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

  1. Jul 22, 2012 #1
    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]
     
  2. jcsd
  3. Jul 22, 2012 #2
    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].
     
  4. Jul 22, 2012 #3

    HallsofIvy

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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: Jul 22, 2012
  5. Jul 22, 2012 #4
    Confuzion solved.

    Thanks guys.
     
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