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)=(√(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)=(√(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\\eq 0`.

`2x-6\geq 0`

`2x\geq 6`

`(2x)/(2)\geq (6)/(2)`

`x\geq 3` ...(i)

And,

`x-3\\eq 0`

Adding 3 on both sides, we get

`x-3+3\\eq 0+3`

`x\\eq 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.