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

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

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

domain

y = \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}

ln (4 - 3 x - x^{2})

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

f (x) = x - 5 + 1

f (x) = ln (x^{2} - 5 x + 6)

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

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