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**1. The problem statement, all variables and given/known data**

Define: [tex]f : D_f \rightarrow \mathbb{R}[/tex] such that

[tex]f(x) = \sqrt{x+2}[/tex] and [tex]Domain(f):=D_f = \{x \in \mathbb{R} : x \geq -2\}[/tex]

Find the range of f(x).

**2. The textbook's solution**

Since the square-root assumes only positive numbers, we conclude that [tex]f(D_f) \subset \{y : y \geq 0\}[/tex]. Further, for every [tex]y \in [0,\inf)[/tex], it follows that [tex]\sqrt{y^2 - 2 + 2} = y [/tex], and hence that [tex]\{y : y \geq 0 \} \subset f(D_f) [/tex] and, finally, that [tex]f(D_f) = \{y : y \geq 0\}[/tex].

**3. My problem**

I can't understand this. I understand that if [tex]A \subset B[/tex] and [tex]B \subset A[/tex], then A = B. I also understand all notation. I just don't understand why they go [tex]\sqrt{y^2 - 2 + 2} = y \Rightarrow \{y : y \geq 0 \} \subset f(D_f) [/tex]