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Help to find range?

  1. Jan 30, 2005 #1
    I really need help how to find range for this question....... thanks
  2. jcsd
  3. Jan 30, 2005 #2
    You cannot take the sqare root of a negative number (at least no without obtaining imaginary numbers...) So the range would correspond to when
    (5-2x) is positive.
  4. Jan 30, 2005 #3
    so, this is the answer........ :shy: really confused. thanks
  5. Jan 30, 2005 #4
    when I graphed in the calculator, it appeared that the range is between 0-7....
  6. Jan 30, 2005 #5
    Actually, you must first isolate x in the equation. Then find for what values of y the function doesn't exist.
  7. Jan 31, 2005 #6


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    4/sqrt(5-2x) is defined only for 5-2x> 0 or x< 5/2. That's the domain, not the range. The range is the set all possible y values when y= 4/sqrt(5-2x). One way to determine that is to solve for x (invert the function) and then think about domain: y= 4/sqrt(5-2x) so y(sqrt(5-2x)= 4 and sqrt(5-2x)= 4/y. Now square both sides: 5-2x= 16/y2 so -2x= 1/y2 - 5 and x= -1/2y2 + 5/2. That's defined for all y except 0 (because y is in the denominator and we can't divide by 0) but we have to be care about that squaring. Looking back at the original function, y obviously must be positive (4 is positive and the square root is never negative). The range is all positive real numbers.
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