What is the domain of sqrt(x^2+x-12) ?

The domain of sqrt(x^2+x-12) is x = 3

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domain of sqrt(x^2+x-12)

domain

x^{2} + x - 12

x \leq - 4 or x \geq 3

Interval Notation

(- \infty, - 4] \cup [3, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - 4 or x \geq 3

The function domain

x \leq - 4 or x \geq 3

- \frac{5}{x + 4}

f (x) = 1 x^{2} + 4 x - 9

f (x) = 13 - x

ln (\frac{x ^{2} + 3}{3 + 5})

f (x) = \frac{x + 8}{x + 7}

What is the domain of sqrt(x^2+x-12) ?
The domain of sqrt(x^2+x-12) is x<=-4\lor x>= 3