Răspuns:
Explicație pas cu pas:
F bijectiva daca f este injectiva si surjectiva
Studiem injectivitatea
x₁≠x₂⇒ f(x₁)≠f(x₂)
[tex]\frac{2x_{1}+5}{4x_{1}+3} \neq \frac{2x_{2}+5}{4x_{2}+3}[/tex]
(2x₁+5)(4x₂+3)≠(2x₂+5)(4x₁+3)
8x₁x₂+6x₁+20x₂+15≠8x₁x₂+6x₂+20x₁+15
6x₁+20x₂≠6x₂+20x₁
6(x₁-x₂)≠20(x₁-x₂)
6≠20⇒ f injectiva
Studiem surjectivitatea
y=f(x), y∈R\{[tex]\frac{1}{2}[/tex] } si x∈R\{[tex]-\frac{3}{4}[/tex]}
[tex]y=\frac{2x+5}{4x+3}[/tex]
2x+5=4xy+3y
2x-4xy=3y-5
x(2-4y)=3y-5
[tex]x=\frac{3y-5}{2-4y}[/tex]
y≠[tex]\frac{1}{2}[/tex]⇒f surjectiva
