# Question #5fa21

Jun 11, 2017

$\left\{x \in \mathbb{R} | x > 3 \vee x < - 2\right\}$

#### Explanation:

We have: $f \left(x\right) = \frac{1}{\sqrt{{\left(x\right)}^{2} - \left(x\right) - 6}}$

The argument of a square root must be greater than or equal to zero.

Also, the denominator of a fraction cannot be equal to zero.

Let's use these conditions to find the largest possible domain of $f \left(x\right)$:

$R i g h t a r r o w \sqrt{{x}^{2} - x - 6} > 0$

Squaring both sides of the equation:

$R i g h t a r r o w {\left(\sqrt{{x}^{2} - x - 6}\right)}^{2} > {0}^{2}$

$R i g h t a r r o w {x}^{2} - x - 6 > 0$

Then, let's factorise the quadratic equation using the "middle-term break":

$R i g h t a r r o w {x}^{2} + 2 x - 3 x - 6 > 0$

$R i g h t a r r o w x \left(x + 2\right) - 3 \left(x + 2\right) > 0$

$R i g h t a r r o w \left(x + 2\right) \left(x - 3\right) > 0$

$R i g h t a r r o w x + 2 > 0 \mathmr{and} x - 3 > 0$

$R i g h t a r r o w x > - 2 \mathmr{and} x > 3$

$\mathmr{and}$

$R i g h t a r r o w x + 2 < 0 \mathmr{and} x - 3 < 0$

$R i g h t a r r o w x < - 2 \mathmr{and} x < 3$

$\therefore x > 3 \mathmr{and} x < - 2$

Therefore, the largest possible domain of $f \left(x\right)$ is $\left\{x \in \mathbb{R} | x > 3 \vee x < - 2\right\}$.