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

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

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

domain

f (x) = 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}

1 + sin (x)

2 x^{3} - 3 x^{2} - 12 x + 24

f (x) = lo g_{x + 3} (- 2 x + 3)

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

f (x) = lo g_{10} (x^{2} + 9 x + 8)

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