# How do you find the domain of f(x) =sqrt(9 - x)?

Apr 10, 2015

To find the (real) domain, we need to make sure we
(1) don't try to take even roots of negative numbers (square roots, 4th, 6th, 8th and so on)

(2) don't try to divide by zero

(and more rarely avoid other undefined expressions like ${0}^{0}$, logarithms of $0$ or of negatives and others)

For $f \left(x\right) = \sqrt{9 - x}$, there's no danger of trying to divide by zero, but there is an even root.
We need to make sure that $9 - x$ is not negative.
So we need to make sure that $9 - x \ge 0$

Solve: $9 - x \ge 0$

(I prefer to avoid negatives in front of the $x$, so I will add $x$ to both sides:)

$9 \ge x$. That means I want all the numbers to the left of $9$.

Let's make it look like what it ays, by writing it this way:

$x \le 9$. That's it, the domain of this function is
all (real) numbers, $x$, with $x \le 9$

There are other ways of expressing the answer, but they all refer to the same set of numbers"
$\left\{x | x \le 9\right\}$,
$\left(- \infty , 9\right]$

Final note

It may seem or 'feel' a little odd that we could use, for example, $x = - 40$ in this function, but it's true.
Before we take the square root, we're going to find $9 - x$ which would be (9-(-40) = 9+40 = 49 So we would get $f \left(- 40\right) = \sqrt{49} = 7$

We could not use $x = 90$, because $9 - 90 = - 81$ does not have a positive or negative square root.