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

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

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

domain

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

x \leq - 1 or x \geq 1

Interval Notation

(- \infty, - 1] \cup [1, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - 1 or x \geq 1

The function domain

x \leq - 1 or x \geq 1

f (x) = \frac{( 2 x ^{2} - 3 x + 3 )}{1 - 2 x}

f (x) = (\frac{8 x ^{2} + 6}{8 x ^{2} - 6})

f (x) = 1 + \frac{3 - x}{5 - x}

f (x) = \frac{x + 5}{x ^{2} - 10 x + 25}

sec (x)

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