What is the domain of sqrt(x^2-x-1) ?

The domain of sqrt(x^2-x-1) is x = (sqrt(5)+1)/2

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domain of sqrt(x^2-x-1)

domain

x^{2} - x - 1

x \leq \frac{- 5 + 1}{2} or x \geq \frac{5 + 1}{2}

Interval Notation

(- \infty, \frac{- 5 + 1}{2}] \cup [\frac{5 + 1}{2}, \infty)

Solution steps

Find non-negative values for radicals:

x \leq \frac{- 5 + 1}{2} or x \geq \frac{5 + 1}{2}

The function domain

x \leq \frac{- 5 + 1}{2} or x \geq \frac{5 + 1}{2}

f (x) = (x - 1)^{2} + 7

f (x) = \frac{2 x - 1}{x - 2}

f (x) = \frac{x - 1}{x ^{2} - 5 x + 6}

f (x) = 3 x^{2} + 5 x - 2

3 x^{2} - 6

What is the domain of sqrt(x^2-x-1) ?
The domain of sqrt(x^2-x-1) is x<= (-sqrt(5)+1)/2 \lor x>= (sqrt(5)+1)/2