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!

Given:

The function is:

To find:

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

Solution:

We have,

This function is defined if the radicand is greater than or equal to 0, i.e., and the denominator is non-zero, i.e., .

            ...(i)

And,

Adding 3 on both sides, we get

            ...(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.

step-by-step explanation:

their are no angles. sorry cant .

for future students the answer is b

hope i

-lavira: )