How do you find the domain of $f(x) = 1/(sqrt(5 - x) - 1)$ ?

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Answer 1 · 2017-05-18T23:39:15

The domain, in this case. is all real numbers except for the values of $x$ that would result in a non-real value.

If $x$ causes the denominator to be $0$ , then the function would be undefined at that point.

$0 = sqrt(5 - x) - 1$

$1 = sqrt(5 - x)$

$1^2 = 5 - x$

$1 = 5 - x$

$-4 = -x$

$x = 4$

The function would also be imaginary at $x$ if there was a square root of a negative value. Therefore...

$sqrt(5 - x) >= 0$

$5 - x >= 0^2$

$5 >= x$

The domain of $x$ is all real values less than or equal to $5$ and not including $4$ .

How do you find the domain of f(x) = 1/(sqrt(5 - x) - 1) f(x) = 1/(sqrt(5 - x) - 1)?