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?
+128475 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)
