How do you find the domain of $\frac{\sqrt{9 - x^{2}}}{x - 1}$ $(sqrt(9-x^2)) / (x-1)$ ?

Question

1662 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-01T11:42:25

$x \in {[- 3, 1) \cup (1, 3]}$ $x in {[-3, 1) uu (1, 3] }$

The domain is all the possible values of $x$ $x$
To find the domain, we must find the values of $x$ $x$ for which the function is undefined

The function,

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

will be undefined if
[1] The denominator becomes 0
[2] The term inside the square becomes negative

Hence,

$x - 1 \neq 0$ $x - 1 != 0$
$\Rightarrow x \neq 1$ $=> x != 1$

$9 - x^{2} > 0$ $9 - x^2 > 0$

$- x^{2} > - 9$ $-x^2 > -9$

$x^{2} \leq 9$ $x^2 <= 9$

$\Rightarrow - 3 \leq x \leq 3$ $=> -3 <= x <= 3$

Hence, the domain is all the real numbers between -3 and 3 except 1

How do you find the domain of √9−x2x−1(sqrt(9-x^2)) / (x-1)?