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

[viendong]

I really need help how to find range for this question... thanks
4/sqrt(5-2x)

 

[christinono]

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.

 

[viendong]

so, this is the answer... :shy: really confused. thanks

 

[viendong]

when I graphed in the calculator, it appeared that the range is between 0-7...

 

[christinono]

Actually, you must first isolate x in the equation. Then find for what values of y the function doesn't exist.

 

[HallsofIvy]

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.