Find the range of this function, given we know its domain is.

[operationsres]

Preface: I can't understand a solution provided in a textbook. Please help me to understand.

Homework Statement

Define: f : D_f \rightarrow \mathbb{R} such that
f(x) = \sqrt{x+2} and Domain(f):=D_f = \{x \in \mathbb{R} : x \geq -2\}

Find the range of f(x).

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



3. My problem

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

 

[Robert1986]

What they want to do is to show that every positive number y is the image of some x in D_f (and that just means there is some x \in D_f such that f(x)=y). To do this, for an arbitrary y \geq 0 they explicitly construct an x such that f(x)=y. In this case, that is x=y^2-2, and f(x)=y and since y \geq 0 then y^2 - 2 \geq -2.

 

[HallsofIvy]

\sqrt{x} is an increasing function, without upper bound, and \sqrt{3- 3}= \sqrt{0}= 0. Those two are enough to tell you that the range is "all positive real numbers".

 

[operationsres]

Confuzion solved.

Thanks guys.