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 Given  \( $f(x) = \frac{\sqrt{2x-6}}{x-3}$\) what is the smallest possible integer value for \(f(x)\) such that \(x\) has a real number \(f(x)\)value?

 Nov 13, 2020
 #1
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If  I understand what  you're  asking

 

Look at the  graph here :  https://www.desmos.com/calculator/mqxce8lgk4

 

This graph  will  approach  0 as x increases from 3   (but never quite gets there)...so.....the lowest integer value  for  f(x)  = y = 1

 

And note that when  y = 1,  then x has the real number value  (5)

 

 

cool cool cool

 Nov 13, 2020

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