Răspunsuri

2014-05-18T11:40:15+03:00
f(x)= log_{3} (2x-5)

2x-5>0
2x>5
x> \frac{5}{2}
⇒ D∈( \frac{5}{2} ; + infinit)

f(x)= \sqrt{ x^{2} -2x-3}
 x^{2} -2x-3 \geq 0
Δ=4+12=16
 x_{1} = \frac{2-4}{2} =-1
 x_{2} = \frac{2+4}{2} =3
faci tabelul semnului si ⇒ D∈(-infinit;-1]U[3;+infinit)

f(x)=arcsin \frac{x-1}{x+1}
-1≤ \frac{x-1}{x+1} ≤1
x+1≠0 ⇒ x≠-1