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.