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}

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

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

- \frac{x - 2}{3} + 1

- 3 x - 1

\frac{5 x}{7 x - 2}

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