Răspuns:
Explicație pas cu pas:
[tex]f(x)=\frac{2}{3} x-2[/tex]
∩OX, y=0 [tex]\frac{2}{3} x-2=0[/tex]
2x-6=0
x=3 A(3,0)
∩OY, x=0 y=0-2=-2 B(0,-2)
b. Fie OD⊥AB
ΔAOB dr in O
[tex]OD=\frac{AO\cdot OB}{AB}[/tex]
Aflam AB din Pitagora
AB²=OB²+AO²
AB²=4+9=13
AB=√13
Deci
[tex]OD=\frac{2\cdot 3}{\sqrt{13} } =\frac{6\sqrt{13} }{13}[/tex]
c. f(x)+f(x+1)≤f(x-2)+4
[tex]\frac{2}{3} x-2+\frac{2}{3}(x+1)-2\leq \frac{2}{3}(x-2)-2+4 \ \ |\cdot 3\\\\2x-6+2x+2-6\leq 2x-4-6+12\\\\2x\leq 12\\\\x\leq 6[/tex]
x∈(-∞,6]
