What is the domain of sqrt(x^2-8x+12) ?

The domain of sqrt(x^2-8x+12) is x = 6

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domain of sqrt(x^2-8x+12)

domain

x^{2} - 8 x + 12

x \leq 2 or x \geq 6

Interval Notation

(- \infty, 2] \cup [6, \infty)

Solution steps

Find non-negative values for radicals:

x \leq 2 or x \geq 6

The function domain

x \leq 2 or x \geq 6

6 x^{2} - 9

f (x) = \frac{x ^{2} - 9}{x ^{2} + x - 6}

f (x) = - 2 x + 5 - 1

f (x) = cos (x^{2})

What is the domain of sqrt(x^2-8x+12) ?
The domain of sqrt(x^2-8x+12) is x<= 2\lor x>= 6