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If $f(x)=\sqrt{x-3}$, what is the smallest real number $x$ in the domain of $f(f(x))$?

 Feb 10, 2019
 #1
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ignore the dollar signs those are for LateX.

 Feb 10, 2019
 #2
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f (f(x) ) =

 

sqrt [  sqrt (x - 3) - 3  ]

 

sqrt (x - 3)  - 3     must be  ≥ 0

 

So...

 

sqrt ( x - 3) - 3 ≥ 0

 

sqrt (x - 3) ≥ 3        square both sides

 

x - 3 ≥ 9

 

x ≥ 12

 

So.....12 is the smallest real number in the domain

 

cool cool cool

 Feb 10, 2019

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