Explicație pas cu pas:
[tex]f(x) = (2 - \sqrt{6})x + 2 \sqrt{6}[/tex]
a)
[tex]f(2) = 2(2 - \sqrt{6}) + 2 \sqrt{6} = 4 - 2 \sqrt{6} + 2 \sqrt{6} = 4 \\ [/tex]
=> M(2; 6) ∉ Gf
b)
[tex]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[/tex]
[tex](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} [/tex]
c)
[tex]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][/tex]
d) intersecția cu axa Ox:
[tex]y = 0 => f(x) = 0 \\ (2 - \sqrt{6})x + 2 \sqrt{6} = 0 \\ = > x = 2(3 + \sqrt{6})[/tex]
intersecția cu axa Oy:
[tex]x = 0 = > f(0) = 2 \sqrt{6}[/tex]
