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?

  • $\begingroup$ 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?" $\endgroup$
    – DanielV
    Oct 1, 2017 at 3:27
  • $\begingroup$ 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? $\endgroup$
    – fleablood
    Oct 1, 2017 at 4:10
  • $\begingroup$ thank you so much @fleablood I do not need anymore help. $\endgroup$
    – Emptymind
    Oct 1, 2017 at 4:21

1 Answer 1


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 $$4-\sqrt{9-x^2} \neq 0$$


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

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)

  • $\begingroup$ 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? $\endgroup$
    – Emptymind
    Oct 1, 2017 at 3:39
  • $\begingroup$ 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 $-3\leq x \leq 3$. $\endgroup$ Oct 1, 2017 at 3:45

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