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

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

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

HallsofIvy

Science Advisor
Homework Helper
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[itex]\sqrt{x}[/itex] is an increasing function, without upper bound, and [itex]\sqrt{3- 3}= \sqrt{0}= 0[/itex]. Those two are enough to tell you that the range is "all positive real numbers".
 
Last edited by a moderator:
Confuzion solved.

Thanks guys.
 

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