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Finding Domain and Range of Functions

  1. Sep 2, 2009 #1
    Find the domain and range of y=sin^2x


    I have a hard type computing this into my graphing calculator. Can someone help in the steps to find the domain?
     
  2. jcsd
  3. Sep 2, 2009 #2

    rock.freak667

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    y=sin2x=(sinx)2.

    since it is squared it is never negative. The domain should be easy to find, given your knowledge of y=sinx
     
  4. Sep 2, 2009 #3
    So the domain would be 0 less than or equal to positive infinity? and the range would be the same?
     
  5. Sep 2, 2009 #4

    rock.freak667

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    What is the domain of sin(x)? The range is the values of 'y' that the graph lies between.

    You know that -1≤ sinx ≤ 1, so where would sin2x lie between?
     
  6. Sep 2, 2009 #5
    the domain of sin (x) is -∞ < x < ∞ and would sin²x lie between -2 ≤ x ≤ 2...sorry i am bit confused.
     
  7. Sep 2, 2009 #6

    rock.freak667

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    Right, sin2x has the same domain as sin(x)


    and would sin²x lie between -2 ≤ x ≤ 2...sorry i am bit confused.[/QUOTE]


    if y=sin2x, what is it's maximum and minimum value?
     
  8. Sep 2, 2009 #7
    Wouldn't its maximum value be 1 and the minimum value be -1?
     
  9. Sep 2, 2009 #8

    rock.freak667

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    yes

    remember, sin2x = (sinx)2 so it is a perfect square, meaning sin2x ≥ 0 . So what's the minimum value going to be?
     
  10. Sep 2, 2009 #9
    So...would it be 1 as well?
     
  11. Sep 2, 2009 #10

    rock.freak667

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    no

    in y=x2, what is the lowest value of y which gives a real value for x?
     
  12. Sep 2, 2009 #11
    I have no clue, honestly. I am graphing it on my calculator. Would it be zero? I do not think the minimum value is a negative number.
     
  13. Sep 2, 2009 #12

    rock.freak667

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    yes it would be zero.
     
  14. Sep 2, 2009 #13
    Okay so the domain would be -∞ < x < ∞ and the range is 0 ≤ y ≤ 1 ?
     
  15. Sep 2, 2009 #14

    rock.freak667

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    That should be correct. You could write [itex] x \epsilon \Re[/itex] as well I believe.
     
  16. Sep 2, 2009 #15
    Okay. Thank you so much for helping, rock.freak667. I really appreciate it :-)
     
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