What is the domain of f(x)=sqrt(x^2-4+3) ?

The domain of f(x)=sqrt(x^2-4+3) is x = 1

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domain of f(x)=sqrt(x^2-4+3)

domain

f (x) = x^{2} - 4 + 3

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

\frac{x}{7}

f (x) = 3 x^{2} + x - 8

f (x) = \frac{x ^{2} + 2 x - 5}{x ^{3} - 4 x ^{2} + 3 x}

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

lo g_{10} (3 x + 6)

What is the domain of f(x)=sqrt(x^2-4+3) ?
The domain of f(x)=sqrt(x^2-4+3) is x<=-1\lor x>= 1