Răspuns:
[tex]f:R->R, f(x)=2x+3[/tex] [tex]A(-1,a) ; B(b,5)[/tex]
[tex]a)[/tex]
[tex]=> f(-1)=a <=> 2*(-1)+3=a <=>-2+3=a => a = 1\\=> f(b)=5 <=> 2*(b)+3=5 <=>2b+3=5 => 2b=2 => b = 1\\[/tex]
[tex]b) A(-1,1) ; B(1,5)[/tex]
[tex]|AB|^2 = (xb-xa)^2+(yb-ya)^2 <=> (1-(-1))^2+(5-1)^2 <=> (2)^2+(4)^2 <=> 4+16 = 20 => |AB| = \sqrt{20} = 2\sqrt{5}[/tex]
Explicație pas cu pas:
