Question

f definita pe R / {-3 supra 4} cu valori in R / {1 supra 2}; f(x) = 2x + 5 supra 4x + 3, demonstreaza ca f e bijectiva

Answer 1

Răspuns:

Explicație pas cu pas:

F bijectiva daca f este injectiva si surjectiva

Studiem injectivitatea

x₁≠x₂⇒ f(x₁)≠f(x₂)

$\frac{2x_{1}+5}{4x_{1}+3} \neq \frac{2x_{2}+5}{4x_{2}+3}$

(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\{$\frac{1}{2}$ } si x∈R\{$-\frac{3}{4}$}

$y=\frac{2x+5}{4x+3}$

2x+5=4xy+3y

2x-4xy=3y-5

x(2-4y)=3y-5

$x=\frac{3y-5}{2-4y}$

y≠$\frac{1}{2}$⇒f surjectiva