Consider

# $$f(x) = \sqrt{2-x} - \frac{1}{9-x^2}$$

Now for the radical to be definable,

$$2-x \ge 0$$

And for the fraction to be definable,

$$9-x^2> 0$$

SO , the number line of solution set looks like: Where red line is for radical's definability & orange for fraction's.

Now,

The result from above stuff is

$$(- \infty , -3)$$

EDIT: I got that the second inequality's solution set was wrong.

Its

$$(-3,3)$$

So final solution plot: Which is $\color{red}{wrong}$

# Why?

• A fraction is defined when the denominator is not equal to zero. May 23, 2018 at 10:28

The fraction is defined if and only if $9-x^2\ne 0\Leftrightarrow x\ne\pm3.$

However, you have done the square root part correctly: $\sqrt{2-x}$ is defined if and only if $x\le 2$.

Because of this, we can conclude that $f(x)$ is defined if and only if $x\le 2$ and $x\ne -3$.

We have to satisfy

• $2-x\ge 0 \implies x\le 2$

and

• $9-x^2\neq0\implies x\neq \pm3$

thus the set of values of $x$ which satisfy both inequalities is

• $x\le2 \land x\neq-3$
• +1 for nice presentation! However, for something this basic you should probably expand "thus" to something like "thus the set of values of $x$ that satisfy both inequalities is". I've certainly had students sometimes correctly obtain multiple inequalities in domain questions such as this, but then answer with the union (not intersection) of the sets involved. May 23, 2018 at 7:14
• @DaveL.Renfro Thanks for your kind suggestion, I’ll add some more details.
– user
May 23, 2018 at 7:17
• The expression $\frac{1}{9 - x^2}$ is defined unless $|x| = 3$. Why are you writing $9 - x^2 > 0$? May 23, 2018 at 10:26

In one line:

$x \le 2$ AND $(x \not = -3$ OR $x \not =3)$.

Hence $x \le 2$ AND $x \not = -3.$

$D= (-\infty, -3)\cup (-3,2]$.