What is the domain of f(x)=(sqrt(6x-2)-3)/(x^2+1) ?

The domain of f(x)=(sqrt(6x-2)-3)/(x^2+1) is x>= 1/3

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domain of f(x)=(sqrt(6x-2)-3)/(x^2+1)

domain

f (x) = \frac{6 x - 2 - 3}{x ^{2} + 1}

x \geq \frac{1}{3}

Interval Notation

[\frac{1}{3}, \infty)

Solution steps

Find non-negative values for radicals:

x \geq \frac{1}{3}

The function domain

x \geq \frac{1}{3}

f (x) = \frac{7}{7 x}

f (x) = \frac{17 x - 6}{6 x + 4}

1 - ∣ x ∣

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

f (x) = \frac{x - 8}{3 x - 4}

What is the domain of f(x)=(sqrt(6x-2)-3)/(x^2+1) ?
The domain of f(x)=(sqrt(6x-2)-3)/(x^2+1) is x>= 1/3