Explicație pas cu pas:
[tex]f(x)=ax+b[/tex]
a)
[tex]f(3) + f(7) = 3a + b + 7a + b \\ = 10a + 2b = 2(5a + b) = 2\cdot f(5)[/tex]
b)
[tex]A\left(0 ; \sqrt{3} \right),B\left( \frac{ \sqrt{3} }{2} ; \frac{3}{2}\right) \in Gf \\ [/tex]
→
[tex]f(0) = \sqrt{3} = > 0 + b = \sqrt{3} => b = \sqrt{3} [/tex]
și
[tex]f\left( \frac{ \sqrt{3} }{2}\right) = \frac{3}{2} = > a\cdot \frac{ \sqrt{3} }{2} + \sqrt{3} = \frac{3}{2} \\ a \sqrt{3} + 2 \sqrt{3} = 3 < = > a \sqrt{3} = 3 - 2 \sqrt{3} \\ a = \frac{3 - 2 \sqrt{3} }{ \sqrt{3} } = > a = \sqrt{3} - 2[/tex]
→
[tex]f(x) = ( \sqrt{3} - 2)x + \sqrt{3}[/tex]
c)
[tex]f(x) \leqslant 2 = > ( \sqrt{3} - 2)x + \sqrt{3} \leqslant 2[/tex]
[tex]( \sqrt{3} - 2)x \leqslant 2 - \sqrt{3} < = > - (2 - \sqrt{3})x \leqslant 2 - \sqrt{3} \\ = > x \geqslant - 1 \\ [/tex]
[tex] = > x\in \left[1; + \infty \right)[/tex]
