[tex]\displaystyle\textrm{a)} (\exists) f \iff (\forall) x \in \mathbb{R}, (\exists) y \in [-1; +\infty) a. i. f(x) = y \iff \\ \\(\forall) x \in (-\infty; 1], (\exists) y \in [-1, +\infty) a. i. f(x) = y \\ \\ \iff x^2 +2x +a \geq -1, (\forall) x \leq 1 \iff x^2 + 2x + 1 + a - 1 \geq -1, (\forall) x \leq 1 \iff (x+1)^2 + a \geq 0, (\forall) x \leq 1\\ \\ \textrm{Minimul functiei se atinge pentru} x = -1 \implies 0 + a \geq 0 \implies a \geq 0[/tex]
[tex]\\\displaystyle\textrm{c)} f \textrm{ surjectiva} \iff (\forall) y \in [-1, +\infty), (\exists) x \in \mathbb{R} a. i. f(x) = y \implies (\exists) x \in \mathbb{R} a. i. f(x) = -1 \iff x^2 +2x +a = -1 \iff (x+1)^2+a=0 \iff x = -1, \boxed{a = 0}\\ \\ \implies f(x) = \begin{cases}x^2+2x, x\leq 1\\ x+2, x > 1\end{cases}\\ \\ \textrm{Cum }x^2+2x, x\leq 1 \textrm{ si } x+2, x > 1\textrm{ sunt functii continue, avem de demonstrat ca}[/tex][tex]\lim\limits_{\substack{x\to 1\\ x < 1}} f(x) = \lim\limits_{\substack{x\to 1\\ x > 1}}f(x)\\ \\ \lim\limits_{\substack{x\to 1\\ x < 1}} f(x) = \lim\limits_{\substack{x\to 1\\ x < 1}} x^2 +2x = 1 + 2 = 3\\ \\ \lim\limits_{\substack{x\to 1\\ x > 1}}f(x) = \lim\limits_{\substack{x\to 1\\ x > 1}} x+2 = 1+2 = 3 \implies f \textrm{ este continua.}[/tex]
[tex]\textrm{La fel e si invers;}\textrm{ daca f este continua, cu limita putem demonstram ca } \\a = 0 \textrm{ si atunci se poate demonstra ca e surjectiva}[/tex]
