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Domain and range question

  1. Apr 2, 2005 #1
    domain and range question!!


    I have a quick domain and range question, I have placed my answer below the question. I'm having doubts that it is right! Can someone please help me verify it. Thanks in advance!
    Let f(x)= x² + 2 and g(x) =√1-x²

    a) Find the domain and range of f and g.
    b) Are the functions f.g and g.f defined? Explain?


    The domain of f is {X|X ε R}
    The range of f is {y|y ≥ 2, yε R}

    f.g= f[g(x)]
    = f ___
    (√ 1-x²)
    = ___
    (√ 1-x²) + 2
    = 1- x² + 2
    = 3-x²

    All of the above is correct. I have also already done g.f as well.

    This is where i need help....what comes next....
    Are the functions f.g and g.f defined? Explain? and what is the domain of g and the range of g?

    Thanks in advance for your help!!
    Last edited: Apr 2, 2005
  2. jcsd
  3. Apr 2, 2005 #2
    They are defined, because they both exist.

    The domain of g is all numbers less than and including 1, because once you get alrger than one, you will be taking the square root of negatives.
  4. Apr 2, 2005 #3
    hey thanks for the quick reply, but how about the range of g?
  5. Apr 2, 2005 #4
    I gave you a huge hint on the lower limit of the range of g, and knowing the domain of g you can figure out the lupper limit on the range of g.
  6. Apr 2, 2005 #5
    well this is what i have so far,

    the doman of g is {x||x| ≤ + 1}
    the range of g is {y|y ≥ 0 ≤ y ≤ 1}

    g of f isn't defined because domain is not real?!!

    is that right
  7. Apr 2, 2005 #6
    It doesnt need to be absolute x if your squaring it.

    [tex]g=\sqrt{1-x^2} [/tex]

    [tex] f=x^2+2 [/tex]

    [tex] g(f(x)) = \sqrt{1-(x^2+2)^2} [/tex]

    So (x^2+2)^2 must be less than 1, work from here.
  8. Apr 2, 2005 #7
    sorry you're losing me...ive been trying for the past 15 minutes....anyone else or whozum can you please clairfy
    Last edited: Apr 2, 2005
  9. Apr 2, 2005 #8
    Actually it does... precisely because you are squaring it! Look again :tongue:

    And while "range of [itex]g = \{ y | y \geq 0 \leq y \leq 1 \}[/itex]" is correct, it's also a pretty strange way of writing it. You can just write [itex]\{ y | 0 \leq y \leq 1\}[/itex].
  10. Apr 2, 2005 #9
    What are you missing
  11. Apr 2, 2005 #10
    You are quite right that [itex]g \circ f[/itex] isn't defined over [itex]\mathbb{R}[/itex] for any real [itex]x[/itex].

    Whozum's anaylsis was right, but then again so was your comment he responded to: [itex](x^2 + 2)^2 \leq 1[/itex] is not satisfied for any real [itex]x[/itex].
  12. Apr 2, 2005 #11
    so is this right?

    the domain of g is {x||x| ≤ + 1}
  13. Apr 2, 2005 #12
    indeed it is.
  14. Apr 2, 2005 #13
    so it's not.......

    the domain of g is {x| x ≤ + 1}
  15. Apr 2, 2005 #14
    well, is [itex]g(-2)[/itex] defined? Is [itex]-2 \leq 1[/itex]? That should answer your question! :tongue:
  16. Apr 2, 2005 #15
    x^2 is always positive. saying |x|^2 is redundant, isnt it.
  17. Apr 2, 2005 #16
    no its not!
  18. Apr 2, 2005 #17
    data no it isn't....
  19. Apr 2, 2005 #18
    Oh ok. I see what you mean by absolute. I was thinking x can go from -1,1 which is the same as |x| from 0 to 1
  20. Apr 2, 2005 #19
    Yes it is. But precisely because of that,

    [tex]g(x) = \sqrt{1 - x^2}[/tex]

    is real if and only if [itex] |x| < 1[/itex]. If [itex]|x| > 1[/itex] then either [itex] x < -1[/itex] or [itex] x > 1[/itex]. In either case, [itex](-x)^2 = x^2 > 1[/itex] so [itex] 1 < x^2 \Longrightarrow 1 - x^2 < 0[/itex].
  21. Apr 2, 2005 #20
    how come your latex looks so sexy
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