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# HELP ASAP!

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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)

Nov 13, 2020