Question

Se considera functia f:R - > R, f(x) = x/2 +1

a) Verificati egalitatea f(rad 3) + f(3rad3)=2f(2rad3)

Answer 1

$f(x) = \dfrac{x}{2}+1\\ \\ f(\sqrt 3) +f(3\sqrt 3) = \Big(\dfrac{\sqrt 3}{2}+1\Big)+\Big(\dfrac{3\sqrt 3}{2}+1\Big) = \\ \\ = \dfrac{\sqrt 3+3\sqrt 3}{2}+2 = \dfrac{4\sqrt 3}{2}+2 = 2\sqrt 3+2 = 2\cdot \Big(\dfrac{2\sqrt 3}{2}+1\Big) = \\ \\ = 2f(2\sqrt 3)\quad (A)$

Answer 2

Răspuns:

Explicație pas cu pas:

2f(2$\sqrt{3}$)=2($\frac{2\sqrt{3} }{2}$+1)

=2$\sqrt{3}$+2

=2(1+$\sqrt{3}$)

f($\sqrt{3}$)+f(3$\sqrt{3}$)=$\frac{\sqrt{3} }{2}$+1+$\frac{3\sqrt{3} }{2}$+1

=$\frac{4\sqrt{3} }{2}$+2

=2$\sqrt{3}$+2

=2(1+$\sqrt{3}$)

Deci da...Se verifica egalitatea