Given:
The function is:
[tex]f(x)=\dfrac{\sqrt{2x-6}}{x-3}[/tex]
To find:
The smallest possible integer value for $x$ such that $f(x)$ has a real number value.
Solution:
We have,
[tex]f(x)=\dfrac{\sqrt{2x-6}}{x-3}[/tex]
This function is defined if the radicand is greater than or equal to 0, i.e., [tex]2x-6\geq 0[/tex] and the denominator is non-zero, i.e., [tex]x-3\neq 0[/tex].
[tex]2x-6\geq 0[/tex]
[tex]2x\geq 6[/tex]
[tex]\dfrac{2x}{2}\geq \dfrac{6}{2}[/tex]
[tex]x\geq 3[/tex] ...(i)
And,
[tex]x-3\neq 0[/tex]
Adding 3 on both sides, we get
[tex]x-3+3\neq 0+3[/tex]
[tex]x\neq 3[/tex] ...(ii)
Using (i) and (ii), it is clear that the function is defined for all real values which are greater than 3 but not 3.
Therefore, the smallest possible integer value for x is 4.
