# Finding the domain of $f(x) = \frac{2}{4-\sqrt{9-x^2}}$.

Finding the domain of $f(x) = \frac{2}{4-\sqrt{9-x^2}}$, how can I deal with the denominator, I have to take into account 2 things: 1- the denominator not equal zero. 2-the value under the square root must be greater than or equal to zero. Could anyone help me please?

• The domain of that $f(x)$ is whatever you say it is. It could be empty or all real numbers or a bitfield. What you are probably intending to ask is "which real numbers, when $f$ is applied to them, also result in a real number?" Oct 1, 2017 at 3:27
• Um..... the denominator can't be equal to zero, and the value under the square root must be greater than or equal to zero. What possible more help could you possibly need? Oct 1, 2017 at 4:10
• thank you so much @fleablood I do not need anymore help. Oct 1, 2017 at 4:21

You are correct (if $f$ is a real-valued function). So you need both of those conditions to hold (if the function is real-valued).
The conditions are $$4-\sqrt{9-x^2} \neq 0$$

and

$$9-x^2 \geq 0.$$

Can you go from here?

(Note that the domain is just the set of numbers that are inputted into the function. This can be chosen freely, so I think you wanted the domain such that $f$ is real-valued)

• yes I wanted the domain such that f is real-valued, but I do not know exactly what to do after taking into account the two conditions mentioned, How can I combine these 2 conditions? Oct 1, 2017 at 3:39
• If you solve the two conditions for $x$, you get $$x^2 \neq -7$$ and $$x^2 \leq 9 \implies -3\leq x\leq 3.$$. If $f$ is real-valued, then the first condition tells us that you have to take real values of $x$. The second condition tells us that $-3\leq x\leq 3$. Combining them you get the domain of $x$ to be the real numbers $x$ such that $-3\leq x \leq 3$. Oct 1, 2017 at 3:45