# How do you find the domain and range of f(x)= sqrt(x^2-x-6)?

Jun 17, 2016

Domain is $\left(- \infty , - 2 \left[\cup\right] 3 , + \infty\right)$
Range is $\left[0 , + \infty\right)$

#### Explanation:

Since under square root the polynomial must be positive, the domain is obtained by solving:

${x}^{2} - x - 6 \ge 0$

you can obtain the zeroes of the polynomial by solving the associated equation:

${x}^{2} - x - 6 = 0$

$x = \frac{1 \pm \sqrt{1 - 4 \cdot \left(- 6\right)}}{2}$

$x = \frac{1 + \sqrt{25}}{2}$

$x = \frac{1 \pm 5}{2}$

$x = - 2 \mathmr{and} x = 3$

so the disequation is solved in the external intervals:

$x < - 2 \mathmr{and} x > 3$
the the domain is:

$\left(- \infty , - 2 \left[\cup\right] 3 , + \infty\right)$

Since f(x) is positive, due to the result of square root, the range includes all positive real numbers: $x \ge 0$