Question

Given $f(x) = \frac{\sqrt{2x-6}}{x-3}$, what is the smallest possible integer value for $x$ such that $f(x)$ has a real number value? please show steps. Thank you!

Answer 1

Given:

The function is:

$f(x)=\dfrac{\sqrt{2x-6}}{x-3}$

To find:

The smallest possible integer value for $x$ such that $f(x)$ has a real number value.

Solution:

We have,

$f(x)=\dfrac{\sqrt{2x-6}}{x-3}$

This function is defined if the radicand is greater than or equal to 0, i.e., $2x-6\geq 0$ and the denominator is non-zero, i.e., $x-3\neq 0$.

$2x-6\geq 0$

$2x\geq 6$

$\dfrac{2x}{2}\geq \dfrac{6}{2}$

$x\geq 3$             ...(i)

And,

$x-3\neq 0$

Adding 3 on both sides, we get

$x-3+3\neq 0+3$

$x\neq 3$             ...(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.