0 => f(0) = -2 => B (0 ; -2)

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b) f : R --> R ; f(x) = x(2x+3)+2(x-6) = 2x²+5x-12

∩ Ox <=> f(x) = 0 => 2x²+5x-12 = 0

a = 2 ; b = 5 ; c = -12 ; Δ=b²-4ac = 25-4·2·(-12) = 25+96 = 121 =>

√Δ = √121 = 11

x₁,₂ =(-5 ± 11)/2 => x₁ = -8 ; x₂ = 3 => A(-8 ; 0) ; B (3 ; 0)

∩ Oy <=> x = 0 => f(0) = -12 => C (0 ; -12)

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c) f : R -{1} --> R ; f(x) = [3x²+2(1-2√2x)]/(x-1)

∩ Oy <=> x = 0 => f(0) = [0+2(1-2√2·0)]/(-1) = -2 => A(0 ; -2)

∩ Ox <=> f(x) = 0 => 3x²+2(1-2√2x) = 0 =>

3x²+2-4√2x = 0 => 3x²-4√2x+2 = 0 =>

a = 3 ; b = -4√2 ; c = 2 ; Δ = (-4√2)²-4·3·2 = 32-32 = 0 =>

x₁=x₂ = 4√2/6 = 2√2/3 => B(2√2/3 ; 0)

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d)  f : R --> R ; f(x) = I3x-1I-2

f(x) = 3x-1-2 = 3x-3 ; x ∈ [1/3 ; +∞)

f(x) = -3x+1-2 = -3x-1 ; x ∈ (-∞ ; 1/3)

∩ Ox <=> f(x) = 0 => 3x-3 = 0 => x = 1 => A(1 ; 0)

                               -3x-1 = 0 => x = -1/3 => B(-1/3 ; 0)

∩ Oy <=> x = 0 => f(0) = -3·0-1 = -1 => C(0 ; -1)

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e) f : [0 ; 4/3] --> R ; f(x) = √(4x-3x²)

∩ Ox <=> f(x) = 0 => 4x-3x² = 0 <=> x(4-3x) = 0 =>

x₁ = 0 ; 4-3x = 0 => x₂ = 4/3 => A(0 ; 0) ; B(4/3 ; 0)

∩ Oy <=> x = 0 => f(0) = 0 => A (0;0)