Why not the domain is just $ ( - \infty , -3) $ of $ \sqrt{2-x} - \frac{1}{9-x^2} $

Question

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?

Answer 1

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 $x\ne -3$.

Answer 2

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

• $x\le2 \land x\neq-3$

Answer 3

In one line:

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

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

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