What is the Range of 4/sqrt(5-2x) and How Can It Be Determined?

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The range of the function 4/sqrt(5-2x) is determined by the condition that the expression under the square root must be positive, leading to the requirement that x must be less than 5/2. The function is defined for all positive values of y, as the numerator is always positive and the square root cannot yield negative results. When analyzing the function, it becomes clear that y cannot equal zero, as this would make the denominator undefined. Therefore, the range of the function is all positive real numbers. Understanding these constraints is crucial for accurately determining the range.
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I really need help how to find range for this question... thanks
4/sqrt(5-2x)
 
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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.
 
so, this is the answer... :shy: really confused. thanks
 
when I graphed in the calculator, it appeared that the range is between 0-7...
 
Actually, you must first isolate x in the equation. Then find for what values of y the function doesn't exist.
 
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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