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

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

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

domain

(x - 1)^{2} + 1

x \geq 1

Interval Notation

[1, \infty)

Solution steps

Find non-negative values for radicals:

x \geq 1

The function domain

x \geq 1

f (x) = \frac{3}{9 - x ^{2}}

f (x) = 2 x + 3

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

f (x) = - \frac{1}{2 7 - x}

\frac{x ^{2} - 5 x - 6}{x + 1}