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find the ramge for   

 y=x/(x-3)^1/2

 Nov 11, 2019
 #1
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y  =   x

       _____

      √[x - 3]

 

This function will not exist   on  ( -infinity , 3 ]

 

When x  approaches  3  from the right, y approaches infinity

 

Using Calculus, we can find the x value where the minimum occurs

 

y '  = (x - 3)^(-1/2)  + (-1/2)x(x - 3)^(-3/2)

 

Set this to 0

 

(x - 3)^(-1/2)  + (-1/2) x (x - 3)^(-3/2)  = 0       factor

 

(x - 3)^(-3/2)  [ (x - 3) - (1/2)x)  ]  = 0

 

(x - 3) - (1/2)x  =  0

 

(1/2)x -  3  =  0

 

(1/2)x  =  3

 

x  = 6

 

So....the minimum occurs at 

 

6 / √[6 - 3 ]   =   6 / √3

 

So.....the range  is    [ 6 / √3, infinity   )

 

 

cool cool cool

 Nov 11, 2019

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