Question

fie functia f : R - R , f(x)=(2-radical din 6)*x + 2 radical din 6 a) stabiliti daca M(2;6) apartine Gf b) rezolvati ecuatia f(1-x) +2f(x)=7 c) determinati x apartine de R stiind ca f(x) > sau egal 4 d) stiind ca f : A - {-2, 0, 1, 4} determinati multimea A si graficul functiei f.​

Answer 1

Explicație pas cu pas:

$f(x) = (2 - \sqrt{6})x + 2 \sqrt{6}$

a)

$f(2) = 2(2 - \sqrt{6}) + 2 \sqrt{6} = 4 - 2 \sqrt{6} + 2 \sqrt{6} = 4$

=> M(2; 6) ∉ Gf

b)

$f(1 - x) + 2f(x) = 7 \\ (2 - \sqrt{6})(1 - x) + 2 \sqrt{6} + 2\left((2 - \sqrt{6})x + 2 \sqrt{6} \right) = 7$

$(2 - \sqrt{6})(1 - x) + 2 \sqrt{6} + 2\left((2 - \sqrt{6})x + 2 \sqrt{6} \right) = 7 \\ 2 - 2x - \sqrt{6} + \sqrt{6}x + 2 \sqrt{6} + 4x - 2 \sqrt{6}x + 4 \sqrt{6} = 7 \\ 2x - \sqrt{6}x = 7 - 2 + 3 \sqrt{6} \\ (2 - \sqrt{6})x = 5 - 5 \sqrt{6} \\ x = \frac{5(1 - \sqrt{6})}{2 - \sqrt{6} } = \frac{5(1 - \sqrt{6})(2 + \sqrt{6})}{(2 - \sqrt{6})(2 + \sqrt{6})} = \frac{5( - 4 - \sqrt{6})}{4 - 6} \\ = > x = \frac{5(4 + \sqrt{6})}{2}$

c)

$f(x) \geqslant 4 = > (2 - \sqrt{6})x + 2 \sqrt{6} \geqslant 4 \\ (2 - \sqrt{6})x\geqslant 4 - 2 \sqrt{6} \\ ( \sqrt{6} - 2)x \leqslant 2( \sqrt{6} - 2) \\ x \leqslant \ \frac{2( \sqrt{6} - 2) }{ \sqrt{6} - 2} \\ x \leqslant 2 = > x\in\left(- \infty ; 2 \right]$

d) intersecția cu axa Ox:

$y = 0 => f(x) = 0 \\ (2 - \sqrt{6})x + 2 \sqrt{6} = 0 \\ = > x = 2(3 + \sqrt{6})$

intersecția cu axa Oy:

$x = 0 = > f(0) = 2 \sqrt{6}$